03 Kunci Jawaban Dan Pembahasan Matematika Ix 4u2c1q

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Sek St olah an : da SMP r /MTs K Kela o s/Se m mes pe te : ter : ns M IX/1 i : at 1. e Pendid *) M m ikanka C e at rakter( er m ik m ah a at a Sila mi bus ke Nilai danMateri yangDiintegrasikan se Bab ensi ba I ng Kes un an eba ba ngu ng un nan da dan tar Kek da n ongr pe uen ng Contoh Instrumen an ngun dataryang sebangun. gu BangunDatar na an angundatar yang kongruen.– Menyebutkan syaratdua bangun dataryang kongruen. (*)– Menjelaskan pe?ngertian skala padagambar, foto, danmodel. – ny at dua bangundatar kongruen.1.1.2 Mampu menen?tukan syaratkesebangunandua bangun datar.1.1.3 Mampu menen?tukan besarskala, ukuransebena a da ngun.Besar ?ADC = . . . .a. 56?b. 86?c. 94?d. 124? 2. Diketahui dua bangundatar berikut sebangun.Tentukan nilai la x, y, dan z.15 cm 10 cmx? 102? z cm 8 cm124?3 6 jp mKelas IXSMP/MTs,Depdiknas4. CD Multi ntanPariwara,halaman 1–202. Buku PGMatematikaKelas IX, IntanPariwara,halaman 1–343. BSEMatematikauntuk pe m ec ah an m as al ah .

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Nilai danMateri yangDiintegrasikan

Contoh Instrumen

buahkarton. Sisa karton disebelah kiri, kanan,dan atas gambarberjarak 5 cm dari tepigambar. Jika gambardan karton sebangun,tentukan sisa karton dib

a segitiga yangkongruen. – Menjelaskan sifat?sifat dua segitigayang kongruen.– Menjelaskan pe?ngertian dua segitigayang sebangun.– Menyebutkan sy nen?tukan sifat-sifatdua segitigayang kongruen.1.2.3 Mampu menen?tukan syarat duasegitiga yang se?bangun. 1.2.4 Mampu menen?tukan sifat-sifatd

an berikutbenar, kecuali . . . . a. BC = ADb. AC = BDc. ?DAO = ?OAB d. ?DAB = ?ABC 2. Perhatikan trapesiumsama kaki ABCD dibawah.a. Tunjukkan bah 6 jp ikaKelas IX, IntanPariwara,halaman 1–343. BSEMatematikauntuk Kelas IXSMP/MTs,Depdiknas4. CD MultimediaMatematika5. Website?website yang rele

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sif 000336465 Bab at 736300000 184000000 ta 6c7774707 II bu 4000001f00Ban ng 000001462 gun pada sebuahkar gambar dibawah ini maka y = . . .mm.a. 305 c. 275 b. 285 d. 250 2. Sebuah gambarberukuran panjang 60cm dan lebar 36 cmditempel 6b7074000 , 002040000 ke 001472585 Rua ru 95a000002 ng 180000001 cu 46758595a Sisi t, 0000022c0 da 000001462 Len 58595a000 n 002400000 gku kongruendan sebangun. bo 0014646d6 Contoh Instrumen n untuk me?nyelesaikan masalah.– Menggunakan sifat?sifat dua segitiga yangsebangun untuk me?nyelesaikanla, masalah. e64000002 ng 540000007 se 0646d6464 rta 000002c40 , danbola jika unsur laindiketahui. – Menggambar jaring?jaring tabung dankerucut. m 000008876 tabungdan kerucut. en 756564000 0034c0000 en 008676696 577000003 tu lengkung ngkaranb. 2 bidang datar ber?bentuk persegi panjangc. 2 bidang datar ber?bentuk setengahlingkarand. 2 bidang 2. Perhatikan gambarberiku 4 jp ka d40000002 46c756d69 epdiknas4. Jangka,penggaris, danalat tulis 5. CD MultimediaMatematika6. Website-websiteyang relevan n 000003f800 uk 0000146d6 ur 561730000 040c00000 an 024746563 ny 680000043 a. 00000000c

matikauntuk Kelas IXSMP/MTs,Depdiknas4. CD MultimediaMatematika5. Website-websiteyang relevan

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Contoh Instrumen

t terpancungmenggunakan jaring?jaring kerucut.– Menyimpulkan rumusluas permukaan bola,luas belahan bolapadat, dan volumebola. – Menghitung lu menen?tukan luas danvolume kerucut.2.2.6 Mampu menen?tukan luas danvolume bola.

rucutterpancung yang mem?punyai lubang ber?bentuk tabung. Jari-jarialas kerucut terpancung12 cm. Jika tinggi danjari-jari alas tabungberturut-turut 1 4 jp

ukaan

4 jp

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Ekono •) mikrea In ovtif( at if

Nilai danMateri yangDiintegrasikan Penilaian

Contoh Instrumen

olumetabung karena per?ubahan jari-jari atautinggi. – Menghitung per?bandingan volumekerucut karena per?ubahan jari-jari atautinggi. – Menghitung .2.3.3 Mampu menen?tukan perban?dingan volumetabung, kerucut,dan bola karenaperubahan jari?jari atau tinggi.

pa buah hiasanberbentuk kerucuttanpa alas. Kerucut?kerucut yang ia buatberukuran sama, yaitutinggi 12 cm dan diam?eter 18 cm. Jika Cikamempunya

te yangrelevan

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Pendid *) ikanka Te rakter( liti


Contoh Instrumen

o m pe te 4.ns Mi : e 3. mM ahel a ak miuk pean lu pe anng g ol keah ja an di da an n se pe de ny rh aji an an a. da ta.

s/Se mes : ter : M IX/1 at e m at ik a

Sila bus Bab III Stat istik a dan Pelu ang

ampel dan popu?lasi. – Menyebutkan sampeldan populasi suatudata. – Menjelaskan pengerti?an rata-rata.– Menghitung rata-ratasuatu data.– Menjelask menen?tukan mediansuatu data.3.1.6 Mampu menen?tukan modussuatu data.3.1.7 Mampu menen?tukan jangkau?an, kuartil, jang?kauan inter?kuarti

c. hasil Ujian Nasionalsiswa SMP di DKIJakartad. hasil Ujian Nasionalsiswa beberapaSMP di DKI Jakarta 2. Sekelompok anak terdiriatas 9 orang ditimbang 6 jp websiteyang relevan


anyak siswa yang ikutbasket dan sepak bola2 : 3. Berapa banyaksiswa yang ikut basket?Basket TenismejaSepakbola Voli Futsal 12%23%

afsir?kan data dalambentuk tabel.3.2.6 Mampu menafsir?kan data dalambentuk diagrambatang, garis, danlingkaran.

besar sudut pusatlingkaran yang me?wakili data.– Menyajikan data yangbelum terurut dalambentuk tabel.– Menyajikan data daritabel dalam bentukdiag



Contoh Instrumen

tudata dalam bentukdiagram lingkaran.

peluang.– Menghitung frekuensirelatif suatu kejadian.– Menjelaskan pengerti?an ruang sampel dantitik sampel.– Menyebutkan ruangsampel dan titik sam 4.1.1 Mampu mende?finisikan frekuensirelatif dan pe?luang.4.1.2 Mampu mende?finisikan ruangsampel dan titiksampel suat

dapat 3calon laki-laki dan2 calon perempuan.Pemilihan dilakukandalam bentuk pasang?an dan dipilih secaraundian.a. Tentukan ruangsampelnya denga 6 jp 1. Buku PRMatematikaKelas IX, IntanPariwara,halaman 49–802. Buku PGMatematikaKelas IX, IntanPariwara,halaman 87–1383. BSEMatematikaun

terjadi. – Menghitung banyak?nya kejadian yang mungkin terjadi.– Menghitung peluangsuatu kejadian. (**)– Menyimpulkan kisar?an nilai peluang.– Men Mampu menen?tukan nilai peluangsuatu kejadiansederhana.4.2.2 Mampu mende?finisikan kisar?an nilai peluang.4.2.3 Mampu mende?finisikan peluang

ebuah uang logamdilempar bersamaan.Peluang muncul matadadu genap atau faktorprima dari 6 adalah . . . .a. 21 c. 32 b. 31 d. 56 2. Sebuah kotak berisi 6 jp 1. Buku PRMatematikaKelas IX, IntanPariwara,halaman 49–802. Buku PGMatematik

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o s/Se m mes pe te : ter : ns M IX/2 i : at 5. e Pendid Sif m *) ikanka at- at RasaIn rakter( ginTah sif ik u at a bil an ga Nilai danMateri yangDiintegrasikan Sila n be bus rp Bab an IV gk at Pan da gkat n be dan nt Ben uk tuk ak ar Akar se Contoh Instrumen rta pe ng gu negatifmenjadi bilanganberpangkat positif.– Menyatakan per?kalian bilangan dalambentuk pangkat.– Menyatakan bilanganberpangkat menjadibilangan na ukan nilai pangkatdan nilai akarsuatu bilangan. an ny a 6 jp da ebsite-websiteyang relevan la m pe m ec ah an m as al ah se de rh an a. ????



Contoh Instrumen

pangkat danbentuk akar.– Menyederhanakanbilangan berpangkatdan bentuk akar.– Menentukan hasil pen?jumlahan dan pengu?rangan bilangan ber?pa 5.2.3 Mampu menen?tukan hasil operasialjabar bilanganberpangkat danbentuk akar.

6 jp 1. Buku PRMatematikaKelas IX, IntanPariwara,halaman 87–992. Buku PGMatematikaKelas IX, Intan

angandalam bentuk baku.– Menentukan nilai darioperasi bilangan ber?pangkat dalam bentukbaku. – Menentukan nilaipangkat dari suatubilangan dalam 5.3.1 Mampu menen?tukan bentukbaku suatu bilang?an. 5.3.2 Mampu menyele?saikan persama?an pangkat se?derhana. 5.3.3 Mampu menyele?sai

ebut154 * 10–6 m 2,jari-jari penampangkawat . . . m.a. 7 * 10–6 b. 7 * 10–5 c. 7 * 10–4 d. 7 * 10–3 2. Nyatakan ke dalambentuk baku dari hasiloperasi hitung 6 jp 1. Buku PRMatematikaKelas IX, IntanPariwara,halaman 87–992. Buku PGMatematikaKelas IX, Intan



Contoh Instrumen

ubarisan. (*)– Melengkapi nilai sukuke-n suatu barisan.– Menentukan barisanbilangan dari suatupola. – Menentukan banyakanggota pada pola ke-n.– Me 6.1.1 Mampu menen?tukan pola suatubarisan.6.1.2 Mampu menen?tukan suku ke?n suatu barisan.6.1

4 dan 125 2. Tentukan rumus sukuke- setiap deret berikut.a. 200 + 196 + 192+ E E E + 8 + 4b. 2 + 6 + 18 + E E E+ 1.458c. 3 + 6 + 12 + E E E+ 1.53 6 jp ra,halaman185–230 3. BSEMatematikauntuk Kelas IXSMP/MTs,Depdiknas4. CD MultimediaMatematika5. Website-websiteyang relevan

aritmetika dan barisangeometri. – Melengkapi nilai sukuke-n barisan aritmetika.

Diketahui suatu baris?an geometri.100, 50, 25, 12 12 , E E E

ra,halaman185–230

6 jp

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000 00146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d4000 000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c000 0080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6 d70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936 362d322e31000000000000000000000000000000000000000000000000000000

Sek St olah an : da SMP r /MTs K Kela o s/Se m mes pe te : ter : ns M IX/2 i : at 6. e Pendid *) M m ikanka Ras e at rakter( aIng Nilai danMateri yangDiintegrasikan m ik in ah a Tah a u mi ba ris an Sila da n bus de Bab ret V bil an Bari ga san Contoh Instrumen n dan se rta Der pe et aksuku barisan aritmetikadan barisan geometri.– Menentukan bedaderet aritmetika.– Menentukan rasioderet geometri. ng gu na an u deretgeometri dengan rumussuku ke-n: Sn = 3n + 2 – 32.a. Tentukan besar sukuke-6.b. Tentukan rasio derettersebut. ny a da la m aritmetika dan deretgeometri. (**)– Menentukan suku ke?n deret aritmetika danderet geometri jikadiketahui rumus pe jumlahn suku pertamanya.– Menent Mampu menen?tukan rumusjumlah n sukupertama deretaritmetika.6.3.2 Mampu menen?tukan jumlah nsuku pertamaderet geometri.6.3.3 Mampu men m ec geometri berikut.1, –3, 9, –27, E E E .Jumlah 8 suku per?tama barisan tersebutadalah . . . .a. –3.280 c. 1.640 b. ah –1.640 d. 3.280 3. Sebuah bola tenisdijat 6 jp anIX, IntanPariwara,halaman105–129 2. B 1. Buku PRMatematikaKelas m as al ah .

ffd8ffe000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 104a46494 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 600010201 00c800c800 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 00ffe20c58 000000000000000000000000000000000000000000000000000000000000000000000000000000000 4943435f50 524f46494c 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 450001010 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0000c484c6 Rencana Pelaksanaan 96e6f02100 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 Pembelajaran Bab II Bangun 0006d6e74 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 725247422 Ruang Sisi Lengkung 058595a20 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 Sekolah : . . . . . . . . . . 07ce00020 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 Kelas/Semester : IX/1 Mata 009000600 310000616 Pelajaran : Matematika 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 373704d53 Alokasi Waktu : 6 × 40 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 465400000 menit 000494543 00000000000000000000000000000000000000000000000 207352474 Standar Kompetensi : 2. Memahami sifat-sifat tabung, kerucut dan bola, serta menentukan 200000000 ukurannya. 000000000 tabung, kerucut, Kompetensi Dasar : 2.1 Mengidentifikasi unsur-unsur 000000000 dan bola. 2.2 Menghitung luas selimut dan volume tabung kerucut dan bola. 00f6d60001 2.3 Memecahkan masalah 00000000d yang berkaitan dengan tabung, kerucut, dan 32d485020 bola. 200000000 Indikator Pencapaian 000000000 K engidentifikasi unsur-unsur tabung. Kompetensi M 000000000 Pendid 000000000 Mengidentifikasi unsur-unsur kerucut. M ikanka ) Nilai danMateri 000000000 yangDiintegrasikan K Mengidentifikasi unsur-unsur bola. M rakter( 000000000 Menentukan luas selimut dan volume tabung. r000000000 M Menghitung luas selimut dan volume kerucut. e M 000000000 000000000 Menentukan luas selimut dan volume bola. M a Menentukan besar perubahan volume tabung.000000000 M 000000116 Menentukan besar perubahan volume kerucut.ti M 370727400 Menentukan besar perubahan volume bola. f000150000

000336465 736300000 Tujuan 184000000 Pembelajaran 6c7774707 Peserta didik mampu: 4000001f00 1. menyebutkan dan menentukan unsur-unsur 000001462 tabung; 2. menyebutkan dan menentukan unsur-unsur 6b7074000 kerucut; 3. menyebutkan dan menentukan unsur002040000 001472585 unsur bola; 4. menghitung luas selimut tabung apabila unsur-unsurnya 95a000002 Contoh Instrumen diketahui; 5. menghitung volume tabung apabila unsur-unsurnya 180000001 diketahui; 6. menghitung luas selimut kerucut apabila unsur-unsurnya 46758595a 0000022c0 diketahui; 7. menghitung volume kerucut apabila unsur-unsurnya 000001462 diketahui; 8. menghitung luas selimut bola apabila unsur-unsurnya 58595a000 diketahui; 9. menghitung volume bola apabila unsur-unsurnya 002400000 diketahui; serta besar perubahan volume bangun 10. menentukan ruang sisi lengkung apabila unsur0014646d6 e64000002 unsurnya berubah. 540000007 Nilai pendidikan karakter yang ditanamkan ke siswa: Peduli terhadap 0646d6464 000002c40 sesama, Inovatif etelah memantulyang ke-5 kalinya. 000008876 Materi 756564000 Pembelajaran 1. Tabung 0034c0000 2. Kerucut 008676696 577000003 3. Bola d40000002 46c756d69 Metode 000003f800 0000146d6 Pembelajaran 1. Model Pembelajaran denganderet aritmetika.6.4.4 Mampu memecah?kan masalah yangberkaitan denganderet geometri. 561730000 Cooperative Learning (CL) 040c00000 Direct Instruction (DI) 024746563 5 c. 3,15 b. 3,20 d. 3,05 2. Setiap bulan Pak Dodimenabung uang dibank. Pada bulanpertama Pak Dodi me?nabung Rp120.000,00,pada bulan keduaRp1 680000043 2. Metode 6 jp 00000000c Tanya jawab 5. Website?website yangrelevan 725452430 Diskusi 000043c00 Tugas 00080c675 452430000 043c00000 80c625452 430000043 c0000080c7 465787400 000000436f 707972696 768742028 632920313 939382048 65776c657 4742d5061 636b61726

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 a. d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 E Eksplorasi 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 Guru 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 menjelask 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 an tentang Langkah-Langkah c. pengertian 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Kegiatan 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 tabung luas Pertemuan 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 Konfirmasi dan ciripermukaa Pertama 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 ciri n bangun tabung 1. Kegiatan Pendahuluan 0000000000000000000000000000000000000000000000 Guru sisi ruang (10 a. menit) lmenjelask lengkung. • unsurGuru memberikan contoh bentuk-bentuk bangun ruang sisi lengkung yang an Guru mempunyai peranan penting dalam iptek. Motivasi unsur menjelask b. an Prasyarat yang tentang pengetahuan • Siswa mengenal bentuk-bentuk benda yang berbentuk tabung, kerucut, terdapat cara dan bola. pada menentuk 2. Kegiatan Inti (60 t luas tabung. an menit) permukaa Guru menjelask n tabung an mengguna pengertian kan rkerucut rumus. dan ciriGuru ciri menjelask k tentang kerucut. an Guru cara menjelask menentuk unsurb. an volume unsur tabung • Guru bersama-sama siswa menggambar jaring-jaring bangun ruang yang mengguna sisi lengkung. Elaborasi c. kan terdapat • Guru menanyakan kepada siswa tentang hasil dari menggambar jaring-jaring rpada rumus. bangun ruang sisi lengkung dengan ukuran tertentu. Konfirmasi k kerucut. Guru Guru menjelask 3. Kegiatan Penutup (10 menit) Guru mengevaluasi hasil pembelajaran siswa dengan menjelask an tentang memberikan soal-soal latihan untuk dikerjakan dan memberikan tugas rumah. an cara pengertian menentuk bola an luas dan Pertemuan ciri-ciri permukaa Kedua 1. Kegiatan Pendahuluan b kerucut bola. n (10 a. menit) Guru mengguna • Guru memberikan contoh tentang manfaat mempelajari luas permukaan dan menjelask kan volume bangun ruang dalam kehidupan sehari-hari. Misalnya dalam menentukan banyak ran rumus. unsurMotivasi bahan untuk membuat robot kaleng dan bola. unsur Guru yang b. menjelask Prasyarat terdapat an pengetahuan • tentang Siswa mengetahui tentang pengertian luas permukaan dan volume bangun pada cara bola. ruang. menentuk 2. Kegiatan Inti (60 Guru volume menit) an menjelask kerucut an bangun mengguna ruang sisi kan lengkung rdan rumus. Guru manfaatny menjelask a terhadap an tentang manusia cara dari menentuk bentukan luas bentuk b. permukaa t tersebut. • bola Guru bersama-sama siswa menurunkan rumus-rumus yang berkaitan dengan luas n Guru permukaan dan volume bangun ruang sisi lengkung. Elaborasi mengguna bersama kan siswa rmenyebut rumus. Guru kan unsurmenjelask unsur dan an tentang bagiancara bagian menentuk pada an volume bangun bola ruang sisi mengguna lengkung.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0000000000000000000000000000000000000000000000

• Guru menanyakan kepada siswa tentang hasil rumus-rumus yang telah ditemukan. 3. Kegiatan Penutup (10 menit) Guru mengevaluasi pembelajaran siswa tentang penguasaan materi yang diajarkan. Setelah itu, guru meminta siswa untuk mengerjakan soal-soal latihan serta memberikan tugas rumah. Pertemuan Ketiga 1. Kegiatan Pendahuluan (10 a. menit) • Guru memberikan gambaran tentang pengaruh besar kecilnya ukuran pada benda berbentuk bangun ruang sisi lengkung terhadap sesuatu yang dikenainya. Motivasi b. Prasyarat pengetahuan • Siswa menguasai tentang volume bangun ruang sisi lengkung beserta rumusrumusnya. 2. Kegiatan Inti (60 menit) a. Eksplorasi E Guru menjelask an tentang b. perubahan volume • Guru bersama siswa membahas tentang cara menentukan perbandingan volume tabung bangun ruang sisi lengkung apabila unsur-unsur mempunyai ukuran yang berbeda. Elaborasi jika jaric. jarinya Guru menanyakan kepada siswa tentang penguasaan materi tentang perubahan b• berubah. volume bangun ruang sisi lengkung. Konfirmasi Guru menjelask Penutup (10 menit) Guru mengevaluasi pembelajaran siswa dengan 3. Kegiatan an tentang meminta kepada siswa untuk mengerjakan soal-soal latihan. Guru juga memberikan tugas perubahan rumah. volume kerucut Alat Sumber jika jariBelajar 1 Buku PR Kelas IX, Intan Pariwara jariMatematika dan 2BSE Matematika Kelas IX, Pusdiknas tingginya 3CD Multimedia Matematika b berubah. 4Website-website yang relevan Guru menjelask Penilaian Hasil an tentang Belajarperubahan 1. Teknik Penilaian dan Bentuk volume Instrumen a. Teknik bola jika Penilaian Tertulis jari-jarinya b. berubah. b Bentuk Instrumen 1) Pilihan Guru ganda 2) menjelask an tentang Uraian 3) cara Tugas 2. Contoh menentuk a. Pilihan Instrumen an Ganda Sebuah tabung padat dibelah menurut garis tingginya. Salah satu bangun yang perbandin dihasilkan adalah bangun ruang yang mempunyai . . . . gan 2 bidang datar berbentuk lingkaran volume 2 bidang datar berbentuk persegi panjang bangun 2 bidang datar berbentuk setengah lingkaran ruang 2 bidang lengkung apabila jari-jari atau tinggi bangun ruang b berubah.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 b. Uraian 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 Perhatikan gambar berikut. 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 ffd8ffe000104a464 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 9460001020100c80 0000000000000000000000000000000000000000000000 0c80000ffe20c5849 43435f50524f46494 c4500010100000c4 84c696e6f02100000 Gambar tersebut memperlihatkan sebuah bola dengan lubang berbentuk kerucut. Jika 6d6e747252474220 diameter bola dibanding tinggi kerucut 6 : 2, tentukan perbandingan jari-jari bola dan jari58595a2007ce0002 jari kerucut. 0009000600310000 616373704d534654 ________, 0000000049454320 ______________ 7352474200000000 0000000000000000 0000f6d600010000 Mengetahui Kepala SMP ______________ Guru Mata Pelajaran 0000d32d48502020 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NIP. 6370727400000150 0000003364657363 ______________________ NIP. ______________________ 000001840000006c 77747074000001f0 00000014626b7074 0000020400000014 7258595a00000218 000000146758595a 0000022c00000014 6258595a00000240 00000014646d6e64 0000025400000070 646d6464000002c4 0000008876756564 0000034c00000086 76696577000003d4 000000246c756d69 000003f800000014 6d6561730000040c 0000002474656368 000004300000000c 725452430000043c 0000080c67545243 0000043c0000080c 625452430000043c 0000080c74657874 00000000436f7079 7269676874202863 2920313939382048 65776c6574742d50 61636b6172642043 6f6d70616e790000 6465736300000000 0000001273524742 2049454336313936 362d322e31000000 0000000000000000 1273524742204945 4336313936362d32 2e31000000000000 0000000000000000 0000000000000000 0000000000

Rencana Pelaksanaan Pembelajaran Bab IV Pangkat dan Bentuk Akar

Sekolah : . . . . . . . . . . Kelas/Semester : IX/2 Mata Pelajaran : Matematika Alokasi Waktu : 10 × 40 menit Standar Kompetensi : 5. Sifat-sifat bilangan berpangkat dan bentuk akar serta penggunaannya dalam pemecahan masalah sederhana. Kompetensi Dasar : 5.1 Mengidentifikasi sifat-sifat bilangan berpangkat dan bentuk akar. 5.2 Melakukan operasi aljabar yang melibatkan bilangan berpangkat dan bentuk 5.3 Memecahkan masalah sederhana yang berkaitan dengan bilangan akar. berpangkat dan bentuk akar. Tujuan Pembelajaran Peserta didik mampu: 1menentukan sifat-sifat bilangan berpangkat; 2menentukan sifat-sifat bentuk akar; 3mengubah pangkat positif menjadi pangkat negatif dan sebaliknya; 4menghitung nilai bilangan berpangkat; 5menghitung nilai bilangan bentuk akar; 6menyederhanakan operasi hitung bentuk aljabar bilangan berpangkat; 7menyederhanakan operasi hitung aljabar bentuk akar; 8mengubah bilangan ke bentuk baku; serta 9melakukan operasi hitung persamaan pangkat sederhana. Indikator Pencapaian K enentukan sifat-sifat bilangan berpangkat. Kompetensi M Menentukan sifat-sifat bentuk akar. M Mengubah pangkat positif menjadi pangkat negatif dan sebaliknya. M Menentukan nilai pangkat dan nilai akar. M Melakukan operasi hitung bentuk aljabar bilangan berpangkat. M Melakukan operasi hitung bentuk akar. M Menentukan bentuk baku. M Menyelesaikan persamaan pangkat sederhana. Nilai pendidikan karakter yang ditanamkan ke siswa: Rasa ingin tahu Materi Pembelajaran 1Bentuk Pangkat 2Operasi Bentuk Akar Metode Pembelajaran 1. Model Pembelajaran Cooperative Learning (CL) Direct Instruction (DI) 2.

Metode Tanya jawab Diskusi

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Langkah-Langkah Kegiatan Pertemuan Pertama 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 1. Kegiatan Pendahuluan (10 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 menit) 0000000000000000000000000000000000000000000000 a. Motivasi • Memberikan contoh penulisan-penulisan bentuk pangkat dan akar dalam kehidupan sehari-hari. b. Prasyarat pengetahuan • Siswa mengetahui tentang pangkat. 2. Kegiatan Inti (60 menit) a. Eksplorasi Guru menjelask an pengertian bentuk b. Elaborasi p pangkat. • Guru membimbing siswa membuktikan sifat-sifat dan rumus-rumus pada bilangan Guru berpangkat. c. Konfirmasi menjelask • Guru an sifat-menanyakan hasil pembuktian rumus-rumus pada bilangan berpangkat. sifat 3. Kegiatan Penutup (10 bentuk menit) Guru p menyuruh siswa untuk mengerjakan soal-soal latihan. pangkat. Guru Pertemuan menjelask Kedua an cara 1. Kegiatan Pendahuluan (10 mengubah menit) a. Motivasi bentuk • Memberikan contoh operasi hitung bilangan berpangkat dalam kehidupan pangkat sehari-hari. b. Prasyarat negatif pengetahuan menjadi • Siswa mengetahui tentang operasi bentuk pangkat dua atau tiga. positif dan 2. Kegiatan Inti (60 sebaliknya menit) a. Eksplorasi . Guru menjelask melakuka an n tentang b. Elaborasi operasi manipulasi • Guru dan siswa melakukan operasi hitung bilangan hitung soal untuk berpangkat. c. Konfirmasi bentuk mengubah • Guru bersama siswa membahas tentang hasil operasi hitung bentuk pangkat. bilangan bentuk berpangka p pangkat. 3. Kegiatan Penutup (10 t t. menit) Guru menyuruh siswa untuk mengerjakan soal-soal latihan. Guru menjelask Pertemuan an tentang Ketiga 1. Kegiatan Pendahuluan (10 operasi menit) hitung a. Motivasi bentuk • Memberikan contoh-contoh bentuk akar dalam kehidupan aljabar sehari-hari. b. Prasyarat bilangan pengetahuan berpangka t t.

b.

• Guru melakukan cara mengubah bentuk akar menjadi bilangan pangkat. • Guru menjelaskan tentang menuliskan bentuk akar suatu bilangan.

• Guru bersama siswa membahas tentang mengubah bentuk akar menjadi bentuk pangkat atau bentuk akar paling sederhana. Elaborasi c. Konfirmasi

• Siswa mengetahui makna bentuk akar suatu bilangan. 2. Kegiatan Inti (60 menit) a. Eksplorasi • Guru menjelaskan tentang bentuk-bentuk akar • Guru menjelaskan tentang sifat-sifat bentuk akar. bilangan.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0000000000000000000000000000000000000000000000

• Guru menanyakan kepada siswa tentang pemahaman materi bentuk akar. 3. Kegiatan Penutup (10 Guru menit) menyuruh siswa untuk mengerjakan soal-soal latihan.

1. Kegiatan (10 menit)

Pendahuluan

Pertemuan Keempat

a. • Memberikan contoh-contoh kegunaan operasi hitung bentuk akar dalam rumusrumus matematika maupun fisika/alam. Motivasi b. Prasyarat pengetahuan • Menanyakan tentang perkalian bentuk akar kepada siswa. 2. Kegiatan Inti (60 menit) a. Eksplorasi E Guru melakuka b. n penjumlah • Guru bersama-sama siswa membuktikan rumus-rumus yang berlaku pada operasi an dan bentuk akar. hitung Elaborasi c. pengurang • bentuk Guru melakukan tanya jawab kepada siswa tentang pemahaman materi yang an dipelajari siswa. Konfirmasi a akar. 3. Kegiatan Penutup Guru (10 Guru menit) memberikan soal-soal latihan dan tugas rumah. melakuka n Pertemuan perkalian Kelima dan 1. Kegiatan Pendahuluan pembagia (10 a. menit) n bentuk • Memberikan contoh-contoh manfaat dari bilangan-bilangan bentuk pangkat dan a akar. akar serta bentuk baku dalam kehidupan sehari-hari. Motivasi Guru b. menentuk Prasyarat pengetahuan • hasilSiswa dapat melakukan operasi hitung bentuk akar dan pangkat a. an sederhana. operasi 2. Kegiatan Inti (60 hitung E Eksplorasi menit) bentuk Guru a akar. menjelask Guru an tentang menentuk bentuk b. an b rumus baku. perkalian Guru • Guru bersama-sama siswa membahas permasalahan-permasalahan yang dan melakuka melibatkan bentuk pangkat dan akar. Elaborasi pembagia c. n n bentuk pengerjaa a akar. n hitung Konfirmasi bentuk b baku. Guru melakuka n pengerjaa n hitung bentuk persamaa n bilangan berpangka t t.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 3. Kegiatan Penutup (10 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 menit)Guru mengevaluasi hasil belajar dan memberikan tugas atau soal-soal 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 latihan. 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 Alat Sumber 0000000000000000000000000000000000000000000000 Belajar 1. Buku PR Matematika Kelas IX, Intan Pariwara 2. BSE Matematika Kelas IX, Pusdiknas 3. CD Multimedia Matematika 4. Website-website yang relevan Penilaian Hasil Belajar 1. Teknik Penilaian dan Bentuk Instrumen a. Teknik Penilaian Tertulis b. Bentuk Instrumen 1) Pilihan ganda 2) Uraian 2. Contoh Instrumen a. Pilihan Ganda −3 −2 xy 1. (xy) 6 diubah ke pangkat positif − menjadi .... x3y4 x4y3 x2y3 x3y2

2. Nilai dari

−

1

 1 3 −  adalah  27 

.... 3

• Guru menanyakan kepada siswa penggunaannya dalam kehidupan sehari-hari.

tentang

penguasaan

materi

serta

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c450 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 00310000616373704d53465400000000494543207352474200000000000000000000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 00000000f6d6000100000000d32d4850202000000000000000000000000000000000 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 00000000000000000000000000000000000000000000000000000000000000116370 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 7274000001500000003364657363000001840000006c77747074000001f000000014 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 626b707400000204000000147258595a00000218000000146758595a0000022c0000 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0000000000000000000000000000000000000000000000 00146258595a0000024000000014646d6e640000025400000070646d6464000002c4

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 00000088767565640000034c0000008676696577000003d4000000246c756d690000 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 03f8000000146d6561730000040c0000002474656368000004300000000c72545243 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465 000000000000000000000000000000000000000000000000000000000000000000000000000000000 787400000000436f70797269676874202863292031393938204865776c6574742d50 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 61636b61726420436f6d70616e790000646573630000000000000012735247422049 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 Bab I Kesebangunan EH FE EH18 × 818 = ⇒ = ⇔ EH = = 12 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 454336313936362d322e310000000000000000000000127352474220494543363139 dan Kekongurenan ADAB 812 12 Jadi, panjang EH = 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 36362d322e31000000000000000000000000000000000000000000000000000000 12 cm. 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 4. Jawaban: 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 c ABCD sebangun dengan 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 EFGH. BCAB AB A. Pilihan =⇔ BC= × FG 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 FGEF EF9 Ganda 1. Jawaban: d Dua bangun datar 00000000000000000000000000000000000000000000000 = ×4=6 dikatakan sebangun jika sudutsudut yang bersesuaian sama besar dan sisi-sisi yang bersesuaian sebanding. Pasangan (i) sebangun, karena semua sudutnya sama 90° dan perbandingan sisi-sisi yang bersesuaian sama yaitu

7 4

. Pasangan (ii)

6Jadi,

panjang BC = 6

cm.

1Jawaban: b NM = KN = 12 cm KLMN kongruen dengan PQRS dan PS bersesuaian dengan NM, maka PS = NM = 12 cm. 2Jawaban: b D 5

tidak sebangun, karena perbandingan sisisisi yang bersesuaian tidak sama yaitu 6 perbandingan sisi yang panjang 4

4 sedangkan perbandingan . 1 sisi yang pendek 3

2

Pasangan (iii) sebangun, karena semua 1 90° dan perbandingan sisisudutnya c. sama – 6 4 3 sisi yang ber sesuaian sama yaitu = = 32Jadi, bangun yang d.pasangan –3 2. sebangun b. Uraian adalah (i) dan (iii). Ubahlah ke bentuk pangkat positif. −13 empat ABCD sebangun dengan 1Jawaban: (2 b−2Segi 3) 1. bersesuaian −4 −5 EFGH. AB dengan FG dan BC 23 bersesuaian dengan GH berarti sudut yang diapit −4 −2 −3 x(ybersesuaian z) sisi AB dan BC dengan sudut yang 2. −12 (xy )GH, yaitu ∠ABC bersesuaian diapit sisi FG dan dengan ∠FGH. Pada bangun datar yang sebangun, sudut-sudut yang bersesuaian sama besar sehingga ________, ______________ ∠ABC = ∠FGH. 2Jawaban: d EFGH sebangun dengan ABCD. EH bersesuaian denganSMP AD dan FE bersesuaian Mengetahui Kepala ______________ Guru dengan Mata Pelajaran AB, sehingga:

...................................................... NIP. ______________________ NIP. ______________________

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe0001 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00 ffd8ffe ffd8ffe 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 000104a 04a4649460 ffd8ffe00010 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0104a464 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 4649460 000104 001020100c 4a464946000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 94600010 0010201 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 800c80000ff 1020100c800 a46494 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 20100c80 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00c800c e20c584943 c80000ffe20c 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 435f50524f4 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0c80000ff 80000ffe 600010 584943435f5 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 20c5849 20100c8 6494c45000 0524f46494c4 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 e20c5849 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 43435f5 10100000c4 50001010000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c450001 ffd8ffe000104a46 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 43435f50 8. Jawaban: b Misalkan x = lebar 2. Perhatikan trapesium ABCF dan trapesium FCDE 00c8000 62d322e31000000000000000000000000000000000000000000000000000000 0524f46 84c696e6f0 0100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003100 0c484c696e6f 49460001020100c 00616373704d5346540000000049454320735247420000000000000000000000000000f 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 524f4649 494c450 21000006d6 sisa karton di bawah foto. ABCD dan EFGH berikut. F 16 cm C E 12 cm D 021000006d6 0ffe20c5 6d6000100000000d32d4850202000000000000000000000000000000000000000000000 0010100 e747252474 800c80000ffe20c5 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 4c450001 sebangun. e7472524742 00000000000000000000000000000000000000000000000000116370727400000150000 000c484 849434 22058595a2 0003364657363000001840000006c77747074000001f000000014626b70740000020400 ffd8ffe000104a4649460001020100c800c800 2058595a200 84943435f50524f4 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0100000c 0000147258595a00000218000000146758595a0000022c000000146258595a000002400 c696e6f 007ce00020 ABAD DC 00ffe20c584943435f50524f46494c45000101 7ce00020009 35f5052 6494c4500010100 = 0000014646d6e640000025400000070646d6464000002c400000088767565640000034c 0000000000000000000000000000000000000000000000 484c696e 20 cm 0210000 0090006003 00060031000 0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c A 00000c484c696e6f021000006d6e747252474 ffd8ffe000104a464946 EF EH 20 +4 30 ++x 000c484c696e6f02 6f021000 06d6e74 4f46494 1000061637 0000002474656368000004300000000c725452430000043c0000080c675452430000043 22058595a2007ce0002000900060031000061 0616373704d c0000080c625452430000043c0000080c7465787400000000436f7079726967687420286 7252474 3704d53465 0001020100c800c8000 2 1000006d6e74725 006d6e74 ⇔ = 6373704d5346540000000049454320735247 53465400000 ffd8ffe000104a c450001 3292031393938204865776c6574742d5061636b61726420436f6d70616e7900006465736 2205859 AB20 5 20 30 24 32 +x 420000000000000000000000000000f6d6000 46494600010204000000004 247422058595a20 AB00049454320 ~ FC dengan = =0ffe20c584943435f505 30000000000000012735247422049454336313936362d322e3100000000000000000000 72524742 010000 5a2007c 9454320735 0012735247422049454336313936362d322e31000000000000000000000000000000000 FC164 FC16 4 100000000d32d48502020000000000000000 73524742000 100c800c80000 07ce00020009000 2058595a e000200 000000000000000000000 ⇔ = 2474200000 24f46494c4500010100 000000000000000000000000000000000000 00000000000 ffe20c58494343 0c484c6 20 30 24 ×30 600310000616373 2007ce00 0900060 FC ~ ED dengan = = 000000000000000000000000000000000000 00000000000 5f50524f46494c0000000000 000c484c696e6f021000 ED123 AF 96e6f02 0310000 704d53465400000 02000900 000000011637072740000015000000033646 000f6d60001 45000101000000000000000 ⇔ = 32 + x 6163737 000f6d6000 006d6e7472524742205 20 57363000001840000006c77747074000001f0 00000000d32 c484c696e6f021 000494543207352 100000 06003100 AF ~ FE dengan AF = FE maka = 04d5346 100000000d FE 00000014626b707400000204000000147258 d4850202000 000006d6e7472 ⇔ x = 36 – 8595a2007ce00020009 474200000000000 00616373 5400000 6d6e74 32d4850202 595a00000218000000146758595a0000022c0 1 00000000000 5247422058595 BC 000000000000000 704d5346 0000000000 32 = 4 cm 0006003100006163737 BC00000000000 ~ CD dengan0004945 BC = CD maka 00000146258595a0000024000000014646d6 a2007ce000200 725247 CDOleh karena perbandingan panjang sisi4320735 0900060031000 0000000000 Jadi, lebar sisa 00f6d60001000000 54000000 e640000025400000070646d6464000002c400 00000000000 =1 04d5346540000000049 422058 2474200 B sisi00000000000 yang bersesuaian tidak sama, maka0000000000 A karton di bawah foto 000088767565640000034c000000867669657 0616373704d53 00d32d485020200 00494543 0000000 0000000000 4543207352474200000 trapesium ABCF dan trapesium FCDE tidak 7000003d4000000246c756d69000003f80000 00000000000 4654000000004 595a200 4 cm. 000000000000000 20735247 0000000 94543207352470000000000 9. Jawaban: b Jarak dua kota pada 00146d6561730000040c00000024746563680 00000000000 0000000000000000000 sebangun. 000000000000000 0000000 7ce0002 42000000 00004300000000c725452430000043c000008 00000000000 42000000000000000000000 peta hasil fotokopi 6 cm. Jarak dua kota 00000f6 0000000000 0000f6d600010000000 000000000000000 00000000 3. 0c675452430000043c0000080c62545243000 00000000000 Jawaban soal ini bersifat 0000000000000 terbuka, artinya 000900 pada peta yang asli d600010 0000f6d600010 0000000000 0043c0000080c7465787400000000436f7079 00001163707 000000000000000 00000000 0d32d48502020000000 ada banyak jajargenjang yang dapat dibuat. Hal 0000000 060031 726967687420286329203139393820486577 27400000150 0000000d32d480000000000 000000000000000 000000f6 yang perlu diperhatikan dalam membuat 0000011637 d32d485 0000000000000000000 = 6 : 120% = 6 100 × = 5 cm 6c6574742d5061636b61726420436f6d70616 00000033646 5020200000000 000061 120Jarak dua kota sebenarnya = 000000000000000 0202000 0727400000 d6000100 jajargenjang sebangun adalah sudut-sudut e79000064657363000000000000001273524 57363000001 0000000000000 0000000000000000000 0000000 1500000003 637370 000116370727400 000000d3 2.500.000 × 5 cm = 12.500.000 cm = 7422049454336313936362d322e310000000 840000006c7 sama besar 0000000000000 yang bersesuaian dan sisi-sisi yang 0000000 3646573630 0000000000000000000 000000000000000127352474220494543363 77470740000 0000000000000 000150000000336 2d485020 4d5346 125 km Jadi, jarak dua kota sebenarnya bersesuaian sebanding. Berikut ini tiga di 0000184000 0000000 13936362d322e31000000000000000000000 01f00000001 0000000000000 0000000000000000000 465736300000184 20000000 0000000 0006c77747 540000 125 km. antara jajargenjang yang dapat dibuat. 000000000000000000000000000000000 4626b707400 0000000000000 0000006c7774707 0000000 00000000 ffd8ffe000104a46494600010201 0000000000001163707 00020400000 0000000000000074000001f 10. Jawaban: 000049 0000000 00000000011630000000146 4000001f00000001 00000000 00c800c80000ffe20c584943435f5 0147258595a 2740000015000000033 a APQD sebangun dengan ABCD, 0000000 454320 00000218000 707274000001526b7074000 4626b7074000002 00000000 0524f46494c4500010100000c484 0000000 6465736300000184000 00014675859 00000003364650020400000 735247 040000001472585 00000000 c696e6f021000006d6e74725247 ffd8ffe000104a4649460001020100c800c80 maka: APPQ AP18= 18⇒ ×18 0000000 73630000018400147258595 = ⇔AP = = 13,5 5a0000022c0 0006c77747074000001 OD= BCAB 1824 24 Jadi, panjang AP 95a000002180000 0000000 00000000 420000 422058595a2007ce00020009000 000ffe20c584943435f50524f46494c4500010 00000146258 000006c777470a000002180 0000000 f000000014626b70740 00146758595a000 adalah 13,5 cm. 00000000 100000c484c696e6f021000006d6e74725247 600310000616373704d53465400 595a0000024 74000001f0000 0000014675 000000 0000000 00014626b70748595a00000 00000001464 0022c0000001462 422058595a2007ce00020009000600310000 000000494543207352474200000 = 00000000 25 − 16 0000000 0000204000000147258 22c0000001 000000 6d6e6400000 0000020400000 B. 58595a000002400 00000000 4. Oleh 616373704d53465400000000494543207352 karena trapesium ABGC dan EFGH46258595a0 sebangun 00000000000000000000000f6d6 0116370 595a000002180000001 25400000070 0147258595a00 B = 000000 0000014646d6e64 47420000000000000000000000000000f6d6 00000000 maka AB 646d6464000 BD = 2berlaku: × 000100000000d32d48502020000 OD =AC 27274000 × 346758595a0000022c00 = 6 0002180000001 cm ABCD dan0000240000 1. Semua sudut persegi panjang sama 0015000 000000f 00014646d6 000002540000007 000100000000d32d48502020000000000000 00000000 000000000000000000000000000 002c4000000 Uraian EFGH maka EG 46758595a0000 = BD = 6 cm AC20 0000336 0000146258595a00000 = kongruen, ⇔ =000000000000000000000000000 besar yaitu 90°. Pada ABCD, panjang = AB = 88767565640 022c000000146e640000025 000000000000000000000000000000000000 0646d6464000002 00000000 6d6000 4000000706 EHEF 2550 dan FH = 20 AC = 84657363 cm Luas 258595a000002 EFGH: 000034c0000 CD = 12 cm dan lebar = AD = BC = 8 cm. 24000000014646d6e64 000000000000000000000000000000000000 c40000008876756 11637072 000000000000000000000000000 0000018 46d6464000 100000 00867669657 ⇔AC= × 25 4000000014646 = 10 Pada PQRS, panjang = QR = PS = 9 cm dan 000000000011637072740000015000000033 5640000034c0000 4000000 74000001 000000000011637072740000015 0000025400000070646 7000003d400 d6e6400000254002c400000 000d32 50HGEF HC +CG 50 6c77747 00000070646d6 0887675656 11 50000000 lebar = PQ = SR = 6 cm. Perbandingan 008676696577000 000000033646573630000018400 0000246c756 L64657363000001840000006c777470740000 = × EG × FH =0740000 × 6d6464000002c4000000 × 8 = 24 cm2 d48502 22 = ⇔ = 01f000000014626b70740000020400000014 d69000003f8 464000002c40040000034c0 panjangnya: 003d4000000246c 33646573 00006c77747074000001f000000 01f0000 0000086766 88767565640000034c0 CGAB CG 20 HC +12 5 000000146d6 020000 7258595a00000218000000146758595a0000 7. Jawaban: c 0001462 Skala = 10000887675656 : 200 756d69000003f800 63000001 014626b70740000020400000014 9657700000 ABCD 12 4 56173000004 40000034c0000 0000086766965770000 == = 022c000000146258595a0000024000000014 ⇔ = 6b70740 3d40000002 0000146d6561730 Panjang kebun pada denah = 7 cm 84000000 000000 7258595a0000021800000014675 0c000000247 PSQR93 12 2 008676696577046c756d690 0000204 646d6e640000025400000070646d64640000 03d4000000246c756d6 000040c00000024 6c777470 Panjang kebun sebenarnya =00003d4000000 200 × 7 8595a0000022c00000014625859 46563680000 000000 50000001 Perbandingan 04300000000 ⇔ HC=9000003f8000000146d × 12246c756d69000 – 12 = 18 00003f8000 746563680000043 74000001 cm 02c400000088767565640000034c00000086 5a0000024000000014646d6e640 4725859 000146d656 000000 276696577000003d4000000246c756d690000 Jadi, panjang AC dan HC berturut-turut 10 c7254524300 003f800000014 00000000c725452 f0000000 lebarnya: 000025400000070646d64640000 5a00000 =6d65617300000 1.400 cm 1730000040 = = ADBC = 84 6561730000040c00000 00043c00000 cm dan 18 cm. 000000 2180000 c000000247 430000043c00000 PQSR63 Perbandingan sisi-sisi yang 14626b70 02c400000088767565640000034 =03f8000000146d6561730000040c000000247 14 m Lebar kebun pada40c0000002474 denah = 80c67545243 0247465636800000430 4656368000004300000000c7254524300000 0014675 4656368000 000000 80c675452430000 74000002 0000043c000 c0000008676696577000003d400 6563680000043 4,5 cm Lebar kebun sebenarnya = bersesuaian sama ABCDADBC 4 8595a00 0043000000 43c0000080c675452430000043c0000080c6 0000000c72545243000 0080c625452 00000000c7254 5. 200 Ketiga bangun sebangun, berarti 043c0000080c625 04000000 000000 0000246c756d69000003f800000 × 4,5 cm 00022c0 52430000043c000c7254524 yaitu = = = = . 430000043c0 25452430000043c0000080c7465787400000 0043c0000080c675452 perbandingan yang bersesuaian tetap 452430000043c00 14725859 3 000000 =000436f70797269676874202863292031393 900 cm 0146d6561730000040c00000024 =sisisisi 9 m 0000014 Luas kebun PSQRPQSROleh karena sudut-sudut yang 000080c7465 000080c67545230000043c0 6258595 000080c675 00080c746578740 dan besar sudut-sudut yang bersesuaian 5a000002 2 74656368000004300000000c725 430000043c0000080c6 78740000000 430000043c000 bersesuaian sama besar dan sisi-sisi yang sebenarnya = 14a000002 × 9 = 000000 126 m 938204865776c6574742d5061636b6172642 4524300000 0000000436f70797 sama. 18000000 0436f707972 0080c62545243 452430000043c0000080c675452 56 25452430000043c0000 Jadi, sebenarnya 126 bersesuaian sebanding, maka persegi a. = xluas ⇔xkebun =5 × = 10 cm 4000000 43c0000080 0436f6d70616e79000064657363000000000 000000 69676874202 0000043c00000 269676874202863 3 96 430000043c0000080c625452430 2 6314675859 014646d m . 080c746578740000000 panjang ABCD dan persegi panjang PQRS 0000012735247422049454336313936362d3 86329203139 80c7465787400c625452430 000000 292031393938204 5a000002 000043c0000080c746578740000 6e64000 3y = 96 ⇔ y = = 32 cm 39382048657 000000436f707 000043c000 22e310000000000000000000000127352474 sebangun. 0436f70797269676874 0025400 865776c6574742d 2c000000 000000 C3 0000436f7079726967687420286 76c6574742d 97269676874200080c74657 22049454336313936362d322e31000000000 23 0000706 4 2028632920313939382 5061636b6172642 z2 14625859 × 5061636b617 28632920313938740000000 = 3292031393938204865776c6574 ⇔z = = 1 cm 000000000000000000000000000000000000 3 6 6 46d6464 000000 26420436f6d 938204865776c0436f70797 0436f6d70616e790 5a000002 742d5061636b61726420436f6d7 04865776c6574742d50 000002c 2696768742 0000000004 cm 5 cm 10 cm000000 206574742d50616 cm 70616e79000 000646573630000 40000000 0616e7900006465736300000000 4000000 61636b61726420436f6 06465736300 36b61726420430286329203 000000 8876756 000000000012735 14646d6e 000000127352474220494543363 A 00000000000 6f6d70616e790 1393938204 d70616e790000646573 5640000 000000 247422049454336 64000002 01273524742 0006465736300865776c657 13936362d322e31000000000000 034c000 6300000000000000127 20494543363 00000000000014742d50616 313936362d322e3 54000000 000000000012735247422049454 0008676 011637 13936362d32 273524742204936b6172642 3524742204945433631 100000000000000 70646d64 336313936362d322e3100000000 6965770 072740 2e310000000 45433631393630436f6d706 00003d4 000000001273524 64000002 000000000000000000000000000 3936362d322e3100000 00000000000 62d322e31000016e7900006 0000002 000015 742204945433631 c4000000 00001273524 00000000000004657363000 0000000000000000000 0000000000000000012 46c756d 000000 3 74220494543 00000127352470000000000 3936362d322e310 88767565 6900000 7352474220494543363 36313936362 42204945433630127352474 000000000000000 64000003 3f80000 033646 d322e310000 13936362d322e2204945433 13936362d322e310000 00146d6 000000000000000 4c000000 573630 00000000000 31000000000006313936362 5617300 0000000000000000000 000000000000000 86766965 00000000000 0000000000000d322e31000 000018 00040c0 2 00000000000 00000000000000000000000 77000003 0000000000000000000 00000000CD − 0000024 400000 00000000000 00000000000000000000001 2 d4000000 7465636 2735247422 000000000000 16 cm 8 cm OC A 000000 0000C

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffeffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffeffd8ffe ffd8ffe ffd8ffe ffd8ffe ffd8ffe000104a4 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 000104a 2 64946000102010 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 4649460 0c800c80000ffe20 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 0010201 c584943435f5052 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 00c800c 4f46494c4500010 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 80000ffe 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 20c5849 3. Jawaban: a Pada gambar 8. Oleh karena ABCD dengan berikut EFGH b. r = 12 cm p3100000c484c696e 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 43435f5 ∆AOB kongruen dengan ∆COD. 312 6f021000006d6e7 sebangun dan 11 × GF =0524f46 BC maka GH = CD. =0524f46 ⇔ p 0524f46 = 0524f46 = 7,2 cm 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 0524f46 125 5 q1 12 47252474220585 44494c450 11 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 494c450 Y = 0010100 ⇔ q 95a2007ce000200 = = 2,4 cm 0000000000000000000000000000000000000000000000 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 0010100 EH = GF= BC = ×0010100 2 = 0,5 cm 1 12 55 A. Pilihan 4 000c484 4 09000600310000 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 000c484 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 Ganda 6. Ketiga foto c696e6f bayi616373704d5346 tersebut berbentuk c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f c696e6f 1Jawaban: c Perhatikan ∆ABD dan ∆BCD. AB = CD EF =c696e6f 1 c696e6f 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 GH= CD = × 14 = 1 cm persegi panjang dan saling sebangun 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 0210000 44 c = BC – b – GF = 2 – 0,7 – 0,5 = (sisi sejajar jajargenjang) AD = BC (sisi sejajar 54000000004945 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 sehingga perbandingan sisisisinya sama. 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74Oleh 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 06d6e74 jajargenjang) BD =06d6e74 BD 43207352474200 (berimpit) karena 0,8 cm d = AB – a – EF = 4 – 0,4 – 1 000000000000000000000000000000000000000000000000000000000000000000000000000000000 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 7252474 xdm 7252474 6dm panjang ketiga sisinya sama, maka ∆ABD dan = 2,6 cm = 00000000000000 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 2205859 1,2m 0,9m ∆BCD kongruen. 9. Perhatikan jajargenjang ffd8ffe000104a4 000000000000f6d 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c 5a2007c ABQP dan 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 2Jawaban: b 1,2 m × 6 dm 64946000102010 jajargenjang PQCD berikut. ⇔ x dm = e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 e000200 0,9 m 6000100000000d 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 Jadi, koordinat titik D yang mungkin adalah (–3, – 0c800c80000ffe2 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0900060 0c584943435f505 32d48502020000 2). Jawaban: b Segitiga 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 O 12 ×60310000 4. 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000 0310000kongruen adalah 24f46494c450001 ⇔ x= = 8 ffd8ffe000104a46494600 00000000000000 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 9 6163737 segitiga yang bentuk dan ukurannya sama. a. 0100000c484c69 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 6163737 ffd8ffe000104a4649460001020 C6163737 F ffd8ffe000104a4649460001020100c800c80 00000000000000 100c800c80000ffe20c58494343 Dari 01020100c800c80000ffe2 gambar di atas diperoleh pasangan 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 04d5346 ycm 6e6f021000006d 16 cm 04d5346 6e747252474220 000ffe20c584943435f50524f46494c4500010 5f50524f46494c4500010100000 = PQ A 0c584943435f50524f4649 segitiga kongruen sebagai berikut. ∆ABF B 00000000000000 0,9m58595a2007ce00 5400000 5400000 5400000 1,2m 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 5400000 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 100000c484c696e6f021000006d6e74725247 c484c696e6f021000006d6e7472 4c4500010100000c484c69 ≅ ∆DEF ∆BCF ≅ ∆CDF ∆ACF ≅ ∆CEF ∆ACD 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 0004945 02000900060031 422058595a2007ce00020009000600310000 5247422058595a2007ce000200 0,9 m0004945 × 1600000000000000 cm 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 6e6f021000006d6e747252 ≅∆BCE Jadi, ada 44320735 pasang segitiga ⇔ 0000616373704d y cm= 616373704d53465400000000494543207352 ∠ PAB dan ∠ DPQ sehadap makayang ∠PAB = 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 4320735 1,2 m –3 09000600310000616373704d53 00000000000000 47422058595a2007ce000 00000000000000000000000000000000000000000000000 53465400000000 A2474200 BD E 47420000000000000000000000000000f6d60 46540000000049454320735247 kongruen. 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 2474200 ∠200090006003100006163 DQP. ∠ ABQ 2474200 dan2474200 ∠ –2PQC sehadap maka 49454320735247 00100000000d32d485020200000000000000 00000000000000 42000000000000000000000000 × –1 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 ⇔ y= Segitiga =916 12 ABC dan DEF kongruen karena ketiga ∠73704d534654000000004 ABQ = ∠PQC. ∠ BQP0000f6d6000100000000d32d485 dan ∠QCD 42000000000000 000000000000000000000000000000000000 12 –1 0 00000001163707 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 sisinya sama panjang dan ketiga sudutnya 00000000000000 sehadap maka ∠ BQP = ∠QCD. ∠APQ dan 000000000000000000000000000000000000 02020000000000000000000000 945432073524742000000 1 Jadi,sama nilai xbesar. = 8 0000000 dan 27400000150000 nilai y = 12. 00f6d600010000 000000000116370727400000150000000336 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 00000000000000000000000000 ∠000000000000000000000 PDC sehadap maka ∠ APQ = ∠PDC. 2 5. Jawaban: c AB = PO 0000d32d485020 4657363000001840000006c7774707400000 00000000000000000000000000 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 00000f6 3 00033646573630 Diperoleh sudut-sudut yang 7. b. B 20000000000000 –3 0f6d6000100000000d32d4 1f000000014626b7074000002040000001472 00000000000000000000011637 maka ∠ ACB = ∠ PTO BC = TO bersesuaian E d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 d600010 sama besar. AB bersesuaian 00000000000000 58595a00000218000000146758595a000002 07274000001500000003364657 00001840000006c 850202000000000000000 maka ∠ BAC = ∠0000000 TPO0000000 AC = PT dengan PQ 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 00000000000000 2c000000146258595a000002400000001464 363000001840000006c7774707 dan AB = PQ. PQ bersesuaian dengan CD 4A 000000000000000000000 77747074000001f maka ∠ ABC = ∠d32d485 POTd32d485 00000000000000 6d6e640000025400000070646d6464000002 4000001f000000014626b707400 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 d32d485 dan PQ = CD. AQ bersesuaian dengan PD 000000000000000000000 00000000000000 c400000088767565640000034c0000008676 000204000000147258595a0000 000000014626b7 A0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 0202000 6. dan Jawaban: c ∆ABC dan ∆DEF kongruen. AQ = PD. 0202000 BP bersesuaian dengan QC 000000000000000000000 00000000000000 696577000003d4000000246c756d69000003f 0218000000146758595a000002 p 0000000 4p 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 07400000204000 D F Oleh karena ∠ A = ∠ F dan ∠ C = ∠ E maka ∠B A00000000000011 C 8000000146d6561730000040c00000024746 2c000000146258595a00000240 dan BP = QC. Diperoleh sisi-sisi yang 000000000000000001163 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 63707274000001 56368000004300000000c725452430000043 D, 0000000 sehingga BC = DE,00000014646d6e640000025400 AB =Oleh DF, dan 000147258595a0 Misal: p = panjang benda diperbesar = ∠ 707274000001500000003 bersesuaian sama panjang. karena Segitiga ABC dan DEFsebelum tidak kongruen 50000000336465 c0000080c675452430000043c0000080c6254 000070646d6464000002c40000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 00002180000001 364657363000001840000 sudut-sudut yang bersesuaian sama besar karena sisi AC ≠ DF. 73630000018400 52430000043c0000080c7465787400000000 0088767565640000034c000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 006c77747074000001f000 00006c77747074 436f707972696768742028632920313939382 dan sisi-sisi yang bersesuaian sama 8676696577000003d400000024 46758595a00000 AC = EF. Jadi, 0000000 pilihan c. 000001f0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 AD 04865776c6574742d5061636b61726420436f 6c756d69000003f8000000146d6 000014626b70740000020 panjang, maka jajargenjang ABPQ dan yang benar c. 22c000000146258 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 6d70616e7900006465736300000000000000 561730000040c0000002474656 ffd8ffe000104a4649460001020100 4000000147258595a0000 a. 14626b70740000 Oleh karena p = 3 jajargenjang PQCD kongruen. 02040000001472 10. Misal: p′ = 0000000 panjang persegi 12735247422049454336313936362d322e31 368000004300000000c7254524 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 595a0000024000 c800c80000ffe20c584943435f50524 7. 0218000000146758595a0 Jawaban: b ∆ABC dan ∆DCB kongruen 58595a00000218 000000000000000000000012735247422049 D 30000043c0000080c675452430 panjang kecil f46494c4500010100000c484c696e6 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 dengan ∠ BAD = ∠ BDC dan ∠ABD = ∠BCD 000014646d6e64 000022c00000014625859 00000014675859 454336313936362d322e3100000000000000 000043c0000080c62545243000 f021000006d6e7472524742205859 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 ⇔ maka BD = BC dan AD = BD. Jadi, panjang BC 5a0000022c0000 000000000000000000000000000000000000 0043c0000080c7465787400000 00000254000000 5a0000024000000014646 5a2007ce000200090006003100006 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 00146258595a00 0000 000436f70797269676874202863 = BD = AD = 4 cm. ⇔ maka: AB = p ′ + 3 16373704d53465400000000494543 d6e640000025400000070 B00024000000014 C E F Segitiga ABC dan DEF kongruen 70646d64640000 292031393938204865776c6574 –2 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 20735247420000000000000000000 646d6464000002c400000 646d6e64000002 karena memenuhi sifat (s – sd – s). 742d5061636b61726420436f6d7 02c400000088767 000000000f6d6000100000000d32d 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 088767565640000034c00 5400000070646d 0616e790000646573630000000 48502020000000000000000000000 8. Jawaban: b ∆ABC ≅∆PQR Oleh 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 565640000034c00 6464000002c400 00000001273524742204945433 00008676696577000003d 00000000000000000000000000000 d. 00008876756564 karena ∠ ACB = ∠ PQR maka AB = PR. A D 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 0116370 L′4000000246c756d690000 =0116370 6313936362d322e31000000000 00008676696577 00000000000000000000000000000 Oleh karena AC = QR maka ∠ 0000034c000000 9 9 216 00000000000001273524742204 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 7274000 ABC = 00000000000000011637072740000 ffd8ffe000104a4649460001020100c800c B 000003d4000000 03f8000000146d65617300 86766965770000 9454336313936362d322e31000 ∠ QPR. 01500000003364657363000001840 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 0015000 80000ffe20c584943435f50524f46494c450 ⇔00040c000000247465636 p′ · 03d4000000246c 00000000000000000000000000 246c756d6900000 000006c77747074000001f0000000 0010100000c484c696e6f021000006d6e74 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 0000336 756d69000003f8 0000000000000000000000000D 8000004300000000c7254 14626b70740000020400000014725 725247422058595a2007ce0002000900060 b. 000000146d6561 Benda diperbesar k kali 3f8000000146d65 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 4657363 8595a00000218000000146758595a 52430000043c0000080c6 0310000616373704d53465400000000494 maka: 730000040c0000 19.200 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 0000018 61730000040c000 0000022c000000146258595a00000 B CE F0000018 Segitiga siku-siku sama kaki 5432073524742000000000000000000000 kp × k0000018 75452430000043c000008 00247465636800 p×A 24000000014646d6e640000025400 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 0000000f6d6000100000000d32d48502020 00024746563680 mempunyai sudut yang sama. Segitiga 0c625452430000043c000 0004300000000c 000070646d6464000002c40000008 0000000000000000000000000000000000 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 6c77747 2 ABC dan DEF kongruen 1karena 19.200 00004300000000c 0080c7465787400000000 8767565640000034c000000867669 k72545243000004 =0740000 = 0000000000000000000000000000000000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 0740000 3c0000080c6754 6577000003d4000000246c756d690 memenuhi sifat 72545243000004 (sd – s – 192 sd). 20000000000000000000000000011637072 p436f707972696768742028 ′= = = 12 k52430000043c00 =01f0000 100 ′ A 2 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 01f0000 00003f8000000146d656173000004 7400000150000000336465736300000184 632920313939382048657 00080c62545243 k= 10 Ukuran benda 3c0000080c67545 setelah diperbesar 10 cm 0c000000247465636800000430000 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0001462 0000006c77747074000001f000000014626 Panjang AB = p ′ + 3 76c6574742d5061636b61 0000043c000008 0000c725452430000043c0000080c kali adalah: panjang = 10 × p = 10 × 24 dm b707400000204000000147258595a00000 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 6b70740 2430000043c0000 0c746578740000 726420436f6d70616e7900 675452430000043c0000080c62545 218000000146758595a0000022c0000001 = 240 dm lebar = 10 × 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000204 0000436f707972 080c62545243000 2430000043c0000080c7465787400 006465736300000000000 46258595a0000024000000014646d6e640 69676874202863 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 0000001 000000436f7079726967687420286 000025400000070646d6464000002c4000 000127352474220494543 0043c0000080c74 29203139393820 3292031393938204865776c657474 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 4725859 00088767565640000034c0000008676696 4865776c657474 36313936362d322e31000 65787400000000 2d5061636b61726420436f6d70616 577000003d4000000246c756d69000003f8 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 5a00000 2d5061636b6172 000000000000000000012 e7900006465736300000000000000 000000146d6561730000040c0000002474 436f70797269676 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 2180000 6420436f6d7061 12735247422049454336313936362 735247422049454336313 656368000004300000000c725452430000 6e790000646573 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 0014675 87420286329203 d322e310000000000000000000000 043c0000080c675452430000043c0000080 A A 936362d322e3100000000 63000000000000 12735247422049454336313936362 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 8595a00 c625452430000043c0000080c7465787400 M2 13939382048657 00127352474220 000000000000000000000 d322e310000000000000000000000 000000436f7079726967687420286329203 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 00022c0 49454336313936 000000000000000000000 76c6574742d5061 00000000000000000000000000000 1393938204865776c6574742d5061636b6 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 0000014 362d322e310000 4 000 0000P Q 4 1726420436f6d70616e7900006465736300 636b6172642043 00000000000000 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 6258595 0000000000001273524742204945433631 00001273524742 6f6d70616e79000 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 a000002 3936362d322e3100000000000000000000 20494543363139 0012735247422049454336313936362d32 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 4000000 06465736300000 36362d322e3100 2e31000000000000000000000000000000 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 014646d 00000000000000 00000000012735 000000000000000000000000 00000000000000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 6e64000 A A 00000000000000 24742204945433 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0025400 0000000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe0001 ffd ffd 4 ffd8ffe000104a4 ffd8ffe000 ffd8ffe00010 ffd8ffe000 ffd8ffe0001 ffd8ffe0 4 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 04a4649460 8ffe 8ffe 64946000102010 104a46494 4a464946000 104a46494 04a4649460 00104a4 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 001020100c 000 000 0c800c80000ffe2 600010201 1020100c800 600010201 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 800c80000ff 104 104 001020100c 6494600 0c584943435f505 00c800c80 c80000ffe20c 00c800c800 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 e20c584943 a46 a46 800c80000ff 0102010 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 435f50524f4 494 494 24f46494c450001 000ffe20c5 584943435f5 00ffe20c58 L 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 6494c45000 600 600 L e20c584943 0c800c80 0100000c484c69 84943435f 0524f46494c 4943435f50 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 10100000c4 010 010 6e6f021000006d 50524f464 45000101000 524f46494c 9. Jawaban: d ∆ABC dan ∆PQR kongruen ffd8ffe000104a4 435f50524f4 000ffe20 62d322e31000000000000000000000000000000000000000000000000000000 84c696e6f02 201 QT =201 6e747252474220 94c450001 00c484c696e 450001010 1000006d6e 00c 00c dengan AC = PR dan AB = PQ sehingga: ∠P = 649460001020100 6494c45000 c584943 7472524742 800 800 58595a2007ce00 0100000c4 2 26f021000006 0000c484c6 ∠A ⇔ 2x° = 60° c800c80000ffe20c TS = c80 3 + 3 = + = 1899Oleh c80 karena 10100000c4 435f5052 2058595a20 02000900060031 84c696e6f0 d6e74725247 96e6f02100 584943435f50524 07ce000200 000 000 panjang ketiga sisi ∆PQR dan ∆QST 84c696e6f02 4f46494c 0000616373704d 21000006d 422058595a2 0006d6e74 f46494c45000101 ⇔ x= 30 ∠Q= ∠B 0900060031 ffe2 ffe2 3 3 sama,0c5 maka ∆PQR ≅∆QST. 1000006d6e 4500010 53465400000000 6e7472524 007ce000200 725247422 0000616373 0c5 00000c484c696e6 = (x + 20)° = (30 704d534654 849 c. Panjang PQ =09000600310 QS maka ∠849 PRQ = 49454320735247 742205859 058595a20 f021000006d6e74 7472524742 100000c + 20)° = 50° 0000000049 434 434 L ∠QTS. Panjang PR = QT maka ∠ PQR 42000000000000 5a2007ce0 00061637370 07ce000200 L = 725247422058595 ∠R = ∠C = 180° – 2058595a20 484c696e 4543207352 35f 35f ∠QST. Panjang RQ = TS maka ∠505 RPQ = 00000000000000 002000900 4d534654000 090006003 a2007ce00020009 4742000000 505 (∠A + ∠B) = 180° – (60° 07ce000200 6f021000 00f6d600010000 060031000 00000494543 100006163 0000000000 24f ∠TQS. 24f 000600310000616 + 50°) = 180° – 110° = 4. Koordinat titik M yang mungkin adalah 0900060031 006d6e7 0000000000 464 464 0000d32d485020 061637370 20735247420 73704d534 373704d53465400 70° Jadi, pernyataan yang salah adalah ∠ R = 80°. M (–4, 2), M (–2, 4), M (1, –3), dan M (3, – 00f6d60001 94c 94c 4 1 2 3 4 654000000 0000616373 4725247 20000000000000 4d5346540 00000000000 000000494543207 00000000d3 450 450 10. 00000000000000 000000049 00000000000 004945432 1). YY 352474200000000 704d534654 4220585 2d48502020 001 001 3 5010 50103 2 Oleh karena AC = CD, ∠ACB = ∠DCE, dan 0000000000 00000000000000 454320735 00000f6d600 073524742 000000000000000 0000000049 95a2007c 2 0000000000 000 000 BC = CE 00000000000000 247420000 0100000000d 000000000 Jawaban: b makaT ∆ABC ≅∆CDE. Sebagai 00000f6d6000100 4543207352 e000200 0000000000 0c4 0c4 M1 akibatnya: ∠CED = ∠ABC = 28° 00000000000000 000000000 32d48502020 000000000 000000d32d48502 0000000000 84c 84c 4742000000 0900060 ∠AEB = ∠AED – 00000000000000 000000000 00000000000 000000000 020000000000000 0000000000 696 696 X X ∠CED= 65° – 0000000000 e6f 0000000000 e6f 0310000 00000000000011 000000f6d 00000000000 0f6d600010 000000000000000 0000000000 021 021 63707274000001 600010000 00000000000 0000000d3 28° = 37° 000000000000000 0000000000 6163737 0000000000 000 000 50000000336465 0000d32d4 00000000000 2d4850202 –3006 –3 YY 000000000000000 0000000000 006 00f6d60001 04d5346 3 73630000018400 850202000 00000000000 000000000 B. Uraian 000000000000000 0000116370 d6e d6e 00000000d3 5400000 5747 5747 7274000001 00006c77747074 000000000 00000000000 000000000 000000000000000 2 5000000033 252 252 2d48502020 0004945 2 1. ∆PQR 000001f0000000 000000000 00000000000 000000000 000000011637072 6465736300 474 474 1 1 ≅∆PRS ∆PUS 0000000000 14626b70740000 000000000 00000000000 000000000 4320735 740000015000000 220 0001840000 220 ≅∆QRV ∆PQV 02040000001472 000000000 00000011637 000000000 033646573630000 006c777470 585 585 0000000000 2474200 X X ≅∆SUR 74000001f0 95a 95a 58595a00000218 000000000 07274000001 000000000 Jadi, ada 3 pasang segitiga yang kongruen. 01840000006c777 0000000000 0000000 0000001462 200 200 00000014675859 000000000 50000000336 000000000 47074000001f000 2. a. Perhatikan ∆KMN dan ∆LMN. 6b70740000 7ce 7ce 0000000000 0000000 M 3 5a0000022c0000 000000000 46573630000 000000000 000014626b70740 –3000 – 000 0204000000 1) KN = LM (trapesium sama kaki) 0000000000 0000000 00146258595a00 000000000 01840000006 000000000 –4∠C dengan 000020400000014 147258595a 3 200 5. a. FD di 200 depan –4 ∠E dan AB di depan 2) ∠KNM = 2 ∠LMN (trapesium sama –3 0000021800 090 090 00024000000014 000000000 c7774707400 000000000 0000000000 00000f6d 7258595a0000021 –3 ∠E = ∠C maka FD = AB = 12 –2 kaki) 3) NM = MN 0000146758 006 006 –2 646d6e64000002 000000000 0001f000000 000001163 800000014675859 0000000000 6000100 cm. 003 595a000002 003 –1 (berimpit) ∆KMN dan ∆LMN memenuhi syarat s–1 1 5400000070646d 001163707 014626b7074 707274000 10 1 5a0000022c00000 2c00000014 100 0000000000 000000d 01 sd-s, sehingga ∆KMN ≅∆LMN. DE= 100 ffd8ffe000104a464946000 6464000002c400 274000001 00000204000 001500000 2 6258595a00 006 0146258595a0000 006 2 1020100c800c80000ffe20c 0000000000 32d4850 3 b. Perhatikan ∆KON dan ∆LOM. 1) KN 0002400000 163 163 00008876756564 500000003 00014725859 003364657 024000000014646 3 584943435f50524f46494c4 4 0014646d6e 737 737 0000116370 2020000 = LM (trapesium sama kaki) 2) ∠NKO = 0000034c000000 364657363 4 5a000002180 363000001 500010100000c484c696e6f d6e640000025400 04d 6400000254 04d 021000006d6e7472524742 ∠MLO (∆KMN ≅∆LMN) 3) ∠NOK = ∠MOL 86766965770000 000001840 00000146758 840000006c 7274000001 0000000 ffd8ffe000104a4649460001020100c800c80000ffe 000070646d64640 0000007064 534 534 2058595a2007ce00020009 03d4000000246c 000006c77 595a0000022 777470740 (bertolak belakang) ∆KON dan ∆LOM 6d64640000 654 654 00002c400000088 5000000033 0000000 20c584943435f50524f46494c4500010100000c484 –4 000600310000616373704d 02c4000000 000– 756d69000003f8 25 = 000 5747074000 cm FD +c0000001462 DE = 00001f0000 memenuhi syarat s-sd-s, sehingga ∆KON 767565640000034 –3 5346540000000049454320 6465736300 0000000 4 –3 c696e6f021000006d6e74725247422058595a2007c 8876756564 000 000000146d6561 00014626b –1 7352474200000000000000 12 + 000 5001f00000 =–2 17 cm 58595a00000 ≅∆LOM. c00000086766965 –2 –1 0000034c00 004 004 0001840000 0000000 e00020009000600310000616373704d5346540000 –1 00000000000000f6d60001 730000040c0000 0014626b7 24000000014 707400000 –1 77000003d400000 0000867669 945 945 3. a. Y 01 00000000d32d4850202000 006c777470 0000000 01 00004945432073524742000000000000000000000 00247465636800 074000002 646d6e64000 204000000 6577000003 432 432 0246c756d690000 0000000000000000000000 2 2 d400000024 073 073 0004300000000c 040000001 00254000000 147258595 74000001f0 0000000 03f8000000146d6 0000000f6d6000100000000d32d48502020000000 C 0000000000000000000000 3 C3 6c756d6900 524 524 72545243000004 47258595a 70646d64640 a00000218 0000000000000000000000 4 561730000040c00 0000001462 0000000 4 00000000000000000000000000000000000000000 0003f80000 742 742 0000000000000000000000 3c0000080c6754 000002180 00002c40000 000000146 000024746563680 K 00146d6561 000 000 6b70740000 0000000 K 0000116370727400000150 00000000000000000000000000000000000000000 52430000043c00 000001467 00887675656 758595a00 730000040c 000 000 00004300000000c 0000003364657363000001 0204000000 0000000 00000011637072740000015000000033646573630 0000002474 000 000 00080c62545243 58595a000 00022c0000 725452430000043 A R40000034c00 840000006c777470740000 6563680000 000 000 147258595a 0000000 0000043c000008 0022c0000 00008676696 001462585 01f000000014626b707400 00001840000006c77747074000001f00000001462 c0000080c675452 –2 –1 0430000000 000 000–1 –2 000204000000147258595a 0c746578740000 001462585 577000003d4 95a000002 0000021800 0000000 430000043c00000 000 0c72545243 000 6b707400000204000000147258595a00000218000 1 0000021800000014675859 0000436f707972 95a000002 000000246c7 400000001 0000043c00 000 000 80c625452430000 0000146758 0000000 5a0000022c000000146258 000146758595a0000022c000000146258595a0000 B 00080c6754 000 000 69676874202863 400000001 56d69000003 4646d6e64 043c0000080c746 595a000002400000001464 595a000002 0000000 C 5243000004 000 000 024000000014646d6e640000025400000070646d6 C 29203139393820 4646d6e64 f8000000146 000002540 6d6e640000025400000070 578740000000043 3c0000080c 0f6 0f6 2c00000014 1163707 M4 Q 646d6464000002c4000000 4865776c657474 000002540 d6561730000 000007064 464000002c400000088767565640000034c000000 6f7079726967687 6254524300 d60 d60K 88767565640000034c0000 K 6258595a00 2740000 2d5061636b6172 000007064 040c0000002 6d6464000 00043c0000 001 001 420286329203139 8676696577000003d4000000246c756d69000003f 008676696577000003d400 080c746578 000 000 6420436f6d7061 6d6464000 47465636800 002c400000 0002400000 0150000 3938204865776c6 0000246c756d69000003f80 8000000146d6561730000040c0000002474656368 7400000000 000 000 6e790000646573 002c40000 00043000000 088767565 00000146d6561730000040 574742d5061636b 0014646d6e 0003364 436f707972 00d 00d –2 000004300000000c725452430000043c0000080c6 c000000247465636800000 –2 63000000000000 008876756 00c72545243 640000034c 61726420436f6d7 –2 6967687420 32d 32d 6400000254 6573630 4300000000c72545243000 –1–1 00127352474220 564000003 0000043c000 000000867 75452430000043c0000080c625452430000043c00 2863292031 485 0616e7900006465 485 0043c0000080c675452430 0000007064 0000184 01 3939382048 020 020 49454336313936 4c0000008 0080c675452 669657700 736300000000000 000043c0000080c6254524 00080c7465787400000000436f707972696768742 2 65776c6574 200 200 6d64640000 0000006 362d322e310000 676696577 430000043c0 0003d4000 30000043c0000080c74657 000127352474220 3 742d506163 000 000 02863292031393938204865776c6574742d506163 87400000000436f7079726 00000000000000 000003d40 000080c6254 000246c756 02c4000000 c777470 494543363139363 000 4 6b61726420 000 9676874202863292031393 6b61726420436f6d70616e7900006465736300000 5 00001273524742 00000246c 52430000043 d69000003f 436f6d70616 000 000 62d322e31000000 7400000 8876756564 938204865776c6574742d5 6 e790000646 000 000 00000000012735247422049454336313936362d32 20494543363139 756d69000 c0000080c74 800000014 000000000000000 061636b61726420436f6d70 0000034c00 1f000000 5736300000 000 000 36362d322e3100 003f80000 65787400000 6d6561730 616e790000646573630000 012735247422049 2e310000000000000000000000127352474220494 0000000001 000 000 0000867669 014626b –2 0000000000127352474220 00000000000000 00146d656 000436f7079 000040c000 454336313936362 2735247422 000 000 54336313936362d322e3100000000000000000000 49454336313936362d322e 6577000003 7074000 00000000000000 173000004 72696768742 000247465 0494543363 000 000 d322e3100000000 3100000000000000000000 0000000000000000000000000000000000Q 13936362d3 000 000 00000000000000 0c0000002 02863292031 636800000 d400000024 0020400 000000000000000 0012735247422049454336 22e3100000 000 000= PQ. 2 dan Akibatnya, ∠BAC = 39393820486 ∠PRQ BC 474656368 430000000 313936362d322e31000000 000000000000000 000000000013 − 6c756d6900 0000147 0000000000 000 000 0000000000000000000000 2 000004300 5776c657474 0c72545243 Jadi, pernyataan yang salah ∠ BCA = 000000000000000 0000000127 000 000 12 = 169 − 144 0003f80000 258595a 0000000000000000000000 2 2 000000c72 2d5061636b6 0000043c00 3524742204 000 ∠PRQ. 000 0EF − FD = 0000 =

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104 ffd8ffe0001 ffd8ffe0001 ffd8ffe00 ffd8ffe00 R 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 a46494600010 04a4649460 04a46494600 0104a4649 0104a4649 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 20100c800c800 001020100c 01020100c80 460001020 460001020 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c ffd8ffe000104a4 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 00ffe20c58494 800c80000ff 0c80000ffe20 100c800c8 100c800c8 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 64946000102010 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 3435f50524f46 e20c584943 c584943435f 0000ffe20c 0000ffe20c 0000000000000000000000000000000000000000000 0c800c80000ffe20

ffd8ffe000104a46 49460001020100c 800c80000ffe20c5 84943435f50524f4 6494c4500010100 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 494c45000101 435f50524f4 50524f46494 584943435 584943435 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 00000c484c696 6494c45000 c4500010100 f50524f464 f50524f464 5. Jawaban: b CEB ∆ABC dan ∆ADE ∠ AEB c584943435f5052 dan ∠ berpelurus, 7. a. Perhatikan ∆ABC dan000c484c696e6f02 ∆DEC 4f46494c4500010 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 e6f021000006d 10100000c4 000c484c696 94c450001 94c450001 sebangun. AD bersesuaian dengan AB dan DE s sehingga: ∠ AEB + ∠ CEB = 180° BC = EC (diketahui) ∠BCA = ∠1000006d6e74725 ECD ffd8ffe000104a46 100000c484c696e 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6e7472524742 84c696e6f02 e6f02100000 0100000c4 0100000c4 bersesuaian dengan BC. (bertolak belakang) CA = CD ∠A AEB 49460001020100c + 69° = 180° 247422058595a20 6f021000006d6e7 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 2058595a2007 1000006d6e 6d6e7472524 84c696e6f 84c696e6f ⇔ ∠AEB = 180° – 69° = 800c80000ffe20c5 (diketahui) Jadi, ∆ABC ≅∆DEC. 07ce00020009000 47252474220585 0000000000000000000000000000000000000000000000 ce0002000900 7472524742 7422058595a 021000006 021000006 111° A. Pilihan 84943435f50524f4 Perhatikan ∆ABE. ∠BAE + ADDE AB ×DE 12 ×6 95a2007ce000200 600310000616373 060031000061 2058595a20 2007ce00020 d6e747252 d6e747252 = ⇔= BC = = = 18 cm ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 6494c4500010100 Ganda b. Sudut B menghadap sisi AC dan ∠ AEB A + ∠ ABE 1. Jawaban: b ∆PQR dengan ∠PQR = ABBC AD 4 Jadi, panjang180° BC = 09000600310000 6373704d5346 07ce000200 0090006003 474220585 474220585 000c484c696e6f02 704d53465400000 EPRQ menghadap sisi CD, sedangkan CA 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 42° + 616373704d5346 4 111° + ∠ABE = 180° 53°sudut dan ∠ = 64°. ∠ QPR = 180° – 18 cm. 540000000049 0900060031 1000061637 95a2007ce 95a2007ce 1000006d6e74725 = CD, ∠ABC =∠ CED = 72°. 000494543207352 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 (∠PQR + akibatnya ∠PRQ) ∠ABE 54000000004945 = 180° – 42° – 111° 454320735247 0000616373 3704d53465 000200090 000200090 6. 247422058595a20 ∠ECD = ∠BCA 474200000000000 = 27° ∠ CAD bersesuaian dengan 000000000000000000000000000000000000000000000000000000000000000000000000000000000 = 180° – (53° + 43207352474200 420000000000 006003100 006003100 ∆ABC dan 07ce00020009000 ∆ADE sebangun maka = 180° – ∠BAC 704d534654 –4000000004 ∠ABC = 180° ∠ ABE maka ∠CAD = ∠ABE = 27°. 000000000000000 64°) = 180° –0000000049 00000000000000 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 006163737 006163737 600310000616373 berlaku:cADDE AD DE Jawaban: – 36° 000000000000 – 72° = 72°9454320735 Jadi, besar = ⇔ =704d53465400000 117° = 63° 4543207352 00f6d6000100000 04d534654 04d534654 ABBC AD + 000000000000f6d DB BF + FC 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 ∠ABC000000f6d6000 dan ∠ECD2474200000 = 72°. 6000100000000d Segitiga yang sebangun dengan segitiga 100000000d32 4742000000 0000000000 000000004 000000004 10. a. Perhatikan ∆BCDAD dan 000494543207352 000d32d48502020 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 14 ∆DEF. ∠BDC = ⇔ = A 8. a. sama kaki dengan 32d48502020000 PQR ∆MON besard48502020000 sudut-sudutnya 0000000000 0000000000 sama 945432073 945432073 dengan 474200000000000 ∠ EDF karena bertolak belakang. AD + 8 14 + 6 ∠DCB = 000000000000000 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 00000000000000 0000000000 000f6d60001 524742000 524742000 besar sudutsudut segitiga PQR. Jadi, 000000000000000 ∠ DEF = 90° BD = DF = 26 cm Oleh karena MO =000000000000 MN maka: 1 2 20AD = 14AD + 112 000000000000000 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 00000000000000 000000000000 00f6d600010 00000000d3 000000000 000000000 segitiga yang sebangun dengan segitiga 00f6d60001000000 besar dua sudutnya sama dan panjang salah 6AD = 112 2 2 MON =∠ MNO0000000d32 =b. (180° – 000000000000000 00000000000000 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 000000000000 2d48502020 000000000 000000000 PQR∠ ada pada pilihan 00d32d485020200 2 satu sisinya sama, maka ∆BCD dan ∆DEF AD= 18 cm 2. Jawaban: d KMKL LM 000000000000000 00000000000000 000000000000 d485020200 0000000000 0000000f6 0000000f6 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 ∠OMN) kongruen. Jadi,000000000000000 ∆BCD ≅∆DEF. 1 00000000000000 000000000000 0000000000 0000000000 d60001000 d60001000 000000000000000 (180° – 000000000000000 ∆KLM ~ ∆ABC = = 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 . = dengan KM ~ AB maka KLM 2ABBCAC = 70° ∠ KON +0000000000 ∠ MON∠ =00000d32d 180°= ∠ACB 00000000000000 000000000000 0000000000 00000d32d 000000000000000 40°) 000000000000000 00000000000000000000000000000000000000000000000 KL ~(sudut ∠KML0000000000 = ∠BAC LM ~ AC 00000001163707 000000000000 0000000000 485020200 485020200 ( BC maka berpelurus) 000000000000000 000011637072740 b. ∆BCD dan ∆DEF kongruen, maka EF 27400000150000 maka ∠LKM = ∠ABC Hubungan yang 000000011637 0000000000 0000000000 000000000 000000000 000000000000000 = BC = 24 cm. Segitiga BCD siku-siku di C, 000015000000033 KON K + 70° =pilihan 180° 00033646573630 072740000015 0000000000 0000000000 000000000 000000000 benar ada pada d yaitu ∠LKM = 000000000000000 maka: ⇔∠ = 110°0000000000 .0000000000 . . (1)646573630000018 00001840000006c 000000033646 000000000 000000000 ∠MKL =KON ∠ ABC. 000116370727400 Perhatikan ∆PLN. ∠NPL40000006c777470 = 573630000018 0000000000 0000000000 000000000 000000000 000150000000336 CD= 77747074000001f 180° –40000006c777 (∠PNL +0000000000 ∠ PLN) 000000000 000000014626b7 0000000000 000000000 465736300000184 74000001f000000 3. Jawaban: d PR3 07400000204000 = 180° – (60° +0000000000 50°) = 70° ∠KPN + 47074000001f0 0000116370 000000000 000000000 0000006c7774707 014626b70740000 000147258595a00 183 ∠ NPL N = 180° (sudut berpelurus) 00000014626b 0001163707 7274000001 000000000 000000000 4000001f00000001 =⇒= SU4 SU4 707400000204 020400000014725 00021800000014 2740000015 5000000033 000000000 000000000 ∠KPN K + 70° = 180° = 4626b7074000002 676 −576 6758595a0000022 000000147258 0000000336 6465736300 000000000 000000000 8595a0000021800 040000001472585 ∠KPN . (2) (1) dan ⇔ = 110° 18 .×. 4 = 3Dari × SU c00000014625859 595a00000218 4657363000 0001840000 000000000 = 95a000002180000 100 (2) diperoleh ∠KON = 0000146758595a0 ∠000000000 KPN. KN 18 ×4 000116370 5a000002400000 000000146758 0018400000 006c7774707 000116370 00146758595a000 = 10 merupakan garis bagi ∠ Kcm maka ⇔ SU= = 24 000022c00000014 0014646d6e64000 595a0000022c 06c7774707 4000001f000 727400000 727400000 3 Jadi, panjang SU 0022c0000001462 cm Segitiga ABC siku-siku di C, maka: ∠OKN = ∠PKN. KN = NK (berimpit) 6258595a0000024 00254000000706 = 24Oleh cm. karena 000000146258 4000001f000 000014626b 150000000 150000000 58595a000002400 ∆KON dan000000014646d6e ∆KPN AC= 46d6464000002c4 595a00000240 7074000002 336465736 0000014646d6e64 4. Jawaban: adua000014626b mempunyai pasang 336465736 sudut yang 00000088767565 00000014646d 7074000002 0400000014 300000184 300000184 640000025400000 000002540000007 Perhatikan ∆ABD dan ∆ABC. bersesuaian sama besar dan 2 6e6400000254 0400000014 7258595a000 0000006c7 0000006c7 C 0646d6464000002 070646d64640000 = 640000034c00000 51 sepasang sisi yang bersesuaian ffd8ffe000104a4649460001020100c800c80000 08676696577000 00000070646d 7258595a00 0021800000 774707400 774707400 2 c40000008876756 sama panjang maka ∆KON dan 02c400000088767 −24 ffe20c584943435f50524f46494c4500010100000 = 5640000034c0000 2.602 −576 003d4000000246c 6464000002c4 0002180000 0146758595a 0001f0000 0001f0000 c484c696e6f021000006d6e74725247422058595 ∆KPN kongruen. 565640000034c00 b. PN000000887675 bersesuaian dengan NO maka 756d69000003f80 0014675859 0000022c000 00014626b 00014626b D a2007ce00020009000600310000616373704d53 008676696577000 PN465400000000494543207352474200000000000 = NO = 8 cm. 5a0000022c = 003d4000000246c 2.025 000086766965770 00000146d65617 65640000034c 0001462585 707400000 707400000 b. ∠ACB = ∠DEF = 64° 00000000000000000f6d6000100000000d32d48 = 45 cm 30000040c000000 000000867669 0000001462 95a00000240 204000000 204000000 00003d400000024 9. a. 502020000000000000000000000000000000000 Perhatikan ∆ACD dan ∆ABE. ∠BAC756d69000003f800 = 180° – (64° + AD = AC – CD = 45 – 10 = 35 cm 24746563680000 6577000003d4 58595a0000 0000001464 147258595 147258595 DC = AE (diketahui) 0000146d6561730 6c756d69000003f 000000000000000000000000000000000000000 90°) = 26° AB memenuhi Luas segitiga ADF: 04300000000c725 000000246c75 6d6e6400000 a00000218 a00000218 B000000000000000000000011637072740000015 S0240000000 ∠DCA = ∠BAE 000040c00000024 8000000146d6561 syara BD 6. a. Perhatikan ∆AEB dan ∆DEF. X 00000003364657363000001840000006c777470 452430000043c00 6d69000003f80 14646d6e64 2540000007 000000146 000000146 = 42° (sudut dalam 1 746563680000043 s t – sd – s 730000040c00000 74000001f000000014626b70740000020400000 L=∠B × AD × EF BAE = ∠ FDE = 90° 00080c675452430 00000146d656 0000025400 0646d64640 758595a00 758595a00 00000000c725452 2 b berseberangan) 0147258595a00000218000000146758595a0000 memenuhi ∠BAD =∠ BAC (berimpit) maka BD ~ BC 024746563680000 000043c0000080c 1730000040c0 000070646d 00002c40000 00022c000 00022c000 DE = EA 430000043c00000 022c000000146258595a0000024000000014646 AC = –3 ABP (diketahui) Jadi, 1 62545243000004 syarat sd – s – sd ∠AEB =∠ 000002474656 6464000002 0088767565 000146258 000146258 04300000000c725 d6e640000025400000070646d6464000002c400 = 80c675452430000 ×DEF 35 × 24 AC ∆ACD dan kongruen. ( 000088767565640000034c00000086766965770 )∠ BDA = ∆ABE ∠ CBAc400000088 (siku-siku) maka AB b. Panjang sisi-sisi ∆PQR:595a00000 2=Sehingga 420 Jadi, luas segitiga ADF 3c0000080c74657 368000004300 640000034c0 595a00000 ∆AEB dan ∆DEF kongruen. 043c0000080c625 BC 452430000043c00 b. 00003d4000000246c756d69000003f800000014 ∆ABC sama kaki maka: 2 AD PQ Oleh = 000000c72545 4 karena 7675656400 adalah 420 cm .cm 87400000000436f 0000086766 240000000 240000000 b. DE = AE = 5 452430000043c00 6d6561730000040c00000024746563680000043 ∠ABD = ∠BCA maka AD1~00080c675452430 AB ( ) AB 2430000043c0 70797269676874 9657700000 14646d6e6 14646d6e6 2 200034c0000 AB +1 00080c746578740 DF bersesuaian dengan AB maka DF = 00000000c725452430000043c0000080c675452 = ∠CBA =2 (180° – ~ AC (∠BCA ) 000043c0000080c 20286329203139 000080c67545 0086766965 3d40000002 400000254 400000254 430000043c0000080c625452430000043c00000 0000000436f70797 AB = 10 cm. ∠2BAC) Diperoleh perbandingan sisi: = = 2 625452430000043 80c7465787400000000436f7079726967687420 3938204865776c6 ADBD AB = 2430000043c0 46c756d6900 000000706 000000706 ABBCAC 269676874202863 QR 3 +sisi-sisi 3 =77000003d4 + = 18 1 99Panjang 2863292031393938204865776c6574742d50616 Luas ∆DEF = 1× DE × DF 574742d5061636 =000000246c (180° –46d646400 42°) = 000080c62545 0003f800000 46d646400 c0000080c746578 292031393938204 36b61726420436f6d70616e7900006465736300 ∆QST: = 4 ∆BEC sama kaki maka 2 Oleh QS karena 2 b61726420436f6d 69°. 2430000043c0 756d690000 0146d65617 0002c4000 0002c4000 865776c6574742d 7400000000436f7 000000000000127352474220494543363139363 ∠CEB000080c74657 = ∠BCE = 69°. 70616e790000646 = 21 × 5 × 10 03f80000001 30000040c00 000887675 000887675 62d322e31000000000000000000000012735247 5061636b6172642 079726967687420 422049454336313936362d322e3100000000000 57363000000000 874000000004 46d6561730 0000247465 656400000 656400000 0436f6d70616e790 = 25 2 286329203139393 000000000000000000000000000000000000000 00000127352474 36f7079726967 000040c000 6368000004 34c000000 34c000000 000646573630000 cm 0000 8204865776c6574 22049454336313 687420286329 0002474656 300000000c7 867669657 867669657 000000000012735 936362d322e3100 203139393820 3680000043 2545243000 7000003d4 7000003d4 742d5061636b617 247422049454336 Kunci Jawaban dan Pembahasan 00000000000000 4865776c6574 00000000c7 0043c000008 000000246 000000246 313936362d322e3 26420436f6d7061 00000012735247 742d5061636b 2545243000 0c675452430 c756d6900 c756d6900 100000000000000 6e7900006465736 42204945433631 61726420436f6 0043c00000 000043c0000 0003f8000 0003f8000 000000001273524 300000000000000 3936362d322e310 d70616e79000 80c6754524 080c6254524 000146d65 000146d65 742204945433631 127352474220494 00000000000000 064657363000 30000043c0 30000043c00 617300000 617300000 3936362d322e310 00000000000000 000000000001 000080c625 00080c74657 40c000000 40c000000 54336313936362d 000000000000000 00000000000000 273524742204 4524300000 8740000000 247465636 247465636 000000000000000 322e31000000000 2 945433631393 43c0000080 0436f707972 800000430 800000430 00000000000BD 000000000000000 000000000000012 2 2 2 2 2 6362d322e310 c746578740 6967687420 0000000c7 0000000c7

−BC = 26 −24 00000000AB −BC

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 ffd8ffe000104a4 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 64946000102010 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 0c800c80000ffe20 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 DB = panjang bayangan siswa c584943435f5052 Segitiga AEH sebangun dengan segitiga 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 = 3,54f46494c4500010 m ∆ABC dan ∆DBE sebangun. AB ABG. 9 =EHAE ⇒AB = EH 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 bersesuaian dengan DB dan AC 100000c484c696e BG 96 8 + 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 bersesuaian dengan DE, sehingga berlaku 6f021000006d6e7 × 89 0000000000000000000000000000000000000000000000 perbandingan: 47252474220585 ⇔EH= = 4,8 cm 15 EF = EH + HF = 4,8 + 16 95a2007ce000200 AB AC = = 20,8 cm Jadi, panjang EF = 09000600310000 DB DE 20,8 cm. 616373704d5346 AC ⇔ 5654000000004945 = 9. Jawaban: 3,5 1,5 43207352474200 1,5 ×56 a KN L ⇔ AC = 00000000000000 ffd8ffe000104a464946000102 3,5 = 24 Jadi, tinggi gedung 0100c800c80000ffe20c584943 000000000000f6d sebenarnya 24 m. 435f50524f46494c4500010100 6000100000000d 000c484c696e6f021000006d6e 32d48502020000 74725247422058595a2007ce0 00000000000000 ffd8ffe000104a4649460 0020009000600310000616373 B. 14 ∆KOL sebangun dengan ∆NOM, 704d534654000000004945432 00000000000000 001020100c800c80000ff 0735247420000000000000000 maka: KO KL KO105 ffd8ffe000104a464946 00000000000000 e20c584943435f50524f4 1. Uraianffd8ffe000104a4649460 000000000000f6d60001000000 0001020100c800c80000 00000000000000 3 2 6494c4500010100000c4 = ⇔ = = 00d32d4850202000000000000 OMNM OM147 001020100c800c80000ff ffe20c584943435f50524 00000000000000 0000000000000000000000000 84c696e6f021000006d6 Jadi, panjang AD = 18 e20c584943435f50524f4 0000000000000000000000000 f46494c4500010100000 000000000000003 e74725247422058595a2 5 ⇔KO = OM 0000000000000000000000000 6494c4500010100000c4 c484c696e6f021000006 00000001163707 007ce000200090006003 7. cm. 7Q titik tengah KM, 0000000011637072740000015 84c696e6f021000006d6 d6e7472524742205859 27400000150000 10000616373704d53465 0000000336465736300000184 maka: 11 Ae74725247422058595a2 5a2007ce000200090006 00033646573630 0000006c77747074000001f000 40000000049454320735 Jawaban: a 007ce000200090006003 000014626b707400000204000 MQ = KM = (KO + OM) 00310000616373704d5 00001840000006c Q 7 P 24742000000000000000 2 2 000147258595a000002180000 10000616373704d53465 3465400000000494543 77747074000001f 0000000000000f6d6000 15 112 00146758595a0000022c00000 a.40000000049454320735 Perhatikan ∆OPQ dan ∆SOR. 1) ∠POQ =( OM + OM) = ( OM) 2073524742000000000 000000014626b7 100000000d32d4850202 0146258595a00000240000000 27 27 6 = 24742000000000000000 ∠SOR karena8bertolak belakang. 2) ∠OPQ 0000000000000000000f 07400000204000 14646d6e64000002540000007 00000000000000000000 = 0000000000000f6d6000 ∠ORS karena merupakan pasangan sudut 0646d6464000002c400000088 6d6000100000000d32d 000147258595a00 = OM 00000000000000000000 C 767565640000034c000000867 B100000000d32d4850202 D dalam berseberangan. 3) ∠ OQP = ∠OSR 761 4850202000000000000 00021800000014 00000000000000000000 6696577000003d4000000246c 00000000000000000000 karena merupakan 0000000000000000000 6758595a0000022 00000000000000000000 OQ = OM – MQ = OM – OM = OM 756d69000003f8000000146d65 ∆CDE siku-siku di E, maka: 00000000000000000000 77∆POQ sebangun dengan 0000000000000000000 c00000014625859 00000000000000011637 61730000040c0000002474656 22 DC = 00000000000000000000 368000004300000000c725452 0000000000000000000 5a000002400000 pasangan sudut dalam berseberangan. ∆NOM, maka: 07274000001500000003 430000043c0000080c6754524 00000000000000000000 0000000000000000000 0014646d6e64000 1 Oleh karena besar ketiga sudutnya 36465736300000184000 OM 22 6 + 8 PQOQ PQ 30000043c0000080c62545243 =00254000000706 00000000000000011637 0000000116370727400 sama, maka ∆OPQ dan ∆SOR = ⇔= 7 0006c77747074000001f 0000043c0000080c746578740 07274000001500000003 NMOM 14 OM 0000000436f707972696768742 =46d6464000002c4 sebangun. 3618 +cm 64 Misalkan 000000014626b7074000 b. 0001500000003364657 SQ = OS = x cm 36465736300000184000 0286329203139393820486577 363000001840000006c7 00000088767565 00204000000147258595 1 ⇔PQ= × 14 = 2 cm Jadi, maka OQ = (18 – x) cm. ∆OPQ dan ∆SOR 6c6574742d5061636b6172642 7 = 0006c77747074000001f 100640000034c00000 = 10 cm ∆ABC sebangun 7747074000001f000000 a0000021800000014675 sebangun, maka: 0436f6d70616e7900006465736 000000014626b7074000 014626b707400000204 08676696577000 dengan ∆CDE, maka: panjang PQ = 2 cm. 8595a0000022c0000001 3000000000000001273524742 SR OR SR 8 00204000000147258595 ffd8ffe000104a4649460001020100c800 000000147258595a000 003d4000000246c 10. 2049454336313936362d322e3 Jawaban: d Persoalan tersebut dapat = ⇒= 46258595a00000240000 a0000021800000014675 1000000000000000000000012 PQOP 18 12 18 00014646d6e640000025 c80000ffe20c584943435f50524f46494c4 digambarkan sebagai berikut. BC0021800000014675859 AC 756d69000003f80 BC15 7352474220494543363139363 8595a0000022c0000001 =×8 ⇒= 500010100000c484c696e6f021000006d6 5a0000022c0000001462 00000146d65617 ⇔SR= = 12400000070646d6464000 cm ECCD 8 10 62d322e310000000000000000 46258595a00000240000 e74725247422058595a2007ce00020009 58595a0000024000000 30000040c000000 12 OSOR x8 002c4000000887675656 0000000000000000000000000 ×815 00014646d6e640000025 000600310000616373704d53465400000 014646d6e6400000254 24746563680000 0000000000000M ⇔= BC= =40000034c00000086766 12 cm =18 −x 12 ⇒ O OQOP 400000070646d6464000 10 00000070646d6464000 BD04300000000c725 = BC – DC = 12 – 10 000494543207352474200000000000000 96577000003d40000002 12x = (18 – x) × 8BD = 00000000000000f6d6000100000000d32 002c4000000887675656 452430000043c00 = 002c4000000887675656 21 cm Jadi, panjang 2 46c756d69000003f8000 40000034c00000086766 12x 1 = 144 – 8x d48502020000000000000000000000000 40000034c00000086766 00080c675452430 cm. 000146d6561730000040 96577000003d40000002 20x 2 = 144 c Pada gambar 000000000000000000000000000000000 96577000003d4000000 000043c0000080c c0000002474656368000 8. Jawaban: di 46c756d69000003f8000 x 62545243000004 = 7,2 Diperoleh: OS004300000000c7254524 = 7,2 cm dan OQ = 000000000000000000000000000000000 246c756d69000003f800 000146d6561730000040 B 18 – 7,2 = 11,8 cm Jadi, SR = 12 cm, OS = 000011637072740000015000000033646 0000146d65617300000 3c0000080c74657 30000043c0000080c675 A D c0000002474656368000 7,2 cm, dan OQ = 11,8 cm. 57363000001840000006c777470740000 40c00000024746563680 87400000000436f 452430000043c0000080 004300000000c7254524 Keteranga 01f000000014626b70740000020400000 00004300000000c72545 70797269676874 c625452430000043c000 30000043c0000080c675 0147258595a0000021800000014675859 2430000043c0000080c6 20286329203139 n: AC = tinggi gedung AB = = 16 cm B 6 cm0080c746578740000000 452430000043c0000080 5a0000022c000000146258595a0000024 75452430000043c00000 3938204865776c6 0436f707972696768742 panjang bayangan gedung GC G= – 16 = = 90° 8 2) 2. BG= a.BC 1)–∠ ABC =24 ∠CDE c625452430000043c000 000000014646d6e640000025400000070 80c625452430000043c0 574742d5061636 = 56 m DE = tinggi siswa = 02863292031393938204 cm ∠ BCD B = ∠ BCA + ∠ ACE + ∠ ECD 0080c746578740000000 646d6464000002c400000088767565640 000080c7465787400000 b61726420436f6d 865776c6574742d50616 1,5 m 0436f707972696768742 000034c0000008676696577000003d400 000436f7079726967687 70616e790000646 180° = ∠BCA + 90° 1 + ∠ECD 36b61726420436f6d706 02863292031393938204 0000246c756d69000003f8000000146d6 4202863292031393938 57363000000000 ∠BCA = 90° – ∠ECD 16e79000064657363000 865776c6574742d50616 561730000040c00000024746563680000 204865776c6574742d50 00000127352474 00000000000127352474 36b61726420436f6d706 04300000000c725452430000043c00000 61636b61726420436f6d 22049454336313 22049454336313936362 16e79000064657363000 80c675452430000043c0000080c625452 70616e7900006465736 936362d322e3100 d322e310000000000000 00000000000127352474 430000043c0000080c746578740000000 3000000000000001273 00000000000000 00000000012735247422 22049454336313936362 0436f7079726967687420286329203139 5247422049454336313 00000012735247 049454336313936362d3 d322e310000000000000 3938204865776c6574742d5061636b617 936362d322e31000000 42204945433631 22e31000000000000000 00000000012735247422 26420436f6d70616e7900006465736300 0000000000000000127 3936362d322e310 00000000000000000000 049454336313936362d3 000000000000127352474220494543363 3524742204945433631 00000000000000 0000000000000000000A 22e31000000000000000 13936362d322e31000000000000000000 3936362d322e3100000 00000000000000 16 cm samping AG D 00000000000000000000 000012735247422049454336313936362 0000000000000000000 00000000000000 9 0000000000000000000 d322e3100000000000000000000000000 0000000000000000000 00000000000DE sejajar CD, sehingga: cm E00000000000 0000000000000000000000000000 + EC SR GC = HF= AD E F

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 ∠E ECD + ∠CDE + ∠CED = 180° 3CF = CF + 12 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 2CF =12 ∠E ECD + 90° + ∠CED = 180° 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 CF= 6 cm ffd8ffe000104a46 ∠CED = 90° – ∠ECD Diperoleh 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 49460001020100c8 ∠BCA = ∠CED. 4. S 12R 0000000000000000000000000000000000000000000000 00c80000ffe20c584 ffd8ffe000104a46494600010 3) ∠943435f50524f464 BAC = 180° – (∠ABC + 20100c800c80000ffe20c58494 94c4500010100000 ∠BCA) = 180° – (∠CDE + ∠CED) 3435f50524f46494c45000101 c484c696e6f02100 00000c484c696e6f021000006 = ∠CED d6e74725247422058595a200 0006d6e74725247 Oleh karena besar sudut-sudut yang 7ce000200090006003100006 P 422058595a2007ce bersesuaian sama maka ∆ABC 2216373704d534654000000004 Segitiga OMN sebangun dengan 000200090006003 sebangun dengan ∆CDE. 945432073524742000000000 segitiga OPQ. MNOM 10 x 10000616373704d 0000000000000000000f6d600 b. Segitiga ABC siku-siku di B. 0100000000d32d4850202000 534654000000004 =x3 ⇒= 22 + 2 PQOP 000000000000000000000000 AC= 945432073524742 000000000000000000000000 10(x + 3) = 22x ⇔10x + 000000000000000 000000000000000000000000 30 = 22x ⇔10x – 22x = – 22 0000000000000f6d 000000000000000000001163 = 20 +15 707274000001500000003364 30 ⇔ –12x = –30 3 6000100000000d3 657363000001840000006c77 x = 2,5 cm 2d4850202000000 = 400 +225 747074000001f000000014626 000000000000000 b70740000020400000014725 = 000000000000000 625 = 25 cm 8595a0000021800000014675 Segitiga OMN sebangun dengan segitiga 8595a0000022c00000014625 000000000000000 c. ∆ABC sebangun dengan segitiga ∆CDE. OSR. 8595a0000024000000014646 =MNOM ⇒= 102,5 000000000000000 SROR 12 y d6e640000025400000070646 CD 15 CE = 000000000000000 × AC = × 25 d6464000002c400000088767 10y = 12 × 2,5 AB20 = 18,75 565640000034c00000086766 000000000000001 10y = 30 96577000003d4000000246c7 cm 163707274000001 y = 3 cm 56d69000003f8000000146d65 500000003364657 CD 15 MR = x + y 61730000040c000000247465 DE = 363000001840000 × BC = × 15 6368000004300000000c7254 AB20= 11,25 = 2,5 + 3 52430000043c0000080c67545 006c777470740000 = 5,5 cm cm 2430000043c0000080c625452 01f000000014626b Jadi, 430000043c0000080c7465787 3. D 9 C 707400000204000 panjang 400000000436f707972696768 4 ffd8ffe000104a464946000102 F 000147258595a00 742028632920313939382048 MR = 5,5 E0100c800c80000ffe20c5849434 65776c6574742d5061636b61 000218000000146 35f50524f46494c450001010000 cm. 726420436f6d70616e7900006 0c484c696e6f021000006d6e74 758595a0000022c0 465736300000000000000127 725247422058595a2007ce0002 00000146258595a A0009000600310000616373704d B 352474220494543363139363 000002400000001 62d322e31000000000000000 53465400000000494543207352 a.47420000000000000000000000 ∆EAG sebangun dengan ∆DAC. 000000012735247422049454 4646d6e64000002 336313936362d322e3100000 000000f6d6000100000000d32d 5400000070646d6 EGAE AE ×DC 89 × 000000000000000000000000 48502020000000000000000000 = ⇔ EG= = 464000002c400000 DCAD 84 AD + 000000000000000000000000 00000000000000000000000000 088767565640000 00000000000000000000000000 0 Q 72 034c000000867669 00000000000000000000000011 = 12 = 6 cm 63707274000001500000003364 6577000003d4000 b.657363000001840000006c7774 ∆EAG sebangun dengan ∆DAC. 000246c756d69000 7074000001f000000014626b70 003f8000000146d6 AGAE 8 82 7400000204000000147258595a =561730000040c000 = = = 8412 ACAD+3 00000218000000146758595a00 00022c000000146258595a0000 ∆CGF 000247465636800 sebangun dengan 024000000014646d6e64000002 0004300000000c72 5400000070646d6464000002c4 ∆CAB GF = EF – EG = 14 – 6 5452430000043c00 00000088767565640000034c00 = 8 cm 00008676696577000003d40000 00080c6754524300 GFGC AC −AG AC AG 00246c756d69000003f8000000 00043c0000080c62 ===– 146d6561730000040c00000024 ABACAC ACAC 5452430000043c00 74656368000004300000000c72 5452430000043c0000080c6754 800080c7465787400 21 ⇔ = 1000000436f707972 –= 52430000043c0000080c625452 AB33 430000043c0000080c74657874 696768742028632 ⇔AB = 3 × 8 = 24 cm 00000000436f70797269676874 920313939382048 c.20286329203139393820486577 ∆CGF sebangun dengan ∆CAB. 65776c6574742d50 6c6574742d5061636b61726420 CFGF CF 81 436f6d70616e79000064657363 61636b617264204 CBAB CF +12 243= ⇔ = = C 00000000000000127352474220 36f6d70616e79000 49454336313936362d322e3100 064657363000000 00000000000000000000127352 000000001273524 47422049454336313936362d32 2e310000000000000000000000 742204945433631 00000000000000000000000000 3936362d322e310 000000 000000000000000 000000127352474 220494543363139 36362d322e31000 000000000000000 000000000000000 000000000000000 2 2 000000AB +BC

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 ffd8ffe000104a4 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 64946000102010 ffd8ffe000104a46 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 9. B 0c800c80000ffe20 ffd8ffe000104a46 AE BE ∆ABC sebangun dengan ∆AEB maka == 49460001020100c 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 . c584943435f5052 49460001020100c ABBCAB AC AEAB AE 8 800c80000ffe20c5 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 4f46494c4500010 800c80000ffe20c5 84943435f50524f4 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 100000c484c696e 84943435f50524f4 6494c4500010100 0000000000000000000000000000000000000000000000 6f021000006d6e7 6494c4500010100 =⇒= 000c484c696e6f02 472524742205859 000c484c696e6f02 ABAC 8 10 88 64 1000006d6e74725 × 5a2007ce0002000 1000006d6e74725 ⇔ AE = = = 6,4 cm CD 20 247422058595a20 m 90006003100006 247422058595a20 10 10BEAB BE 8 07ce00020009000 16373704d534654 07ce00020009000 600310000616373 =⇒= 00000000494543 600310000616373 Misal:704d53465400000 BC = tinggi704d53465400000 tiang BCAC 6 10 20735247420000 000494543207352 × bendera B ED = tinggi tiang 00000000000000 000494543207352 ⇔ BE = =68 = 48 4,8 cm 474200000000000 bendera D ∆ABC dan ∆AED 0000000000f6d60 474200000000000 10 10 000000000000000 CE = AC – 00100000000d32 000000000000000 sebangun. = 00f6d6000100000 ⇔ = BC CA BC20 EDDA 2 5 AE = 10 – d4850202000000 00f6d6000100000 000d32d48502020 6,4 = 3,6 cm ⇔ BC000d32d48502020 =4×2=8m 00000000000000 000000000000000 00000000000000 000000000000000 ∆ABE sebangun dengan ∆CDE, 22Panjang tali : AB = 000000000000000 00000000000000 000000000000000 maka: DECE DE3,6 000000000000000 =⇒= 00000000000000 000000000000000 000000000000000 =2220 + 8 BEAE 4,8 6,4 00000000000000 000000000000000 000000000000000 4,8 00000000000000 000000000000000 ⇔ DE= = 3 × 0,9 = 2,7 ×3,6 6,4 = 464000000000000000 ≈ 21,5 m Jadi, tinggi tiang bendera 8 00000116370727 000000000000000 cm 000011637072740 m dan40000015000000 panjang tali 21,5 m. Jadi, panjang DE = 2,7 ffd8ffe000104a464946000 cm. 000011637072740 000015000000033 A 1020100c800c80000ffe20c 03364657363000 000015000000033 7. Misalkan garis 646573630000018 584943435f50524f46494c4 001840000006c77 646573630000018 tinggi ∆ABC adalah 40000006c777470 10. A 500010100000c484c696e6f 747074000001f00 40000006c777470 ffd8ffe000104a464946000 H AD.Titik D di tengah74000001f000000 021000006d6e7472524742 B1020100c800c80000ffe20c 0000014626b707 74000001f000000 tengah BC. ∆AIH 014626b70740000 2058595a2007ce00020009 584943435f50524f46494c4 40000020400000 014626b70740000 D 020400000014725 dan ∆HGC 000600310000616373704d 5.BE=500010100000c484c696e6f 0147258595a0000 020400000014725 021000006d6e7472524742 8595a0000021800 sebangun. 5346540000000049454320 B9F D G 9C 6 m2058595a2007ce00020009 4 m02180000001467 8595a0000021800 2 2 = 0000146758595a0 25 −15 7352474200000000000000 58595a0000022c0 0000146758595a0 000600310000616373704d AI IH =AI⇔6= 000022c00000014 5346540000000049454320 HGGC 12 9 12 ×6 00000000000000f6d60001 00000146258595a = 6258595a0000024 625 −225 000022c00000014 7352474200000000000000 00000000d32d4850202000 00000240000000 6258595a0000024 ⇔ AI = = 8 cm 00000000000000f6d60001 000000014646d6e 0000000000000000000000 = 400 = 20 cm Perhatikan 9 11 14646d6e6400000 000000014646d6e 00000000d32d4850202000 640000025400000 0000000000000000000000 ∆ADE dan ∆FBE. ∆ADE dan 25400000070646 640000025400000 0000000000000000000000 L∆AEH22= × EH × ∆AEH AI = 0000000000000000000000 × 12 × 8 = 48 cm2 070646d64640000 Jadi, luas 0000000000000000000000 Misalkan d6464000002c400 BC = x m 070646d64640000 dan PQ = t m. ∆FBE 02c400000088767 sebangun maka berlaku 0000000000000000000000 48 cm2. 0000000000000000000000 ∆PQC 00008876756564 02c400000088767 sebangun dengan ∆ABC, perbandingan sebagai 0000000000000000000000 565640000034c00 0000116370727400000150 0000034c0000008 565640000034c00 0000116370727400000150 8. Misalkan panjang sisi persegi adalah x berikut. ED AE ED50 maka: PQQC tQC 0000003364657363000001 = 000086766965770 ⇒= 67669657700000 000086766965770 0000003364657363000001 cm. A 8 – 6x ∆BQP sebangun dengan 00003d400000024 840000006c777470740000 ABBC (i) 840000006c777470740000 =BEEF ⇔ = ffd8ffe000104a464946000102010 3d4000000246c75 00003d400000024 2025 6c756d69000003f8 01f000000014626b707400 maka: PQ BQ xQC 0c800c80000ffe20c584943435f505 01f000000014626b707400 ∆BCD, x 6d69000003f8000 6c756d69000003f8 000204000000147258595a t − ⇔ED = 2 × 20 000000146d65617 D24f46494c4500010100000c484c69 000204000000147258595a 000146d6561730 000000146d65617 0000021800000014675859 ==30000040c000000 6e6f021000006d6e7472524742205 40 4⇒ x =cm AD AE AD50 0000021800000014675859 CDBC 000040c00000024 30000040c000000 5a0000022c000000146258 8595a2007ce000200090006003100 x 247465636800000 5a0000022c000000146258 (ii)595a000002400000001464 =BFEF ⇔ =1525 t xQC 00616373704d5346540000000049 74656368000004 247465636800000 4300000000c7254 6d6e640000025400000070 ⇔ = – 4300000000c7254 595a000002400000001464 C B4543207352474200000000000000 4= 2 × 15 4xx 300000000c72545 1 ⇔ AD 646d6464000002c4000000 52430000043c000 00000000000000f6d600010000000 6d6e640000025400000070 2430000043c0000 52430000043c000 88767565640000034c0000 0d32d48502020000000000000000 = 30 cm Jadi, t panjang tt ED = 40 cm dan 5 0080c6754524300 Segitiga ADE sebangun dengan segitiga 646d6464000002c4000000 008676696577000003d400 080c67545243000 = 1 – 0080c6754524300 0000000000000000000000000000 AD = 30 cm. 4 6 00043c0000080c6 88767565640000034c0000 0000246c756d69000003f80 ABC, 0000000000000000000000000000 sehingga: AD DE − 0043c0000080c62 00043c0000080c6 3t = 12 – 2t 3 25452430000043c 00000146d6561730000040 0000000000000000000000011637 008676696577000003d400 6. Segitiga ABC siku-siku di B, maka: 5452430000043c0 25452430000043c 8x x =0727400000150000000336465736 ⇒= c000000247465636800000 5t = 12 5 0000080c7465787 0000246c756d69000003f80 AC= 22 000080c74657874 0000080c7465787 A 3000001840000006c777470740000 ABBC 8 15 4300000000c72545243000 12 400000000436f70 ((8 – x) × 15 00000146d6561730000040 = 8x t= = 400000000436f70 2,4 0043c0000080c675452430 00000000436f707 01f000000014626b7074000002040 797269676874202 22 000043c0000080c6254524 120 – 15x =c000000247465636800000 1 8x 00000147258595a0000021800000 797269676874202 = 972696768742028 8 +6 863292031393938 30000043c0000080c74657 0146758595a0000022c0000001462 63292031393938 863292031393938 23x = 120 120 14300000000c72545243000 5 87400000000436f7079726 204865776c65747 58595a0000024000000014646d6e6 = 64 +36 0043c0000080c675452430 204865776c65747 204865776c65747 x= = 5 9676874202863292031393 40000025400000070646d6464000 42d5061636b6172 000043c0000080c6254524 938204865776c6574742d5 42d5061636b617 42d5061636b6172 002c400000088767565640000034c 2323 Jadi, panjang sisi persegi = 6420436f6d70616e 100 061636b61726420436f6d7 30000043c0000080c74657 0000008676696577000003d40000 26420436f6d7061 6420436f6d70616e 0616e79000064657363000 terbesar yang dapat 587400000000436f7079726 = 790000646573630 10 cm 00246c756d69000003f8000000146 6e7900006465736 790000646573630 0000000000012735247422 d6561730000040c00000024746563 000000000000012 9676874202863292031393 30000000000000 000000000000012 dibuat Arif adalah 5 cm. 049454336313936362d322 68000004300000000c72545243000 735247422049454 23 938204865776c6574742d5 e310000000000000000000 012735247422049 735247422049454 0043c0000080c675452430000043c 336313936362d32 0001273524742204945433 061636b61726420436f6d7 0000080c625452430000043c00000 45433631393636 336313936362d32 2e3100000000000 6313936362d322e3100000 80c7465787400000000436f707972 0616e79000064657363000 2d322e310000000 2e3100000000000 0000000000000000000000 6967687420286329203139393820 000000000001273 0000000000012735247422 00000000000000 000000000001273 0000000000000000000000 4865776c6574742d5061636b61726 524742204945433 049454336313936362d322 00000Q C 012735247422049 524742204945433 420436f6d70616e79000064657363 6313936362d322e e310000000000000000000 0000000000000012735247422049 45433631393636 6313936362d322e 310000000000000 454336313936362d322e31000000 0001273524742204945433 2d322e310000000 310000000000000 0000000000000000127352474220 000000000000000 6313936362d322e3100000 00000000000000 000000000000000 49454336313936362d322e310000 000000000000000 0000000000000000000000 00000000000000 000000000000000 0000000000000000000000000000 00000000000 EF2 0000000000000000000000 00000000000000 00000000000AC + 0000000000000000000000 E 2 00000 E −BF 00000AB +BC BC

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 ffd8ffe000104a4649460001020100c800c8 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 0000ffe20c584943435f50524f46494c45000 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 10100000c484c696e6f021000006d6e74725 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 247422058595a2007ce0002000900060031 0000000000000000000000000000000000000000000000 A. Pilihan 0000616373704d534654000000004945432 Ganda 1. Jawaban: d Dua jajargenjang dikatakan 07352474200000000000000000000000000 00f6d6000100000000d32d4850202000000 sebangun jika sudutsudut yang bersesuaian ▲ 00000000000000000000000000000000000 sama besar dan sisi-sisi yang bersesuaian 00000000000000000000000000000000000 sebanding. Besar sudut lancip jajargenjang = 00000000000000000001163707274000001 42° Besar sudut tumpul jajargenjang = 180° – 500000003364657363000001840000006c7 42° 7747074000001f000000014626b70740000 = 138° 0204000000147258595a000002180000001 Jajargenjang yang sebangun ada pada pilihan d karena sudut-sudut yang bersesuaian sama besar 46758595a0000022c000000146258595a00 dan perbandingan sisi-sisi yang bersesuaian sama,00024000000014646d6e640000025400000 070646d6464000002c40000008876756564 8 6 yaitu 12 = . 0000034c0000008676696577000003d4000 9 000246c756d69000003f8000000146d65617 2. Jawaban: c Trapesium sama kaki 30000040c00000024746563680000043000 KLMN dan PQRS dapat digambarkan sebagai 00000c725452430000043c0000080c67545 berikut. 2430000043c0000080c625452430000043c 0000080c7465787400000000436f70797269 N P M ffd8ffe000104a464946000102 676874202863292031393938204865776c6 ffd8ffe000104a4649460001020100c800c80000ffe2 0100c800c80000ffe20c5849434 35f50524f46494c450001010000 574742d5061636b61726420436f6d70616e 0c584943435f50524f46494c4500010100000c484c6 0c484c696e6f021000006d6e74 79000064657363000000000000001273524 725247422058595a2007ce0002 7422049454336313936362d322e31000000 96e6f021000006d6e74725247422058595a2007ce0 0009000600310000616373704d 00000000000000001273524742204945433 53465400000000494543207352 K0020009000600310000616373704d5346540000000 47420000000000000000000000 6313936362d322e31000000000000000000 000000f6d6000100000000d32d 00000000000000000000000000000000000 049454320735247420000000000000000000000000 48502020000000000000000000 0 000f6d6000100000000d32d4850202000000000000 1Jawaban: a Dua buah persegi selalu sebangun 00000000000000000000000000 00000000000000000000000000 karena besar sudut-sudutnya selalu sama 90° dan 000000000000000000000000000000000000000000 panjang setiap sisi persegi00000000000000000000000011 sama sehingga 63707274000001500000003364 000000000000000000000000000000000000000001 perbandingannya dengan 657363000001840000006c7774 sisi persegi lainnya pun sama. 7074000001f000000014626b70 163707274000001500000003364657363000001840 7400000204000000147258595a 2Jawaban: d Diketahui jenis bangun persegi 00000218000000146758595a00 000006c77747074000001f000000014626b7074000 panjang maka sudutnya sama besar. Selanjutnya, 00022c000000146258595a0000 00204000000147258595a000002180000001467585 dibuktikan sisi-sisi yang bersesuaian sebanding. 024000000014646d6e64000002 5400000070646d6464000002c4 Misal: p = panjang 95a0000022c000000146258595a000002400000001 00000088767565640000034c00 00008676696577000003d40000 4646d6e640000025400000070646d6464000002c40 00246c756d69000003f8000000 146d6561730000040c00000024 0000088767565640000034c0000008676696577000 74656368000004300000000c72 5452430000043c0000080c6754 003d4000000246c756d69000003f8000000146d656 52430000043c0000080c625452 1730000040c0000002474656368000004300000000 430000043c0000080c74657874 00000000436f70797269676874 c725452430000043c0000080c675452430000043c0 20286329203139393820486577 6c6574742d5061636b61726420 000080c625452430000043c0000080c74657874000 436f6d70616e79000064657363 00000000000000127352474220 00000436f707972696768742028632920313939382 49454336313936362d322e3100 00000000000000000000127352 04865776c6574742d5061636b61726420436f6d706 47422049454336313936362d32 5Jadi, jarak titik potong kedua 16e790000646573630000000000000012735247422 2e310000000000000000000000 permukaan tanah 2,4 m. 00000000000000000000000000 049454336313936362d322e3100000000000000000 000000S 0000012735247422049454336313936362d322e310 000000000000000000000000000000000000000000 00000000000LQ R

Dari gambar tersebut

tampak ∠LKN = ∠QRS.

tali dengan

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe ffd8ffe0 ffd8ffe0 ffd8ffe d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000104a 00104a400104a4 000104a 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 4649460 64946006494600 4649460 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 0010201 01020100102010 0010201 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 00c800c 0c800c80c800c8 00c800c 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 80000ffe 0000ffe20000ffe2 80000ff 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 20c5849 0c584940c58494 e20c584 18. 6. Jawaban: Jawaban:da∆KLM Pada dua ∆PQR bangun Oleh datar karena 10. Jawaban: 98?. ≅ 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 43435f5 3435f505 3435f505 943435f LM yang = PQ sebangun, maka ∠ PRQ perbandingan = ∠ LKM = sisinya sama c Kedua persegi panjang sebangun, K 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0524f46 24f46494 24f46494 50524f4 dengan perbandingan keliling. Sehingga maka:=y494c450 400 5 ⇔ y= × 330 = 275 mm 6494c45 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 c450001 c450001 perbandingan keliling KLMN dan OPQR 6 330480 0000000000000000000000000000000000000000000000 0010100 01000000100000 0001010 R 11. Jawaban: c Ukuran sudut persegi 000c484 c484c69 c484c69 0000c48 adalah 2. L 3 panjang semuanya sama, yaitu 90°. Jika c696e6f 6e6f0210 6e6f0210 4c696e6 ukuran panjang dan lebarnya sama berarti 2 0210000 00006d600006d6 f021000 Keliling KLMN = × keliling kedua persegi panjang kongruen. 06d6e74 e747252e747252 006d6e7 3 M OPQR 7252474 4742205 4742205 4725247 12. Jawaban: b Pada segitiga (i) besar 2 = × 42 =P 28 cm 8595a20 8595a20 4220585 sudut yang 2205859 belum diketahui, adalah 180° – (45° 3 Belah ketupat mempunyai empat sisi yang 5a2007c 07ce000membandingkan 07ce000 95a2007 + 55°) = 80°. Dengan segitiga Jadi, besar ∠sama PQR = 41°. panjangnya yaitu: 2000900 2000900 ce00020 (i) dan (iii), e000200 diketahui kedua segitiga memenuhi Q 0900060 0600310 0600310 0090006 syarat kongruen, yaitu sd – s – sd. Jadi, segitiga 19. Jawaban: d Pada dua segitiga ?PQR = MN = 11 K = × 28 = 7 cm 0310000 00061630006163 0031000 (i) dan (iii) kongruen. yang kongruen berlaku: 1 2 (180? – 98?) = 44 6163737 73704d573704d5 0616373 sisi-sisi di hadapan 1 2sudut (82?) yang = 41?sama besar 04d5346 34654003465400 704d534 Jadi, panjangpanjang MN = 7 sama; cm. mempunyai 13. Jawaban: b Y Pada gambar di 6540000 5400000 0000004 0000004 sudut-sudut yang menghadap sisi yang sama 7. Jawaban: b Diketahui trapesium samping 0004945 94543209454320 0000494 panjang mempunyai besar sama. ABCD dan EFCG sebangun sehingga ∆MNO kongruen dengan 4320735 73524747352474 5432073 Sisi AB di hadapan sudut C dan sisi EF di hadapan diperoleh: 20000002000000 5247420 ∆PQR. 2474200 sudut D. Oleh karena ∠C = ∠D maka panjang 0000000 0000000 0000000 0000000 Jadi, koordinat titik R yang AB =EFEF. = EG 0000000 0000000 0000000 0000000X mungkin adalah (0, 4). Sudut AB A enghadap sisi BC dan sudut E 0000000 00000000000000 0000000 menghadap sisi DF. Oleh karena panjang 00000f6 0f6d6000 0f6d6000 000000f 14 8 ⇔ 14. Jawaban: c M P BC==42DF maka besar ∠A = ∠E. d600010 10000001000000 6d60001 AD Oleh karena ∆ABC dan ∆DEF kongruen ∆OBC dan ∆AOD kongruen maka berlaku 0000000 00d32d400d32d4 0000000 18 ∠C = ∠D dan ∠A = ∠E maka ∠B ⇔ = dengan hal-hal berikut. ∠ DOA = ∠ BOC maka d32d485 85020208502020 0d32d48 3 AD = ∠F. Jadi, pernyataan yang benar pilihan panjang0202000 sisi0000000 di depannya sama yaitu AD = 0000000 5020200 ⇔ AD = d. BC. AO 0000000 = BO dan OC = OD, maka AC = 00000000000000 0000000 20. 24 cm EF CG BD. ∠DAO = ∠CBO dan ∠OAB =0000000 ∠OBA 0000000 0000000 0000000 Segitiga=sama kaki dengan sudut (∆AOB sama kaki), maka AB CD 0000000 00000000000000 0000000 puncakc62°. 1 Jawaban: Besar sudut yang lain = (180° – 0000000 00000000000000 0000000 2 ⇔ = 14 CG 0000000 00000000000000 0000000 62°) 42 30 1 0000000 0000000 0000000 ⇔ CG = 10 cm = 15. Jawaban: c Segitiga kongruen0000000 yaitu 2 0000000 0000000 0000000 0000000 DG = CD – segitiga yang bentuk dan ukurannya sama. (118°) = 59° Segitiga yang sebangun dengan 0000000 0000000 0000000 CG = 30 – Dari gambar di 0000000 atas diperoleh pasangan segitiga pertama adalah segitiga sama kaki 0000000 0000000 0000000 0000000 10 = 20 cm segitiga kongruen sebagai berikut. ∆KLN yang besar sudut puncaknya 62° dan besar 0000000 0000000 0000000 Jadi, panjang AD = 24 cm dan DG = 20 cm. ≅∆LMN ∆KON ≅0000000 ∆MON ∆KOL ≅∆MOL Jadi, ada 3 sudut kakinya 59°. Jadi, segitiga yang 0000000 0000000 0000000 0000000 pasang segitiga yang kongruen. 8. sebangun Jawaban: b Segi empat ABCD dan dengan segitiga pertama ada 0000000 00000000000000 0000000 PQRS sebangun maka dari gambar diperoleh: pada pilihan c. 0000000 00011630001163 0000000 0116370 70727407072740 0011637 21. Jawaban: a Perhatikan ∆DFC dan 7274000 0000150 0727400 16. Jawaban: c ∠A0000150 = ∠D = 41° maka BC ∆EFA. ∠ DFC = ∠EFA (bertolak belakang) PSPQ 15= p1 =a⇒ 0015000 0000003 bersesuaian dengan EF0000003 sehingga BC =0001500 EF = 3 ∠ FCD = ∠ FAE (sudut dalam berseberangan) Akan dibuktikan bahwa = . p2 A2 ABBC 24 20 0000336 3646573 3646573 0000033 cm. ∠B = ∠ F= 83° maka AC bersesuaian ∠CDF = ∠AEF (sudut dalam berseberangan) 8 = 6 cm. 61 ∠C = 24sudutnya ×15 4657363 6300000 ⇔ a= = 18 dengan DE pilihan sehingga AC6300000 =A1DE Oleh karena ketiga pasang sama Pada d diperoleh: = = .6465736 20 Jadi, a = 18 cm. 16 122 3000001 0000018 1840000 1840000 ∠E = 56° maka AB bersesuaian dengan DF besar maka ∆DFC dan ∆EFA sebangun. Jadi, persegi panjang ABCD cm × 6 006c777 8400000 sehingga AB4000000 = 006c777 DF = 5 cm. Jadi, berukuran DE = 6 cm,8EF sebangun dengan persegi panjang06c7774 berukuran 9. 4707400 4707400 Jawaban: d Skala = 1 : 3.500 Panjang = 3 cm cm, dan6c77747 DF = 5 cm. 22. Jawaban: c= 1,4 km = 140.000 cm 16 cm × 120001f000 cm. 0740000 0001f000 7074000 jalan sebenarnya Perhatikan ∆ABC dandenah ∆CDE.= 140.000 : 3.500 01f0000 0000146 0000146 001f000 Panjang jalan pada 5. Jawaban: c Dari gambar 17. 26b7074 26b7074 0000146 AB diperoleh: 0001462 ∠BAD ~ ∠EFG maka ∠BAD = ∠ACB = ∠DCE, maka AB ~ DE 6b70740 0000020 0000020 26b7074 = 40 cm Jadi, panjang jalan pada denah ∠EFG = 86° ∠ADC = 180° – ∠BAD = ( ) Jawaban: 4000000 yang digambar DE Andi 40 cm. 180° – 86°d0000204 = 94° Jadi, 4000000 besar ∠ADC =0000020 0000001 1472585 1472585 4000000 94°. 4725859 95a000095a0000 1472585 5a00000 02180000218000 95a0000 2180000 00014670001467 0218000 0014675 58595a058595a0 0001467 8595a00 000022c000022c 58595a0 00022c0 00000010000001 000022c 0000014 46258594625859 0000001 6258595 5a000005a00000 4625859 a000002 24000002400000 5a00000 4000000 00146460014646 2400000 014646d d6e6400d6e6400 0014646 6e64000 00025400002540 d6e6400 0025400 00000700000070 0002540

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 104a46494 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 600010201 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 00c800c80 ffd8ffe000104a4 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 000ffe20c5 64946000102010 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 84943435f 0c800c80000ffe20 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 50524f464 Oleh karena sudut-sudut yang bersesuaian c584943435f5052 ffd8ffe000104a46 BC ∠BAC49460001020100c = ∠ CDE, maka BC ~ CE ( 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 94c450001 sama besar maka ∆PQR dan ∆BAR 4f46494c4500010 ) 800c80000ffe20c5 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0100000c4 sebangun. Oleh karena PQ : AB = 2 : 5 100000c484c696e CE AB AC BC 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 84c696e6f0 maka PR : RB = 2 : 5. 84943435f50524f4 6f021000006d6e7 Diperoleh perbandingan sisi: == 27. Jawaban: c Pada gambar di samping DECDCE 0000000000000000000000000000000000000000000000 21000006d 6494c4500010100 47252474220585 AG 6e7472524 000c484c696e6f02 95a2007ce000200 23. Jawaban: b Perhatikan bahwa ∆DCA ffd8ffe000104a4649 742205859 1000006d6e74725 09000600310000 dan ∆DAB sebangun sehingga berlaku: 460001020100c800c 5a2007ce0 247422058595a20 616373704d5346 80000ffe20c5849434 002000900 07ce00020009000 54000000004945 35f50524f46494c450 BD AD = 15 – 8 = 7 cm 060031000 600310000616373 43207352474200 6 cm EH = EF – HF 0010100000c484c69 A = 704d53465400000 ADCD 061637370 00000000000000 B 6e6f021000006d6e74 GC = 12 – 8 = 4 cm Segitiga AEH sebangun AD2= 000494543207352 BD × CD = 4 × (BC – 4d5346540 000000000000f6d 725247422058595a2 dengan segitiga ABG. EHAE 4x BD) =474200000000000 4 ×000000049 (13 – 4) = 4 × 9 6000100000000d 007ce000200090006 454320735 = 36 000000000000000 = 32d48502020000 = ⇒ = 00310000616373704 7+ B BGAB x6 247420000 00f6d6000100000 00000000000000 ⇔ AD= d5346540000000049 4(x + 6) = 7x 4 000000000 000d32d48502020 00000000000000 45432073524742000 4x + 24 = 7x 4 000000000 000000000000000 00000000000000 000000f6d 4x – 7x = –24 ⇔ –3x = –00000000000000000 000000000000000 00000000000000 00000000f6d6000100 24. Jawaban: 600010000 d Panjang sisi-sisi ∆ABC dan 2 24 000000000000000 00000000000000 000000d32d4850202 0000d32d4 ∆DEF berbanding 000000000000000 00000000000000 x= 8 Jadi, nilai x = 8. 00000000000000000 850202000 000000000000000 00000001163707 1 : 2 maka: AB1 81 00000000000000000 000000000 000000000000000 27400000150000 00000000000000000 = 000011637072740 ⇔ = 000000000 DE2 DE00033646573630 2 00000000000000000 000000000 000015000000033 00001840000006c 00000000000000000 ⇔ DE 646573630000018 = 16000000000 EF= 77747074000001f 00000000001163707 000000000 40000006c777470 000000014626b7 27400000150000000 000000000 74000001f000000 07400000204000 2 2 33646573630000018 = 20 014626b70740000 − 16000000000 = 12 cm 000147258595a0 40000006c77747074 000000000 020400000014725 00002180000001 Luas segitiga DEF: 000001f00000001462 8595a0000021800 46758595a00000 1 000000000 6b707400000204000 = × EF × DE L∆DEF 0000146758595a0 000000000 22c000000146258 2 000147258595a0000 001163707 000022c00000014 595a0000024000 1 2 02180000001467585 = × 12 × 16 = 96 cm 274000001 6258595a0000024 000014646d6e64 2 95a0000022c000000 500000003 000000014646d6e 00000254000000 25. Jawaban: b ∆PRS 146258595a0000024 364657363 640000025400000 70646d64640000 000000014646d6e64 siku-siku di R, 000001840 maka: 070646d64640000 02c400000088767 00000254000000706 RS= 02c400000088767 000006c77 565640000034c00 46d6464000002c400 747074000 565640000034c00 00008676696577 2 00008876756564000 001f00000 =000086766965770 13 − 000003d4000000 0034c000000867669 2 246c756d6900000 1200003d400000024 = 1690014626b7 − 144 6577000003d400000 074000002 6c756d69000003f 3f8000000146d65 0246c756d69000003f 040000001 8000000146d6561 61730000040c000 = 25 = 5 cm ∆OPQ sebangun 8000000146d656173 47258595a 730000040c00000 00024746563680 dengan ∆PRS, maka: OQ OP 612 0000040c000000247 000002180 024746563680000 00004300000000c OQ + 46563680000043000 =04300000000c725 ⇒ = 000001467 72545243000004 RSRP 5 12 58595a000 00000c72545243000 452430000043c00 3c0000080c67545 0043c0000080c6754 0022c0000 518 cm 00080c675452430 2430000043c0000 ⇔ OQ= = ×7,5 52430000043c00000 12 001462585 000043c0000080c 080c62545243000 80c62545243000004 95a000002 Jadi, panjang OQ = 7,5 cm. 625452430000043 0043c0000080c74 3c0000080c7465787 400000001 c0000080c746578 65787400000000 26. Jawaban: b PC : AC = 4 : (4 + 6) = 4 : 400000000436f70797 436f70797269676 10 = 2 : 57400000000436f7 Oleh4646d6e64 karena ∆PQC dan ∆ABC 26967687420286329 000002540 87420286329203 sebangun079726967687420 maka PQ : AB = 2 : 5. Perhatikan 20313939382048657 000007064 13939382048657 ∆PQR dan286329203139393 ∆BAR. ∠PRQ = ∠ARB (bertolak 76c6574742d506163 AC 8204865776c6574 = CE (diketahui) 6d6464000 belakang) 76c6574742d5061 ∠PQR = ∠BAR (sudut dalam memenuhi 6b61726420436f6d70 ∠BCA = ∠ DCE (bertolak belakang) 002c40000 742d5061636b617 636b6172642043 berseberangan) ∠RPQ = ∠ABR (sudutsyarat dalams 616e7900006465736 ∠ABC = ∠ EDC = 90° (diketahui) – sd – sd 008876756 26420436f6d7061 6f6d70616e79000 ∠ABC = ∠DEC, maka AC ~ACD ( ) berseberangan) 30000000000000012 C CD Jadi, 6e7900006465736 pasangan segitiga yang kongruen 564000003 06465736300000 73524742204945433 4c0000008 yaitu 300000000000000 ∆CDE dan ∆CBA. 00000000012735 6313936362d322e31 676696577 127352474220494 24742204945433 00000000000000000 000003d40 54336313936362d 6313936362d322 00000127352474220 00000246c 322e31000000000 e3100000000000 49454336313936362 756d69000 000000000000012 00000000000127 d322e310000000000 003f80000 735247422049454 35247422049454 00000000000000000 00146d656 336313936362d32 336313936362d3 00000000000000000 173000004 2e3100000000000 22e31000000000 0000000000A 8 cm D 0c0000002 000000000000000 00000000000000 sejajar CD, sehingga: 474656368 000000000000000 00000000000000 GC= HF = AD = 8 cm 000004300 0000000000000DF 00000000000000 x cm 2 22 2 000000c72 BG= BC – GC E F 000SP − DE − RP

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 ffd8ffe000104a46 41 + x =42 4 30. 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 49460001020100c x= 1 Jadi, sisa karton di bawah gambar adalah 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 800c80000ffe20c5 ffd8ffe000104a4649460001020100c800c80000ff 1 cm. Jawaban: dC 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 e20c584943435f50524f46494c4500010100000c48 84943435f50524f4 4c696e6f021000006d6e74725247422058595a200 0000000000000000000000000000000000000000000000 6494c4500010100 7ce00020009000600310000616373704d53465400 000c484c696e6f02 0000004945432073524742000000000000000000 1000006d6e74725 0000000000f6d6000100000000d32d48502020000 0000000000000000000000000000000000000000 247422058595a20 0000000000000000000000000000000000000000 B 07ce00020009000 A D 1,2 m 15 m 0000000000011637072740000015000000033646 600310000616373 57363000001840000006c77747074000001f00000 ▲ 704d53465400000 0014626b707400000204000000147258595a0000 Perhatikan bahwa ∆ABC dan ∆DBE 0218000000146758595a0000022c0000001462585 000494543207352 95a0000024000000014646d6e640000025400000 sebangun. Diperoleh: 474200000000000 070646d6464000002c40000008876756564000003 000000000000000 AB AC 4c0000008676696577000003d4000000246c756d6 = 00f6d6000100000 9000003f8000000146d6561730000040c00000024 DB DE 15 74656368000004300000000c725452430000043c0 000d32d48502020 AC ⇔ = 000080c675452430000043c0000080c6254524300 000000000000000 1,2 2 00043c0000080c7465787400000000436f7079726 000000000000000 9676874202863292031393938204865776c657474 2 × 15 ⇔ AC= 000000000000000 2d5061636b61726420436f6d70616e79000064657 1,2= 25 Jadi, tinggi menara 3630000000000000012735247422049454336313 000000000000000 sebenarnya 25 m. 936362d322e31000000000000000000000012735 000000000000000 247422049454336313936362d322e31000000000 000000000000000 0000000000000000000000000000000000000000 28. Jawaban: b ∆BED dan 000011637072740 00000 ∆EAF sebangun maka berlaku B. 000015000000033 perbandingan: 646573630000018 1. BDEB BD 6 UraianDua bangun yang sebangun mempunyai 40000006c777470 =⇔ = besar sudut sama dan panjang sisi sebanding. EFAE 12 83 74000001f000000 Diperoleh: x = 124 3y = 102 ⇔ y = 34 014626b70740000 ⇔ BD = × 12 = 9 4Jadi, panjang BD = 9 020400000014725 BC GH cm. 8595a0000021800 z = 8 15 = ⇔×8 z = = 12 29. Jawaban: a Segitiga QRS 0000146758595a0 AB EH 1510 10 Jadi, nilai x = 124, y = 34, siku-siku di R, maka: ? 000022c00000014 dan z = 12. BC 18 QS= 6258595a0000024 2. Trapesium ABCD dan EFGH sebangun. = 000000014646d6e 6 15 2 2 = 640000025400000 12 + 9 070646d64640000 = 02c400000088767 144 + 81 565640000034c00 = 225 000086766965770 = 15 cm ∆QRS sebangun ? BC = dengan ∆QRT, maka: QTQR QT12 00003d400000024 18 6 15 * = 7,2 cm 6c756d69000003f =⇒= QRQS 12 15 8000000146d6561 16 ×15 FE AD ⇔ FE= = 40 = cm 730000040c00000 12 ×12 48 6 Jadi, panjang BC = 7,2 cm dan FE EH AB ⇔ QT= = = 9,6 cm 024746563680000 15 5 = 40 cm. ? 04300000000c725 ST = QS – QT = 15 – 9,6 = 5,4 16 Lebar 3. Panjang gambar = 60FEcm. 452430000043c00 cm ∆PQT sebangun dengan = gambar = 36 cm. Misalkan sisa karton di 00080c675452430 15 6 ∆RST, maka: bawah gambar x cm maka panjang karton 70 PQQT 000043c0000080c =PQ9,6 ⇒ = RSST 9625452430000043 5,4 cm dan lebar karton (41 + x) cm. Gambar dan karton sebangun maka: c0000080c746578 996× ⇔ PQ = = =9616 cm 7400000000436f7 54 6Jadi, panjang PQ = 16 079726967687420 lebar karton panjang karton cm. = 286329203139393 lebar gambarpanjang gambar 8204865776c6574 41+ x 70 742d5061636b617 ⇒ = 36 60 26420436f6d7061 6e7900006465736 36 × 7 ⇔ 41 + x = 300000000000000 6 127352474220494 54336313936362d 322e31000000000 000000000000012 735247422049454 336313936362d32 2e3100000000000 000000000000000 000000000000000 0000000000000Q 2 2 R + RS

ffd8ffe0001 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00010 ffd8ffe000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c450 04a4649460 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4a464946000 104a46494 001020100c 0010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 1020100c800 800c80000ff 600010201 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 c80000ffe20c e20c584943 00310000616373704d53465400000000494543207352474200000000000000000000 00c800c80 435f50524f4 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 584943435f5 00000000f6d6000100000000d32d4850202000000000000000000000000000000000 6494c45000 000ffe20c5 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 0524f46494c 10100000c4 00000000000000000000000000000000000000000000000000000000000000116370 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 7. 45000101000 Perhatikan ∆RUS dan 84943435f 84c696e6f02 a. Segitiga OAB sebangun dengan 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 00c484c696e 1000006d6e ∆UST. Diperoleh: ∠SRU = 7274000001500000003364657363000001840000006c77747074000001f000000014 50524f464 x segitiga OCD. ABOB 3020 −+10 = ⇒ = 7472524742 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6f021000006 ∠UTW = 90° CDOD 15 x 94c450001 2058595a20 626b707400000204000000147258595a00000218000000146758595a0000022c0000 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 ∠d6e74725247 USR = ∠TUS (sudut – sudut – sisi) 07ce000200 4. ∆BCM dan ∆BDE sebangun. 2 30 −x 0100000c4 0000000000000000000000000000000000000000000000 422058595a2 00146258595a0000024000000014646d6e640000025400000070646d6464000002c4 ⇔= 0900060031 SR = UT Jadi, ∆RUS dan 1 1x BCCM 9 CM 84c696e6f0 007ce000200 = ⇔ = 0000616373 00000088767565640000034c0000008676696577000003d4000000246c756d690000 ∆UST kongruen. Perhatikan BDDE 18 9 704d534654 2x 2 = 30 – x 09000600310 21000006d ∆VSU dan ∆WUS. × 3x =0000000049 3 30 99 03f8000000146d6561730000040c0000002474656368000004300000000c72545243 00061637370 6e7472524 4543207352 ⇔xCM= = 4,5 cm Diperoleh: = 10 cm 4d534654000 18 ∆AEH dan ∆KEG4742000000 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465 ∠USV = ∠WUS 742205859 00000494543 0000000000 sebangun. Jadi, panjang OD = 10 cm. US = SU 787400000000436f70797269676874202863292031393938204865776c6574742d50 5a2007ce0(sudut – 0000000000 AHHESegitiga 9 27 20735247420 b. OMN sebangun sisi – sisi) = ⇔ = 00f6d600010 dengan segitiga OCD. VS = WU (karena002000900 ∆RUS ≅∆UTW) KGGE KG18 00000000000 61636b61726420436f6d70616e790000646573630000000000000012735247422049 Oleh karena sisi ON dan sisi OD yang 0000000d32 Jadi, disimpulkan bahwa ∆VSU dan ∆WUS ×918 00000000000 060031000 d485020200 bersesuaian panjangnya 454336313936362d322e310000000000000000000000127352474220494543363139 ⇔ KG= = 6 sama cm maka kongruen. 00000f6d600 0000000000 27 DAEH dankongruen DLEF segitiga OMN dengan segitiga OCD. Jadi, 061637370 0000000000 36362d322e31000000000000000000000000000000000000000000000000000000 Perhatikan ∆RVU dan ∆TWS. 0100000000d sebangun. panjang MN = CD = 15 4d5346540 0000000000 cm. Diperoleh: 32d48502020 AHHE 9 27 = ⇔ =dan AEF0000000000 000000049 ∆ADC 00000000000 RU = TS 10. a. ffd8ffe000104a46 sebangun (sisi – sudut – sisi) ∠RUV = ∠TSW 454320735 00000000000 00000000000 UV U = SW 247420000 00000000000 Jadi, ∆RVU dan ∆TWS kongruen. 000000000 00000000000 Disimpulkan bahwa pasangan segitiga000000000 00000000000 segitiga yang kongruen yaitu: 000000f6d 00000000000 ∆RUS ≅∆UST 00000000000 600010000 ∆VSU ≅∆WUS 00000011637 ∆RVU ≅∆TWS 0000d32d4 07274000001 850202000 8. 50000000336 a. Perhatikan ∆ADE dan ∆ECB. ∠ADE = 000000000 ∠BCE (siku-siku) ∠ AED = ∠ CBE (diketahui) 46573630000 000000000 ∠DAE = ∠ BEC (karena dua pasang sudutnya 01840000006 samac7774707400 besar maka sepasang sudut yang 000000000 lain pasti 0001f000000 sama besar). Oleh000000000 karena sudut-sudut bersesuaian pada ∆ADE dan ∆ECB sama besar 014626b7074 000000000 maka00000204000 ∆ADE dan ∆ECB sebangun. 000000000 00014725859 000000000 5a000002180 00000146758 000000000 b. Sisi AD bersesuaian dengan EC dan 595a0000022 sisi DE bersesuaian000000000 dengan BC, sehingga c0000001462 berlaku perbandingan: 000000000 58595a00000 AD DE AD 12 001163707 24000000014 = =AD ECBC 3⇔ E 274000001 646d6e640002 AD = = 36 00254000000 500000003 AD = 6364657363 cm 1 70646d64640 00002c40000 L∆ABE 2= × 000001840 AB × AD 00887675656 000006c77 40000034c00 1 = 2 × 747074000 15 × 6 00008676696 2= 45 cm Jadi, luas segitiga 577000003d42 001f00000 ABC 45 cm . 000000246c7 0014626b7 56d69000003 074000002 f8000000146 9. D15C 040000001 d6561730000 47258595a 040c0000002 47465636800 000002180 00043000000 000001467 00c72545243 58595a000 0000043c000 0022c0000 0080c675452 001462585 430000043c0 000080c6254 95a000002 52430000043 400000001 c0000080c74 4646d6e64 65787400000 000002540 000436f7079 000007064 72696768742 02863292031 6d6464000 39393820486 002c40000 5776c657474 008876756 2d5061636b6

LFFE LF 9 49460001020100c80000000000 ∆ 0000000000 maka: 99 00c80000ffe20c584 ADDC AD 5× ⇔0000000000 LF = = 3 cm ⇔ = 943435f50524f464= 27 AD +3 6 AEEF 0000000000 94c4500010100000 LM = CF – CM 6AD – LF0000000000 = 5AD + 15 c484c696e6f02100 = 0006d6e747252474 (9AD – 4,5 – 3) cm 0001163707 = 15 Jadi, lebar = 22058595a2007ce0 1,5 cm 15 m. 2740000015 sungai 0000000336 002000900060031 (KB +∆ABC LM) × dan BC ∆FEC 0000616373704d5 4657363000 LKBLM =2 sebangun, AD0018400000 garistinggi ∆FEC 346540000000049 ∆ABC dan2 DE garis (3 + 1,5) × 06c7774707 454320735247420 tinggi 4000001f000 = 9 ABAD AB15 maka: 000000000000000 ⇔ × 4,5 = 000014626b 2 ==4,5 EFDE2 63 000000000000f6d6 7074000002 =5 20,25 000100000000d32 ⇔cm AB2 = ×6= 30 Jadi, jarak 0400000014 d48502020000000 cm antara kedua pohon 30 m. 7258595a00 000000000000000 Jadi, luas segi empat KBML 20,25 cm2. 0002180000 000000000000000 0014675859 5. a. Perhatikan ∆ABC dan ∆ABD. AD = BC 000000000000000 5a0000022c 000000000000000 (trapesium sama kaki)0000001462 ∠DAB = ∠ABC 000000000000000 (trapesium sama kaki) AB = AB (berimpit) Oleh 000000000000011 58595a0000 0240000000 karena panjang dua sisi dan besar sudut yang 637072740000015 14646d6e64 diapit000000033646573 kedua sisi tersebut sama, maka ∆ABC dan 630000018400000 0000025400 ∆ABDffd8ffe000104a4649460001020100c kongruen. 000070646d 06c7774707400000 800c80000ffe20c584943435f50524f4 1f000000014626b7 6464000002 c400000088 074000002040000 b. Pasangan segitiga kongruen yang 6494c4500010100000c484c696e6f02 7675656400 A. Pilihan 00147258595a0000 lain adalah ∆AOD 00034c0000 ≅∆BOC dan ∆ACD 1000006d6e74725247422058595a20 021800000014675 Ganda 0086766965 07ce00020009000600310000616373 ≅ ∆BCD. Jawaban: 8595a0000022c000 6.1.Perhatikan ∆EGD dan ∆CBD. Diketahui DC = 77000003d4 704d53465400000000494543207352 000146258595a000 d DE. Oleh karena ∆ABG000000246c kongruen dengan ∆EFG 002400000001464 474200000000000000000000000000 756d690000 maka6d6e640000025400 GE = AG. Oleh karena AG = BC maka BC = 00f6d6000100000000d32d48502020 03f80000001 GE. ∠000070646d64640 BGD = ∠GBD maka ∆BGD sama kaki dan 46d6561730 000000000000000000000000000000 DB =00002c4000000887 DG. 000000000000000000000000000000 67565640000034c0000040c000 000008676696577 0002474656 000000000000000000000000000000 000003d40000002 3680000043 000011637072740000015000000033 46c756d69000003f 00000000c7 2545243000 64657363000001840000006c777470 8000000146d6561 D D bangun ruang yang dihasilkan Salah satu 74000001f000000014626b70740000 730000040c0000000043c00000 80c6754524 mempunyai 2 bidang datar berbentuk 247465636800000 0204000000147258595a0000021800 30000043c0 setengah 4300000000c72545 lingkaran, 1 bidang datar 0000146758595a0000022c00000014 000080c625 2430000043c00000 berbentuk persegi panjang, dan 1 bidang 6258595a0000024000000014646d6e 80c67545243000004524300000 43c0000080 lengkung. 43c0000080c62545 2. 640000025400000070646d64640000 2430000043c00000c746578740 02c400000088767565640000034c00 G80c74657874000000000000436f 2 00008676696577000003d400000024 7079726967 10 = 62c+s2t22 Jawaban: = 000436f707972696 karena 6874202863 yang bersesuaian sama 6c756d69000003f8000000146d65617 2 Oleh 2 2 2 768742028632920 2 2 sisi-sisi r +t t= 10 – 6 = 100 – 36 = 64 2920313939 panjang 30000040c000000247465636800000 maka ∆EGD dan ∆CBD kongruen 313939382048657 3820486577 76c6574742d50616 4300000000c725452430000043c000 (terbukti). 36b61726420436f6 6c6574742d 0080c675452430000043c0000080c6 d70616e7900006465061636b61 25452430000043c0000080c7465787 573630000000000 726420436f6 400000000436f70797269676874202 000012735247422 d70616e790 049454336313936 0006465736 863292031393938204865776c65747 362d322e310000003000000000 42d5061636b61726420436f6d70616e 000000000000000 0000012735 790000646573630000000000000012 012735247422049 2474220494 735247422049454336313936362d32 454336313936362 5433631393 d322e310000000006362d322e3 2e3100000000000000000000001273 000000000000000 1000000000 5247422049454336313936362d322e 000000000000000 0000000000 310000000000000000000000000000 000000000000000 0001273524 7422049454 00000000000000000000000000 EB

Bab II Bangun Ruang Sisi Lengkung

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe0001 ffd8ffe0001 ffd8ffe000104 ffd8ffe000 ffd8ffe000 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 04a4649460 04a4649460 a464946000102 104a464946 104a46494 ffd8ffe000104a 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe000104a46494600010 ffd8ffe000104a4 001020100c 001020100c 0100c800c8000 0001020100 600010201 46494600010201 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 64946000102010 800c80000ff 800c80000ff 0ffe20c5849434 c800c80000 00c800c80 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00c800c80000ffe 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 0c800c80000ffe2 e20c584943 e20c584943 35f50524f4649 ffe20c58494 000ffe20c5 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20100c800c80000ffe20c5849 20c584943435f5 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 0c584943435f505 435f50524f4 435f50524f4 4c45000101000 3435f50524f 84943435f 0524f46494c450 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 2 24f46494c450001 6494c45000 6494c45000 00c484c696e6f 46494c4500 50524f464 3. Jawaban: d Diketahui: L1 : L2 = πr : 0010100000c484 62d322e31000000000000000000000000000000000000000000000000000000 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 43435f50524f46494c4500010 0100000c484c696

10100000c4 10100000c4 021000006d6e7 010100000c 94c450001 garis pelukis = s = 20 cm tinggi c696e6f0210000 πR2 = r2484c696e6f :0100000c4 R2 = e6f021000006d6e 84c696e6f0 84c696e6f0 472524742205 =t 06d6e747252474 2 2 18 :8595a2007ce00 24021000006d = 324 74725247422058 21000006d6 21000006d6 84c696e6f 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000 100000c484c696e6f02100000 22058595a2007c : 576020009000600 = 6e74725247 9021000006 : 16 595a2007ce00020 595a2007ce0002 e747252474 e0002000900060 = 16 cm Sisi e747252474 belahan kerucut yang 00900060031000 00090006003100 ffd8ffe000104a464 22058595a2 22058595a2 310000616373 422058595a d6e747252 03100006163737 berbentuk segitiga merupakan segitiga 00146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d4000 6d6e74725247422058595a20 0616373704d534 00616373704d53 9460001020100c80 007ce00020 704d53465400 2007ce0002 474220585 04d53465400000 siku-siku sehingga007ce00020 diperoleh: 6. Jawaban: a Misalkan 65400000000494 46540000000049 0c80000ffe20c58494 0090006003 0090006003 000000494543 0009000600 95a2007ce 00049454320735 d = diameter tabung ffd8ffe000104a4649 54320735247420 45432073524742 3435f50524f46494c 1000061637 1000061637 207352474200 3100006163 000200090 r=07ce00020009000600310000 24742000000000 000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c000 460001020100c800c t = tinggi 00000000000000 4500010100000c48 3704d53465 3704d53465 000000000000 73704d5346 006003100 00000000000000 2 80000ffe20c5849434 tabung 1 = 00000f6d600010 20 0000000000000f6 00000000000000f 4c696e6f021000006 4000000004 4000000004 000000000000 5400000000 006163737 Jari-jari kerucut = × diamater 35f50524f46494c450 2 2 0080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6 616373704d53465400000000 d6000100000000 6d600010000000 d6e7472524742205 − 0000000d32d485 16 9454320735 9454320735 00f6d60001000 4945432073 04d534654 0010100000c484c69 1 kerucut = 400 − 256 = 2474200000 d32d4850202000 0d32d485020200 8595a2007ce000200 2474200000 00000d32d485 5247420000 000000004 ⇔ r= × diameter tabung 02020000000000 6e6f021000006d6e7 2 144 = 12 cm L = 00000000000000 0900060031000061 0000000000 0000000000 020200000000 0000000000 945432073 00000000000000 d70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936 494543207352474200000000 4725247422058595a 00000000000000 6373704d53465400 0000000000 000000000000 0000000000 524742000 = 1× 12 2 ×00000000000000 luas segitiga 0000000000 2007ce00020009000 2 = 6 cm Tinggi kerucut = 00000000000000 0000004945432073 000f6d6000 000f6d6000 000000000000 0000f6d600 000000000 00000000000000 60031000061637370 1 00000000000000000000f6d60 362d322e31000000000000000000000000000000000000000000000000000000Kunci Jawaban dan Pembahasan = 00000000000000 2 × × r × t 100000000d tinggi tabung 00000000000000 5247420000000000 100000000d 000000000000 0100000000 000000000 2 4d534654000000004 00000000000000 0000000000000000 32d4850202 32d4850202 000000000000 d32d485020 0000000f6 2 00000000000000 = 1294543207352474200 × 16 = 192 cm Jadi, luas sisi belahan ⇔ t = 8 cm Misalkan garis 00000000000000 00f6d600010000000 0000000000 0000000000 000000000000 2000000000 d60001000 2 00000000000000 00100000000d32d485020200 00000000000000000 kerucut yang berbentuk segitiga 192 cm . 00000000116370 00000000011637 0d32d48502020000 pelukis kerucut = s. 0000000000 0000000000 000000000000 0000000000 00000d32d 01163707274000 000000000f6d60001 72740000015000 07274000001500 0000000000000000 0000000000 0000000000 000000000000 0000000000 485020200 00150000000336 00000000d32d48502 s= 0000000000000000 00003364657363 00000336465736 4. 000000000000000000000000 Jawaban: a Misalkan α = besar sudut 0000000000 0000000000 000116370727 0000000000 000000000 46573630000018 02000000000000000 00000184000000 30000018400000 0000000000000000 pusat juring rebahan 0000000000 0000000000 400000150000 0000000000 000000000 40000006c77747 00000000000000000 6c777470740000 06c77747074000 0000000000000000 0000000000 0000000000 000336465736 0000000000 000000000 selimut kerucut 074000001f0000 000000000000000000000000 00000000000000000 01f000000014626 001f00000001462 0000000000000000 0000000000 0000000000 300000184000 0000000000 000000000 = 36 + 64 = 10 cm Jadi, panjang garis 00014626b70740 00000000000000000 α = r× 360°0000000000 b7074000002040 6b707400000204 0000000000011637 0000000000 0006c77747074 0000000000 000000000 00002040000001 s pelukis kerucut tersebut 10 cm. 00000000000000000 000000000000000000000000 00000147258595 00000014725859 0727400000150000 0000000000 0000000000 000001f000000 0000000000 000000000 8 47258595a00000 00000000000011637 ⇔ 120° = × 360° a0000021800000 5a000002180000 0003364657363000 0000011637 0000011637 014626b70740 0000001163 000000000 s18 21800000014675 07274000001500000 0146758595a000 00146758595a00 001840000006c777 0727400000 0727400000 000020400000 7072740000 000000000 8595a0000022c0 7. Jawaban: dr=7 ⇔000000000000000000000116 = 00336465736300000 0022c000000146 00022c00000014 47074000001f00000 1500000003 1500000003 0147258595a00 0150000000 000000000 2 3s 00000146258595 cm t = 24 cm s = t2 + r2 = 1840000006c777470 258595a0000024 6258595a000002 0014626b70740000 3646573630 3646573630 000218000000 3364657363 000000000 a0000024000000 2 ⇔ s 74000001f00000001 = 24 cm t= 242 + 70204000000147258 = 0000018400 625 370727400000150000000336 000000014646d6 4000000014646d 0000184000 0000184000 146758595a000 000116370 014646d6e64000 4626b707400000204 e6400000254000 6e640000025400 595a000002180000 0006c77747 0006c77747 0022c00000014 00006c7774 727400000 00254000000706 000000147258595a0 2 00070646d64640 000070646d6464 ⇔ s 00146758595a0000 = 7074000001 074000001f 074000001f 6258595a00000 150000000 46d6464000002c = 24 4657363000001840000006c7 00002180000001467 00002c40000008 000002c4000000 022c000000146258 0000000146 0000000146 240000000146 f000000014 336465736 2 Besar sudut pusat lingkaran selimut − 40000008876756 8 58595a0000022c000 87675656400000 88767565640000 595a000002400000 26b7074000 26b7074000 46d6e64000002 626b707400 300000184 5640000034c000 = 576 − 64 = kerucut: r 000146258595a0000 α = 0014646d6e640000 × 360° 34c00000086766 034c0000008676 7747074000001f00000001462 0020400000 0020400000 540000007064 0002040000 0000006c7 00086766965770 024000000014646d6 s 96577000003d40 696577000003d4 025400000070646d 0147258595 0147258595 6d6464000002c 0014725859 774707400 00003d40000002 e6400000254000000 7 = 6464000002c40000 × 360° = 100,8° 00000246c756d6 000000246c756d a000002180 a000002180 400000088767 5a00000218 0001f0000 46c756d6900000 6b7074000002040000001472 70646d6464000002c 25 9000003f8000000 69000003f800000 0088767565640000 0000014675 0000014675 565640000034c 0000001467 00014626b 3f8000000146d6 40000008876756564 8. 5. 146d6561730000 0146d656173000 034c000000867669 8595a00000 8595a00000 000000867669 58595a0000 707400000 561730000040c0 0000034c000000867 Jari-jari alas kerucut pertama 040c0000002474 0040c000000247 6577000003d40000 58595a000002180000001467 22c0000001 22c0000001 6577000003d4 022c000000 204000000 00000247465636 r= 6696577000003d400 = r.80000043000000 65636800000430 46563680000043 00246c756d690000 Jawaban: c 46258595a0 46258595a0 000000246c756 146258595a 147258595 r= a 0000246c756d69000 Jawaban: 0000000c725452 00000000c72545 = 03f8000000146d656 27 (9 5) 0000240000 0000240000 d69000003f800 0000024000 a00000218 00c72545243000 003f8000000146d65 58595a0000022c0000001462 2 430000043c0000 2430000043c000 1730000040c00000 00014646d6 00014646d6 0000146d6561 000014646d 000000146 = 0043c0000080c6 30 61730000040c00000 2 = 0247465636800000 729 − 080c6754524300 0080c675452430 e640000025 e640000025 730000040c000 6e64000002 758595a00 − 75452430000043 24 02474656368000004 00043c0000080c6 000043c0000080c =58595a000002400000001464 900 − 576 = 324 = 18 4300000000c72545 4000000706 4000000706 000247465636 5400000070 00022c000 c0000080c62545 405 = 324 300000000c7254524 25452430000043 62545243000004 2430000043c00000 46d6464000 46d6464000 800000430000 646d646400 000146258 cm 2430000043c000 Jari-jari alas kerucut 30000043c0000080c = 80c6754524300000 18 cm r c0000080c746578 3c0000080c74657 002c400000 002c400000 0000c72545243 0002c40000 595a00000 0080c746578740 kedua = R. 675452430000043c0 6d6e64000002540000007064 R= 0887675656 7400000000436f7 87400000000436f 43c0000080c625452 0887675656 0000043c00000 0088767565 240000000 α = × 360° 0000000436f707 000080c6254524300 s 07972696768742 70797269676874 430000043c000008 40000034c0 40000034c0 80c6754524300 640000034c 14646d6e6 97269676874202 00043c0000080c746 1 02863292031393 20286329203139 0c74657874000000 0000086766 0000086766 00043c0000080 0000008676 400000254 6d6464000002c40000008876 86329203139393 = × 360° = 240° 2 5787400000000436f 27 =8204865776c657 40 8 938204865776c6 3938204865776c 00436f70797269676 9657700000 9657700000 c625452430000 6965770000 000000706 70797269676874202 2 574742d5061636 6574742d506163 8742028632920313 3d40000002 3d40000002 043c0000080c7 03d4000000 46d646400 −4742d5061636b6 32 86329203139393820 7565640000034c0000008676 b61726420436f6d 6b61726420436f6 =1726420436f6d7 1.600 − 93938204865776c6 46c756d690 46c756d690 465787400000 246c756d69 0002c4000 4865776c6574742d5 –x 70616e79000064 d70616e7900006 574742d5061636b6 00003f8000 2000003f8000 000436f707972 000003f800 000887675 0616e790000646 061636b6172642043 1.024 = 576 = 65736300000000 46573630000000 1726420436f6d7061 000146d656 000146d656 696768742028 0000146d65 656400000 M696577000003d4000000246c A57363000000000 6f6d70616e7900006 ⇔ t= 00000012735247 00000001273524 6e79000064657363 1730000040 1730000040 632920313939 6173000004 34c000000 2400000127352474 cm 46573630000000000 42204945433631 74220494543363 0000000000000012 c000000247 c000000247 38204865776c6 0c00000024 867669657 22049454336313 00001273524742204 756d69000003f8000000146d6 30 3936362d322e31 13936362d322e3 7352474220494543 4656368000 4656368000 574742d50616 7465636800 7000003d4 936362d322e310 9454336313936362d 00000000000000 10000000000000 36313936362d322e 0043000000 0043000000 36b617264204 0004300000 000000246 00000000000000 322e3100000000000 00000000127352 00000000012735 3100000000000000 00c7254524 00c7254524 36f6d70616e79 000c725452 c756d6900 00000001273524 561730000040c00000024746 00000000000127352 47422049454336 24742204945433 0000000012735247 30000043c0 30000043c0 000064657363 430000043c 0003f8000 74220494543363 47422049454336313 313936362d322e 6313936362d322 4220494543363139 000080c675 000080c675 000000000000 0000080c67 000146d65 13936362d322e3 936362d322e310000 56368000004300000000c725 31000000000000 e3100000000000 36362d322e310000 4524300000 4524300000 001273524742 5452430000 617300000 10000000000000 00000000000000000 00000000000000 0000000000000000 43c0000080 43c0000080 204945433631 043c000008 40c000000 00000000000000 00000000000000000 00000000000000 0000000000000000 c625452430 c625452430 3936362d322e3 0c62545243 247465636 00000000000000 452430000043c0000080c675 0000000000000000 00000000000000 S 0000000000000000 000043c000 000043c000 100000000000 0000043c00 800000430 0000000000000s 2 2 0 22 2 222 2 0080c74657 0080c74657 000000000001 00080c7465 0000000c7 00 s − t2 2 − 2 s− s−−r− Tt t

ffd8ffe0001 ffd8ffe000 ffd8ffe0001 ffd8ffe000 ffd8ffe000104a4 04a4649460 104a46494 04a4649460 104a46494 64946000102010 001020100c 600010201 001020100c 600010201 ffd8ffe000104a4 0c800c80000ffe2 800c80000ff 00c800c80 800c80000ff 00c800c80 64946000102010 0c584943435f505 e20c584943 000ffe20c5 e20c584943 000ffe20c5 0c800c80000ffe2 24f46494c450001 435f50524f4 84943435f 435f50524f4 84943435f 0c584943435f505 0100000c484c69 6494c45000 50524f464 6494c45000 50524f464 9. Jawaban: 24f46494c450001 6e6f021000006d 10100000c4 94c450001 10100000c4 94c450001 0100000c484c696 b Bidang alas tabung berupa 6e747252474220 84c696e6f0 0100000c4 84c696e6f0 0100000c4 e6f021000006d6e llingkaran. K= 2πr 58595a2007ce00 21000006d6 84c696e6f 21000006d6 84c696e6f0 74725247422058 100,48 = 2 × 3,14e747252474 × r 21000006d 100,48 1 02000900060031 e747252474 021000006 r= =0000616373704d 16595a2007ce0002 cm 22058595a2 d6e747252 22058595a2 6e7472524 00090006003100 53465400000000 007ce00020 474220585 007ce00020 742205859 6,28t= 00616373704d53 49454320735247 0090006003 95a2007ce 0090006003 5a2007ce0 46540000000049 42000000000000 1000061637 000200090 1000061637 002000900 45432073524742 3704d53465 3704d53465 060031000 2000000000000000 16 00000000000000 = 006003100 00f6d600010000 4000000004 006163737 4000000004 061637370 00000000000000f 144 = 12 cm Jadi, tinggi 0000d32d485020 9454320735 04d534654 9454320735 4d5346540 6d600010000000 kerucut tersebut 12 cm. 20000000000000 2474200000 000000004 2474200000 000000049 0d32d485020200 00000000000000 0000000000 945432073 0000000000 454320735 00000000000000 00000000000000 0000000000 524742000 0000000000 247420000 10. 00000000000000 00000000000000 000f6d6000 000000000 000f6d6000 000000000 00000000000000 00000000000000 100000000d 000000000 100000000d 000000000 Jawaban: d00000000000000 s 00000000000000 32d4850202 0000000f6 32d4850202 000000f6d = 10 cm rs2=00000000000000 =5d60001000 r2 + t22 0000000000 00000000000011 0000000000 600010000 00000000000000 cm 63707274000001 102 = 5200000000011637 +00000d32d t22 0000000000 0000000000 0000d32d4 0000000000 0000000000 850202000 t2 = 50000000336465 10207274000001500 – 485020200 52 = 100 – 25 = 75 73630000018400 0000000000 000000000 0000000000 000000000 00000336465736 ⇔ t= 00006c77747074 0000000000 000000000 0000000000 000000000 30000018400000 000001f0000000 0000000000 000000000 0000000000 000000000 06c77747074000 14626b70740000 0000000000 000000000 0000000000 000000000 001f00000001462 02040000001472 0000000000 000000000 0000000000 000000000 6b707400000204 B. 58595a00000218 0000000000 000000000 0000000000 000000000 00000014725859 00000014675859 0000011637 000000000 0000011637 000000000 1. a. Misalkan sudut pusat juring selimut 5a000002180000 5a0000022c0000 0727400000 000000000 0727400000 000000000 kerucut = α 00146758595a00 .r Uraian 00146258595a00 1500000003 000000000 1500000003 000000000 α = s× 360° 00022c00000014 00024000000014 3646573630 000000000 3646573630 000000000 6258595a000002 r 646d6e64000002 0000184000 000116370 0000184000 001163707 ⇔ 4000000014646d 270° = × 360° 24270° r 5400000070646d 0006c77747 727400000 0006c77747 274000001 6e640000025400 6464000002c400 074000001f 150000000 500000003 ⇔ =000070646d6464 360 ° 074000001f 00008876756564 0000000146 336465736 0000000146 364657363 000002c4000000 24 3r 0000034c000000 26b7074000 300000184 26b7074000 000001840 ⇔ =88767565640000 4 86766965770000 0020400000 0000006c7 0020400000 000006c77 r 034c0000008676 r= 03d4000000246c 0147258595 774707400 0147258595 747074000 696577000003d4 24 24 2 ×43jari-jari kerucut r= 18 cm Jadi, 756d69000003f8 a000002180 0001f0000 a000002180 001f00000 000000246c756d 18 cm. 000000146d6561 0000014675 00014626b 0000014675 0014626b7 69000003f800000 730000040c0000 8595a00000 707400000 8595a00000 074000002 20146d656173000 Lπr b. =00247465636800 alas0040c000000247 =204000000 3,14 × 22c0000001 182 = 3,14 × 324 22c0000001 040000001 2 0004300000000c 46258595a0 147258595 46258595a0 47258595a = 1.017,36 cm Jadi, luas alas 46563680000043 2 72545243000004 0000240000 a00000218 0000240000 000002180 kerucut 1.017,36 cm . 00000000c72545 3c0000080c6754 00014646d6 000000146 00014646d6 000001467 2430000043c000 52430000043c00 e640000025 758595a00 e640000025 58595a000 0080c675452430 00080c62545243 4000000706 00022c000 4000000706 0022c0000 000043c0000080c c. t=0000043c000008 46d6464000 000146258 46d6464000 001462585 62545243000004 0c746578740000 002c400000 595a00000 002c400000 95a000002 3c0000080c74657 = 0887675656 242 −0887675656 182 400000001 0000436f707972 240000000 87400000000436f 69676874202863 40000034c0 14646d6e6 40000034c0 4646d6e64 = 576 −70797269676874 324 = 0000086766 29203139393820 0000086766 400000254 000002540 20286329203139 4865776c657474 9657700000 000000706 9657700000 000007064 3938204865776c 2d5061636b6172 3d40000002 46d646400 3d40000002 6d6464000 6574742d506163 6420436f6d7061 46c756d690 0002c4000 46c756d690 002c40000 6b61726420436f6 6e790000646573 00003f8000 000887675 00003f8000 008876756 d70616e7900006 63000000000000 000146d656 656400000 000146d656 564000003 46573630000000 00127352474220 1730000040 34c000000 1730000040 4c0000008 00000001273524 49454336313936 c000000247 867669657 c000000247 676696577 74220494543363 362d322e310000 4656368000 7000003d4 4656368000 000003d40 13936362d322e3 00000000000000 0043000000 000000246 0043000000 00000246c 10000000000000 00001273524742 00c7254524 c756d6900 00c7254524 756d69000 00000000012735 20494543363139 30000043c0 0003f8000 30000043c0 003f80000 24742204945433 36362d322e3100 000080c675 000146d65 000080c675 00146d656 6313936362d322 00000000000000 4524300000 617300000 4524300000 173000004 e3100000000000 00000000000000 43c0000080 40c000000 43c0000080 0c0000002 00000000000000 00000000000000 c625452430 247465636 c625452430 474656368 00000000000000 0000000000 s2000043c000 − 000004300 000043c000 800000430 00000000000000 0080c74657 0000000c7 0080c74657 000000c72 r2 =0 2 2 2 2 s− − r

ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe0001 104a46494 104a464946 104a46494 104a46494 04a4649460 600010201 0001020100 600010201 600010201 001020100c 00c800c80 c800c80000 00c800c80 00c800c80 800c80000ff 000ffe20c5 ffe20c58494 000ffe20c5 000ffe20c5 e20c584943 84943435f 3435f50524f 84943435f 84943435f 435f50524f4 50524f464 46494c4500 50524f464 50524f464 2. a. Luas seluruh bagian segitiga = 6494c45000 L. 94c450001 010100000c 94c450001 94c450001 10100000c4 L =2 × luas segitiga siku0100000c4 484c696e6f 0100000c4 0100000c4 84c696e6f0 siku 1 ⇔ 84c696e6f 150= 2 × × a × 84c696e6f t 021000006d 84c696e6f 21000006d6 2 021000006 6e74725247 021000006 021000006 e747252474 150= 2422058595a ×× r 1× 15 d6e747252 d6e747252 d6e747252 22058595a2 2 474220585 2007ce0002 474220585 474220585 007ce00020 150 = r × 15 95a2007ce 0009000600 95a2007ce 95a2007ce 0090006003 r= 10 cm Jadi, jari-jari alas 000200090 3100006163 000200090 000200090 1000061637 tabung 10 006003100 cm. 006003100 73704d5346 006003100 3704d53465 Luas006163737 alas tabung = π × 006163737 r2 = 4000000004 5400000000 006163737 3,1404d534654 ×4945432073 10204d534654 = 3,14 × 10004d534654 = 9454320735 2 000000004 5247420000 000000004 ffd8ffe000104a46494 000000004 2474200000 314 cm 945432073 0000000000 94543207360001020100c800c80 945432073 0000000000 524742000 0000000000 524742000000ffe20c584943435f 524742000 0000000000 Jadi, 000000000 luas alas tabung 50524f46494c450001 314 cm2000f6d6000 0000f6d600 000000000 000000000 000000000 0100000000 0000000000100000c484c696e6f 000000000 100000000d 3. OT = 7 cm 0000000f6 d32d485020 0000000f6 021000006d6e747252 0000000f6 R32d4850202 rbolaJari-jari = OQd60001000 = ORd60001000 = 11 cm 2000000000 d60001000 0000000000 kerucut 47422058595a2007ce 0000000000 00000d32d000200090006003100 00000d32d 0000000000 = TQ. 00000d32d 2 0000000000 2 485020200 485020200 0000000000 TQ2 = OQ – OT485020200 00616373704d534654 2 0000000000 000000000000000004945432073 000000000 0000000000 = 112 – 7000000000 = 121 000000000 0000000000 000000000 000000000 0000000000 – 49 = 72 524742000000000000 000000000 0000000000 0000000000000000000000000f6 000000000 0000000000 ⇔ TQ= 000000000 0000000000 000000000d6000100000000d32d 000000000 0000000000 000000000 0000000000 000000000485020200000000000 000000000 0000000000 000000000 0000000000 000000000 000000000 0000000000 ffd8ffe000104a46494600010 000000000000000000 000000000 0000001163 000000000 000000000 0000011637 20100c800c80000ffe20c5849 000000000000000000 000000000 7072740000 000000000 000000000 0727400000 43435f50524f46494c4500010 000000000000000000 = 7 +000000000 11 = 000000000 18 cm QR2= 000000000 TR2 +1500000003 TQ2 0150000000 100000c484c696e6f02100000 000000000000000000 = 182000000000 +3364657363 ( 000000000 000000000 3646573630 6d6e74725247422058595a20 000000000000116370 000116370 0000018400 000116370 000116370 0000184000 07ce00020009000600310000 727400000150000000 727400000 00006c7774 727400000 727400000 0006c77747 616373704d53465400000000 336465736300000184 150000000 7074000001 150000000 150000000 074000001f 494543207352474200000000 0000006c7774707400 336465736 f000000014 336465736 336465736 0000000146 00000000000000000000f6d6 0001f000000014626b 300000184 626b707400 300000184 300000184 26b7074000 11 000100000000d32d48502020 707400000204000000 4. Alas kerucut = d = × 60 = 30 cm 0000006c7 0002040000 0000006c7 0000006c7 0020400000 22Tinggi kerucut = 40 cm 147258595a00000218 Perhatikan 000000000000000000000000 774707400 0014725859 774707400 774707400 0147258595 000000000000000000000000 000000146758595a00 gambar di 0001f0000 5a00000218 0001f0000 0001f0000 a000002180 000000000000000000000000 00022c000000146258 00014626b 0000001467 00014626b 00014626b 0000014675 000000000000000000000011 595a00000240000000 samping.707400000 58595a0000 707400000 707400000 8595a00000 E ∆ABC sebangun 637072740000015000000033 14646d6e6400000254 204000000 022c000000 204000000 204000000 22c0000001 64657363000001840000006c 00000070646d646400 dengan ∆ADE 147258595 146258595a 1472585950002c4000000887675 147258595 46258595a0 sehingga berlaku: AB77747074000001f000000014 a00000218 0000024000 a00000218 a00000218 0000240000 626b70740000020400000014 65640000034c000000 = BC AD DE 000000146 000014646d 000000146 000000146 00014646d6 7258595a0000021800000014 8676696577000003d4 40 30 758595a00 6e64000002 758595a00 758595a00 e640000025 ⇔= 1 6758595a0000022c00000014 000000246c756d6900 10 r 00022c000 5400000070 00022c000 00022c000 4000000706 40 × r =000146258 30 × 000146258 10 3 30 6258595a0000024000000014 ×10 300003f8000000146d65 646d646400 000146258 46d6464000 646d6e640000025400000070 61730000040c000000 r= = =595a00000 7,5 cm 595a00000 0002c40000 595a00000 002c400000 40 4 646d6464000002c400000088 247465636800000430 240000000 0088767565 240000000 240000000 0887675656 767565640000034c00000086 0000000c7254524300 14646d6e6 640000034c 14646d6e6 14646d6e6 40000034c0 76696577000003d400000024 00043c0000080c6754 400000254 0000008676 400000254 400000254 0000086766 6c756d69000003f800000014 52430000043c000008 000000706 6965770000 000000706 000000706 9657700000 6d6561730000040c00000024 0c625452430000043c 46d646400 03d4000000 46d646400 46d646400 3d40000002 74656368000004300000000c 0000080c7465787400 0002c4000 246c756d69 0002c4000 0002c4000 46c756d690 725452430000043c0000080c 000000436f70797269 000887675 000003f800 000887675 000887675 00003f8000 675452430000043c0000080c 676874202863292031 656400000 0000146d65 656400000 656400000 000146d656 625452430000043c0000080c 393938204865776c65 34c000000 6173000004 34c000000 34c000000 1730000040 7465787400000000436f7079 74742d5061636b6172 867669657 0c00000024 867669657 867669657 c000000247 726967687420286329203139 6420436f6d70616e79 7000003d4 7465636800 7000003d4 7000003d4 4656368000 3938204865776c6574742d50 000064657363000000 000000246 0004300000 000000246 000000246 0043000000 61636b61726420436f6d7061 000000001273524742 c756d6900 000c725452 c756d6900 c756d6900 00c7254524 6e7900006465736300000000 204945433631393636 0003f8000 430000043c 0003f8000 0003f8000 30000043c0 000000127352474220494543 2d322e310000000000 000146d65 0000080c67 000146d65 000146d65 000080c675 36313936362d322e31000000 000000000000127352 617300000 5452430000 617300000 617300000 4524300000 000000000000000012735247 474220494543363139 40c000000 043c000008 40c000000 40c000000 43c0000080 422049454336313936362d32 36362d322e31000000 247465636 0c62545243 247465636 247465636 c625452430 2e3100000000000000000000 000000000000000000 800000430 0000043c00 800000430 800000430 000043c000 000000000000000000000000 000000000000000000 0000000c7 00080c7465 0000000c7 0000000c7 0080c74657 0000000000 000000000000 A

ffd8ffe0001 ffd8ffe0001 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00010 ffd8ffe000104a ffd8ffe0001 ffd8ffe0001 ffd8ffe00010 ffd8ffe000104a ffd8ffe000104 ffd8ffe000 ffd8ffe000 ffd8ffe 04a4649460 04a4649460 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4a4649460001 4649460001020 04a4649460 04a4649460 4a4649460001 4649460001020 a46494600010 104a464946 104a464946 000104a 001020100c 001020100c 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 020100c800c8 100c800c80000f 001020100c 001020100c 020100c800c8 100c800c80000f 20100c800c80 0001020100 0001020100 800c80000ff 800c80000ff ffd8ffe000104a4 4649460 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 0000ffe20c584 fe20c584943435 800c80000ff 800c80000ff 0000ffe20c584 fe20c584943435 000ffe20c5849 c800c80000 c800c80000 e20c584943 e20c584943 64946000102010 0010201 435f50524f4 435f50524f4 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 943435f50524 f50524f46494c4 e20c584943 e20c584943 943435f50524 f50524f46494c4 43435f50524f4 ffe20c58494 ffe20c58494 0c800c80000ffe2 6494c45000 6494c45000 00c800c 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 f46494c45000 500010100000c 435f50524f4 435f50524f4 f46494c45000 500010100000c 6494c4500010 3435f50524f 3435f50524f 0c584943435f505 10100000c4 10100000c4 2 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 10100000c484 484c696e6f0210 6494c45000 6494c45000 10100000c484 484c696e6f0210 100000c484c6 46494c4500 46494c4500 5. Perhatikan bahwa ∆ABD sebangun V1d ffd8ffe000104a 7. Diketahui: dalas kerucut = 30 = cm 80000ff 84c696e6f0 24f46494c450001 11 84c696e6f0 alas4649460001020 Banyak tabung kecil = = 2 21000006d6 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 c696e6f02100 00006d6e74725 10100000c4 10100000c4 c696e6f02100 00006d6e74725 96e6f0210000 010100000c 010100000c 21000006d6 V ffd8ffe000104a dengan ∆ECD. Misalkan diameter bola = d, 100c800c80000f 2 0100000c484c696 22 2 e20c584 rtabung = 15 cm 4649460001020 fe20c584943435 e747252474 e747252474 alas247422058595a 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0006d6e74725 247422058595a 84c696e6f0 84c696e6f0 0006d6e74725 06d6e7472524 484c696e6f r1 ⋅ t1 484c696e6f tinggi kerucut = t. 100c800c80000f f50524f46494c4 e6f021000006d6e 943435f kerucut 22058595a2 22058595a2 fe20c584943435 = πrs 500010100000c = 2 021000006d L D2007ce0002000 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 247422058595 21000006d6 21000006d6 247422058595 2007ce0002000 7422058595a2 021000006d selimu rf50524f46494c4 74725247422058 484c696e6f0210 007ce00020 2 ⋅ t2007ce00020 ffd8ffe000104a46494600010 50524f4 500010100000c 00006d6e74725 ffd8ffe000104a4 tt 0000000000000000000000000000000000000000000000 a2007ce00020 9000600310000 e747252474 e747252474 a2007ce00020 9000600310000 007ce0002000 6e74725247 6e74725247 1.177,5 1 = 3,14 × 15 × s 0090006003 0090006003 595a2007ce0002 484c696e6f0210 A. Pilihan 20100c800c80000ffe20c58494 247422058595a = 649460001020100 = 00006d6e74725 6494c45 009000600310 616373704d534 22058595a2 22058595a2 009000600310 616373704d534 900060031000 422058595a 422058595a 1000061637 1000061637 2007ce0002000 c800c80000ffe20c 1.177,5 = 47,1 × s ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 00090006003100 Ganda 3435f50524f46494c45000101 247422058595a 8 9000600310000 584943435f50524 3704d53465 3704d53465 000616373704 6540000000049 007ce00020 007ce00020 000616373704 6540000000049 0616373704d5 2007ce0002 2007ce0002 2007ce0002000 00000c484c696e6f021000006 616373704d534 ⇔ s00616373704d53 = 25 cm t= f46494c45000101 0001010 1. Jawaban: a1 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 9000600310000 4000000004 4000000004 6540000000049 00000c484c696e6 d6e74725247422058595a200 Ad53465400000 4543207352474 0090006003 0090006003 d53465400000 4543207352474 346540000000 0009000600 0009000600 616373704d534 4543207352474 f021000006d6e74 46540000000049 0000c48 9454320735 9454320735 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 2 6540000000049 7ce0002000900060031000061 2000000000000 725247422058595 5. 000494543207 Volume kerucut 2000000000000 = × π1000061637 × rk 1000061637 × 000494543207 2000000000000 049454320735 3100006163 3100006163 4543207352474 2474200000 2474200000 45432073524742 0000000000000 a2007ce00020009 6373704d5346540000000049 31 22 3 4c696e6 2000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000 000f6d60001000 000600310000616 352474200000 0000000000000 3704d53465 3704d53465 352474200000 0000000000000 247420000000 73704d5346 73704d5346 0000000000 0000000000 1 tk 454320735247420000000000 00000000000000 0000000000000 00000d32d4850 373704d53465400 rb==25 d2k− ⇒15 rb2 = f021000 0000000000 0000000000 000f6d6000100 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 2020000000000 000f6d6000100 4000000004 4000000004 000000000000 000f6d60001000 000000000000 5400000000 5400000000 Vk=000000000000 × × 142 × 18 000000494543207 000000000000000000f6d6000 Jawaban: d 2 41 00000000000000f 000000d32d485 37 0000000000000 352474200000000 000f6d6000 000f6d6000 100000000d32d48502020000 0202000000000 006d6e7 r 00000000000f 000000d32d485 9454320735 9454320735 00000000000f 00000d32d4850 000000000f6d 4945432073 4945432073 0000000000000 000000000000000 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 3 100000000d = 6d600010000000 625 − Vk= = 6000000000000000000000000 × 22 × 2 × 14 0000000000000 2 Vkbola : Vkerucut 0000000000000 =3 πr : 3πr:kr2 00000f6d6000100 ·5247420000 t 100000000d 32020000000000 = b4r600010000000 6d6000100000 0202000000000 2474200000 2474200000 6d6000100000 5247420000 0000000000000 0000000000000 4725247 32d4850202 32d4850202 k000000d32d48502 0d32d485020200b 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 000000000000000000000000 3 0000000000000 d3.696 d : t= 6cm :2 0000000000000 020000000000000 Vk =000d32d48502 Volume tabung = V 225 = 400 3 3 0000000000000 0000000000 0000000000 000d32d48502 0000000000000 0d32d4850202 0000000000 0000000000 0000000000 0000000000 0000000000000 000000000000000000000000 0000000000000 00000000000000 000000000000000 · rk = 4rk : rk = 4 : 1 4220585 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 0000000000000 0000000000 0000000000 0000000116370 000000000000000 2r : 020000000000 t= 3 : 1 2r 2 0000000000000 0000000000 0000000000 020000000000 0000000000000 000000000000 0000000000 0000000000 000000000000000000011637 0000000000000 t 7274000001500 000000000000000 = 00000000000000 20 cm 95a2007 0000000000 0000000000 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 0000000011637 072740000015000000033646 0000033646573 = 3 000000000000 000000000000000 0000000000000 000f6d6000 000f6d6000 000000000000 0000000000000 000000000000 0000f6d600 0000f6d600 2 00000000000000 0727400000150 V= π×r ×t 0000000000 0000000000 6300000184000 000000000000000 57363000001840000006c7774 perbandingan volume bola dan kerucut 4 : 1. t tt 22 0000000000000 ce00020 0000003364657 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 000000000000 100000000d 100000000d 8. Jadi,00000000000000 000000000000 Bola menyinggung 0000000000000 000000000000 semua 0100000000 sisi 0100000000 kubus 0006c77747074 000000011637072 t t 0000000000 0000000000 7074000001f000000014626b7 3630000018400 000001f0000000 740000015000000 2 00006c7774707 0000000000 0000000000 0090006 000000000000 0000000000000 32d4850202 000000000000 0000000000000 d32d485020 d32d485020 dari dalam, berarti bola sama ⇔ 3.696= × 21 × tt 32d4850202 6. Jawaban: bdiameter d=000000000000 28 033646573630000 14626b7074000 2r= 3t 3 3 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 074000002040000001472585 00000000000000 4000001f00000 0020400000014 70000000000000 7 01840000006c777 0000000000 0000000000 000000000000 0000000000 0000000000 000000000000 0000000000000 000000000000 2000000000 2000000000 95a000002180000001467585 dengan panjang rusuk kubus, yaitu 42 cm. Jari0014626b70740 cm ⇔ r = 14 cm 0031000 r= t 7258595a00000 47074000001f000 00000000000000 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 0000000000 0000000000 0000204000000 3.696 = 22 × 0000000011637 3 × 21 × ttt 0000000000 2180000001467 000014626b70740 95a0000022c00000014625859 000000000000 0000000000 000000000000 0000000116370 000000000000 0000000000 0000000000 jari bola tersebut 42 147258595a000 t = 50 00000000011637 58595a0000022 000020400000014 23 0000000000 0000000000 0616373 00000000000000000000000000000000000000000000000 5a0000024000000014646d6e = 21 cm. 7274000001500 0021800000014 c000000146258 7258595a0000021 3.696 0727400000150 0000000000 0000000000 000000000000 000000000000 0000000000 0000000000 tt = 000000000000 24 cm 422 0000011637 0000011637 2 V = 6758595a00000 07274000001500 595a000002400 640000025400000070646d64 800000014675859 704d534 botol = πr t ⇔ AB= BD =1.386 CD 22c0000001462 0000014646d6e 0727400000 0727400000 000000000000 0000003364657 0000000000 0000000000 000000000000 0000033646573 000000000001 0000000000 0000000000 5a0000022c00000 3 2 64000002c4000000887675656 00000336465736 220 ml 58595a0000024 2 Dari kesebangunan segitiga 6400000254000 = × × 21 = ×0146258595a0000 211500000003 Vair Vπr 924 3630000018400 1500000003 6540000 bola= 3=37 = 4 × 22 × 441 38.808 000000014646d 011637072740 0000000000 0000000000 011637072740 6300000184000 163707274000 0000000000 0000000000 tabung 22 40000034c0000008676696577 00070646d6464 024000000014646 30000018400000 ⇔ tt= 2 6e64000002540 diperoleh: AB 1.38 BD 3646573630 3646573630 000002c400000 d6e640000025400 3000015000000 000003d4000000246c756d690 000015000000 00006c7774707 0000000000 0000000000 0006c77747074 001500000003 0000000000 0000494 0000000000 2 0000070646d64 cm 06c77747074000 0887675656400 = × 14 × 50 000070646d64640 0000184000 0000184000 62 6 64000002c4000 00003f8000000146d65617300 00034c0000008 00002c400000088 033646573630 = 4000001f00000 0000000000 0000000000 033646573630 000001f0000000 364657363000 0000000000 0000000000 7= 30.800 0008876756564 0006c77747 0006c77747 5432073 6766965770000 001f00000001462 00040c0000002474656368000 tt = EC CD 9. Misalkan panjang sisi 767565640000034 0000034c00000 03d4000000246 c00000086766965 074000001f 074000001f 3 000018400000 0014626b70740 0000000000 0000000000 000018400000 14626b7074000 001840000006 0000000000 0000000000 3 CD 004300000000c725452430000 0867669657700 6b707400000204 cmpermukaan = c756d69000003f 77000003d400000 5247420 AB 2 kubus = a. Luas 0000000146 0000000146 0003d40000002 2 20000204000000 8000000146d65 0246c756d690000 06c777470740 0000000000 0000000000 06c777470740 0020400000014 c77747074000 0000000000 0000000000 043c0000080c6754524300000 ⇔ tinggi = Jadi, tabung 2 00000014725859 46c756d690000 3 2 2 61730000040c0 03f8000000146d6 ml 26b7074000 26b7074000 0000000 60000001f000000 =6 6××a22 a30.800 kubus= 43c0000080c62545243000004 3 03f8000000146 EC CD AB 00001f000000 147258595a000 0000011637 0000011637 7258595a00000 001f00000001 0000024746563 0000001163 0000001163 561730000040c00 cm. 3c0000080c746578740000000 5a000002180000 0020400000 0020400000 d656173000004 30.800 6800000430000 000024746563680 0000000 Banyak = = 3 a2 = =00146758595a00 100botol 2180000001467 0c00000024746 0021800000014 0727400000 0727400000 014626b70740 4626b7074000 7072740000 7072740000 0000c72545243 2. ⇔ =014626b70740 00004300000000c 0147258595 0147258595 220 5636800000430 0436f70797269676874202863 0000043c00000 725452430000043 a000002180 a000002180 000020400000 6758595a00000 1500000003 1500000003 000020400000 58595a0000022 002040000001 0150000000 0150000000 0000000 0000000c72545 EC 2 2 a= 10 cm 1 80c6754524300 292031393938204865776c657 c0000080c675452 14000022c00000014 buah 2430000043c00 00043c0000080 0000014675 0000014675 430000043c00000 V== 322c0000001462 πr2 tJadi, perbandingan 7. a = 6258595a000002 Jawaban: c = 0147258595a0 3646573630 3646573630 0147258595a0 c000000146258 47258595a000 3364657363 3364657363 4742d5061636b61726420436f ⇔ AB : EC : jari00080c6754524 000000f 10 cm ⇒ r= × a 80c625452430000 c625452430000 3 1 22 Jawaban: d13 8595a00000 8595a00000 30000043c0000 3 6d70616e7900006465736300 043c0000080c7 043c0000080c746 V 000021800000 58595a0000024 0000184000 0000184000 000021800000 595a000002400 002180000001 0000018400 0000018400 2 = 5 cm Panjang garis 231 liter = 231 dm 080c625452430 4000000014646d ffd8ffe000104a464946 ai 22c0000001 22c0000001 4657874000000 6d60001 jari bola dan 3 : 2. 578740000000043 2 000000000000127352474220 000043c000008 1.188= × × rkerucut × 14 0001020100c800c8000 00436f70797269 0146758595a0 000000014646d 0006c77747 0006c77747 0146758595a0 0000014646d6e 46758595a000 6f7079726967687 00006c7774 00006c7774 = s rpelukis 1 46258595a0 46258595a0 s = 6e640000025400 0c74657874000 49454336313936362d322e31 37 2 6768742028632 420286329203139 0000000 Vair = 6400000254000 × volume bangun 6. 3.564 Misalkan pertama berjari-jari r1 00000436f7079 0000240000 0000240000 000022c00000 6e64000002540 074000001f 074000001f 000022c00000 0022c0000001 7074000001 7074000001 9203139393820 = 22 × tabung r × 2 0ffe20c584943435f505 3938204865776c6 000000000000000000000012 2 000070646d6464 24f46494c4500010100 7269676874202 4865776c65747 574742d5061636b 00014646d6 00014646d6 0d32d48 2 0146258595a0 0000070646d64 0000000146 0000000146 0146258595a0 00070646d6464 46258595a000 f000000014 f000000014 735247422049454336313936 8632920313939 danr tinggi t1. Tabung kedua berjari-jari r2 dan 000c484c696e6f02100 42d5061636b61 ruang 61726420436f6d7 = = 000002c4000000 rt = 1 r +×a(volume e640000025 e640000025 38204865776c6 726420436f6d70 3.564 ⇔ 231 tabung + volume 0616e7900006465 362d322e3100000000000000 0006d6e747252474220 000024000000 64000002c4000 26b7074000 26b7074000 000024000000 000002c400000 002400000001 626b707400 626b707400 5020200 574742d506163 2keruc 616e790000646 736300000000000 tinggi .000000000000000000000000 Diperoleh: 88767565640000 4000000706 4000000706 r2 =t281 58595a2007ce0002000 44 6b61726420436 5736300000000 000127352474220 22 0020400000 0020400000 014646d6e640 0002040000 0002040000 t1 =014646d6e640 t2 r1 = 2r20008876756564 Misalkan luas lingkaran = 46d6464000 46d6464000 ut)0887675656400 1 4646d6e64000 1 f6d70616e79000 0000000 9000600310000616373 034c0000008676 0000001273524 494543363139363 r= 9 0000000000000000 cm 0646573630000 7422049454336 704d534654000000004 62d322e31000000 002c400000 002c400000 000025400000 0000034c00000 0147258595 0147258595 000025400000 00034c0000008 002540000007 0014725859 0014725859 696577000003d4 0000000000127 pada tabung pertama L1 462 = 25 + 100 0000000 313936362d322 000000000000000 9454320735247420000 = = 0887675656 0887675656 3524742204945 e310000000000 070646d64640 0867669657700 a000002180 a000002180 070646d64640 6766965770000 0646d6464000 5a00000218 5a00000218 012735247422049 2 000000246c756d 0000000000000000000 t31 454336313936362 4336313936362 πr tb + 0000000000001 40000034c0 40000034c0 0000000 22 s = 00002c400000 d322e31000000 0003d40000002 0000014675 0000014675 00002c400000 03d4000000246 002c40000008 0000001467 0000001467 00000f6d60001000000 2735247422049 d322e3100000000 2 0000086766 0000086766 Jadi, panjang garis pelukis kerucut 0000000000000 2πr2 × 10 4543363139363 ⇔ 69000003f800000 462 = + πr ×0000000 658595a0000 000000000000000 dan088767565640 luas lingkaran pada 00d32d4850202000000 tabung πr t 46c756d690000 8595a00000 8595a00000 088767565640 c756d69000003f 876756564000 58595a0000 0001273524742 9657700000 9657700000 62d322e310000 3 22 22 000000000000000 3 0146d656173000 0000000000000000000 2 2049454336313 0000000000000 000000000000000 kedua L2. L1 : 03f8000000146 L2 = 2 0000000000000000000 × (πr122c0000001 ) : 2 22c0000001 × 3d40000002 000034c00000 8000000146d65 0034c0000008 022c000000 022c000000 0000000 10. Jari-jari tabung 936362d322e31 = 000034c00000 0000000000000 0040c000000247 2 A = rA = 2 ⋅3d40000002 2 0 70 462= × 10 × r + 2 × × r146258595a 0000000000000 46c756d690 46c756d690 0000000000000 0000000000000000000 2 2 46258595a0 46258595a0 9 cm. Jari-jari 086766965770 61730000040c0 676696577000 146258595a 2 0000000 (πr2 086766965770 ) 2= r1 : rd656173000004 0000000000000 46563680000043 tabung B =22rB =0000000000000 0000000000000000000 00003f8000 00003f8000 2 00000000000 πr 3.234 = 220r20000024746563 + 44r003d40000002 00003d400000 0000240000 0000240000 00003d400000 0000024000 0000024000 20c00000024746 0000000000000116370 00000000c72545 000146d656 000146d656 0000000000000 t 2 2 0000000 = (2r ) : L : 0246c756d690 LB = = 264r πrA 22 :6800000430000 2 5636800000430 7274000001500000003 3. 0246c756d690 00014646d6 00014646d6 12 cm. 46c756d69000 000014646d 000014646d 3.234 A 2430000043c000 00 πrt1730000040 1730000040 2r 2 0000000 3646573630000018400 2 2 2 00003f800000 c000000247 c000000247 2 :2r = 40000000c72545 = 4r : 1 Jadi, perbandingan luas 00003f800000 e640000025 e640000025 0000c72545243 003f80000001 6e64000002 6e64000002 2πr 2 = rA : 2rB = 9 : 2 × r = =B0080c675452430 12,25 L= πr 2 + 22πrt 22 22 00006c7774707400000 4656368000 4656368000 0000000 0146d6561730 2430000043c00 4000000706 4000000706 0146d6561730 0000043c00000 46d656173000 5400000070 5400000070 1f000000014626b7074 Jawaban: c pada kedua tabung tersebut 4 : 1 2 Jadi, 2 2 2 0043000000 lingkaran r = 000043c0000080c 3,512 dm ruang 0043000000 = (380c6754524300 ) jari-jari : 2(3 ×bangun 4)646d646400 0000020400000014725 = 000040c00000 × 282 + 200080c6754524 × × 28 × 100 46d6464000 46d6464000 000040c00000 0040c0000002 646d646400 0000000 00c7254524 00c7254524 8595a00000218000000 4 2 2 3,5 62545243000004 dm. 77=12.464 + atau : 4. = 3 : 200043c0000080c ×3 × 4 =0002c40000 024746563680 30000043c0000 002c400000 002c400000 024746563680 474656368000 0002c40000 30000043c0 30000043c0 146758595a0000022c0 0000000 3c0000080c74657 17.600 = 20.064 00000146258595a0000 000080c675 000080c675 000043000000 080c625452430 0887675656 0887675656 000043000000 6254524300000 004300000000 0088767565 0088767565 9 : 32 87400000000436f 2 0000000 024000000014646d6e6 4524300000 4524300000 cm00c725452430 000043c000008 40000034c0 40000034c0 00c725452430 43c0000080c746 c72545243000 640000034c 640000034c 4000002540000007064 Jadi,70797269676874 perbandingan kedua jari-jari bidangluas atasbagian kerucut terpancung 7,5 43c0000080 43c0000080 0000000 6d6464000002c400000 4. 000043c00000 0c74657874000 0000086766 0000086766 000043c00000 5787400000000 0043c0000080 0000008676 0000008676 tabung yang berbentuk lingkaran adalah 20286329203139 c625452430 c625452430 088767565640000034c cm. 80c675452430 0011637 80c675452430 00000436f7079 9657700000 9657700000 436f7079726967 c67545243000 6965770000 6965770000 000043c000 000043c000 Diketahui: jari-jari tabung besar = r1 3938204865776c 0000008676696577000 9 : 32. 0080c74657 0080c74657 7269676874202 000043c00000 6874202863292 0043c0000080 03d4000000 03d4000000 0727400 Jawaban: 6574742d506163 = 70000043c00000 cmdt1 = 100 cm003d4000000246c756d r2 = 3d40000002 35 cm3d40000002 t2 = 50 8740000000 8740000000 69000003f8000000146 80c625452430 8632920313939 46c756d690 46c756d690 80c625452430 0313939382048 c62545243000 246c756d69 246c756d69 0001500 6b61726420436f6 d6561730000040c0000 0436f70797 0436f70797 cm 000043c00000 38204865776c6 00003f8000 00003f8000 000043c00000 65776c6574742 0043c0000080 000003f800 000003f800 0024746563680000043 d70616e7900006 2696768742 2696768742 0000033 00000000c7254524300 80c746578740 574742d506163 000146d656 000146d656 80c746578740 d5061636b6172 c74657874000 0000146d65 0000146d65 0286329203 0286329203 46573630000000 00043c0000080c67545 6465736 1393938204 1393938204 0000000436f7 6b61726420436 1730000040 1730000040 0000000436f7 6420436f6d7061 00000436f707 6173000004 6173000004 2430000043c0000080c 00000001273524 865776c657 865776c657 3000001 625452430000043c000 079726967687 f6d70616e7900 c000000247 c000000247 079726967687 6e79000064657 972696768742 0c00000024 0c00000024 74220494543363 4742d50616 4742d50616 0080c74657874000000 420286329203 0064657363000 4656368000 4656368000 420286329203 3630000000000 028632920313 7465636800 7465636800 8400000 13936362d322e3 36b6172642 36b6172642 00436f7079726967687 139393820486 0000000000012 0043000000 0043000000 139393820486 0000127352474 939382048657 0004300000 0004300000 4202863292031393938 0436f6d706 0436f6d706 06c7774 10000000000000 204865776c6574742d5 16e7900006 16e7900006 5776c6574742 7352474220494 00c7254524 00c7254524 5776c6574742 2204945433631 76c6574742d5 000c725452 000c725452 00000000012735 061636b61726420436f 7074000 4657363000 4657363000 d5061636b617 5433631393636 30000043c0 30000043c0 d5061636b617 3936362d322e3 061636b61726 430000043c 430000043c 6d70616e79000064657 24742204945433 0000000000 0000000000 001f000 3630000000000000012 26420436f6d7 2d322e3100000 000080c675 000080c675 26420436f6d7 1000000000000 420436f6d706 0000080c67 0000080c67 0127352474 0127352474 6313936362d322 7352474220494543363 0000146 0616e7900006 0000000000000 4524300000 4524300000 0616e7900006 0000000000127 16e790000646 5452430000 5452430000 2204945433 2204945433 13936362d322e310000 e3100000000000 6313936362 6313936362 0000000000000000001 465736300000 0000127352474 43c0000080 43c0000080 465736300000 3524742204945 573630000000 043c000008 043c000008 26b7074 00000000000000 2735247422049454336 d322e31000 d322e31000 000000000127 2204945433631 c625452430 c625452430 000000000127 4336313936362 000000012735 0c62545243 0c62545243 0000020 00000000000000 313936362d322e31000 0000000000 0000000000 352474220494 3936362d322e3 000043c000 000043c000 352474220494 d322e31000000 247422049454 0000043c00 0000043c00 0000000000000000000 0000000001 0000000001 00000000000000 4000000 0000000000000000000 543363139363 1000000000000 0080c74657 0080c74657 543363139363 0000000000000 336313936362 00080c7465 00080c7465 2 2 2735247422 2735247422 0s − r 00000000000002

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 104a464946 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 0001020100 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 c800c80000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 ffe20c58494 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 3435f50524 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 f46494c450 Vtabun = πr2t 1 Luas segitiga= 2 × × r 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 0010100000 2 g 2 2 22 198= × 3 × 7t 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 c484c696e6 ×t = r × t = 1.386 = 1 22 × 9 × t 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 f021000006 30 × 18 tt= 0000000000000000000000000000000000000000000000 d6e7472524 1.386 = 540 t = 7 cm Jadi,198 tinggi 7422058595 Jadi, luas penampang berbentuk segitiga tabunga2007ce000 7 cm. cm2 pada belahan kerucut tersebut 540 cm2. 2000900060 0310000616 1 373704d534 2. Tinggi kerucut = × tinggi tabung 9. Jawaban: d Misalkan: r1 = jari2 6540000000 1 jari tabung pejal = 35 cm t = tinggi 0494543207 tk= × × 24 tabung = 3524742000 60 cm r2 = jari-jari tabung tk = 12 cm besar = 20000000000 × r1 = 70 cm 0000000000 Volume air = V2 – 2V1 = 00000f6d60 2 0010000000 πr2 t – 2 × πr12t = πt(r22 – 0d32d48502 2r12) 0200000000 22 0000000000 = × 60 × (702 – 2 × 7 0000000000 352)×2260 × (4.900 – 2 × = 0000000000 7 0000000000 1.225) 22 × (4.900 – = × 60 0000000000 7 0000000000 2.450) 22 = × 60 × 2.450 0000000000 7= 462.000 cm3 Jadi, 0000000000 volume air 462.000 cm3. 0000000116 3707274000 10. 0015000000 0336465736 Vkerucut = πr2t 3000001840 Jawaban: c1 31 3 000006c777 314= × 3,14 × 52 × t 34707400000 ⇔ t = 12 cm 1f00000001 Panjang garis 4626b70740 0000204000 pelukis: s2= r2 + t2 0001472585 = 5295a0000021 + 122 = 8000000146 169 758595a000 1 0022c00000 ⇔s= 0146258595 7 7 B. Uraian a000002400 0000014646 1. Misalkan V1 = volume 1 belahan d6e6400000 kerucut V2540000007 = volume kerucut utuh V=4× V1 1 0646d64640 00002c4000 ⇔ πr2t = 4 × 4.239 3 10008876756 3 8. Jawaban: b Volume 5640000034 plastisin = p × A × t × 3,14 × r2 × 18= 16.956 3 c000000867 = 6 × 3 × 11 = 198 cm3 Plastisin dibentuk 16.956 3,14r2 6696577000 = menjadi tabung sehingga volume tabung 6 6 003d400000 yang dihasilkan sama dengan volume 3,14r2 0246c756d6 = 2.826 = 2 balok. r = 900 9000003f80 r= 30 00000146d6 5617300000 40c0000002 4746563680 0000430000 0000c72545 2430000043 c0000080c6 7545243000 0043c00000 80c6254524 30000043c0 000080c746

Harga bahan plastik = 10,1736 × 40.000 = Rp406.944,00 Jadi, harga bahan plastik yang digunakan untuk membuat bola tersebut Rp406.944,00. 6. Misalkan r1 = jari-jari tabung besar r2 = jari-jari tabung kecil t1 = tinggi tabung besar t2 = tinggi tabung kecil 1 r3 r = ⇒ r = 1 1

2

3r t122= t

2

1

⇒ t1 =

2t V1 2: V2 = πr12t1 :

πr22t2 = (3r2)2(2t2) : r22t2 = 9r22 · 2t2 :

r22t2 = 18r22t2 : r22t2 = 18 : 1 Jadi, perbandingan volume kedua tabung 18 : 1 atau 1 : 18. 7. Volume air dalam tabung pertama = 2VtA ⎛⎛ 2 V = πr A

A

A

10

= 3,14 × ⎛ ⎛ × 3 ⎛2⎛ = 31,4 × 25 = 785 cm Volume air yang 10 dituangkan ke dalam kerucut = Vk 1

Vk= 23tkπrk 1

= × 3,14 × 52 × 3 = 4 × 3,14 12 × 25 = 314 cm3 Volume sisa air = V sV = V – V A k s

= 785 – 314 = 471 cm3 Misalkan rB = jari-jari tabung besar = tinggi air dalam tairbesar V= πrB2 tabung ×sstair 471 = 3,14 × 102 × tair a 471 = 314 × tair a t=air1,5 cm Jadi, tinggi air dalam tabung besar 1,5 cm.

8. Permasalahan tersebut dapat diubah menjadi sketsa berikut. C

ffd8ffe000104a464946000 1020100c800c80000ffe20c5 84943435f50524f46494c450 0010100000c484c696e6f02 1000006d6e7472524742205 8595a2007ce000200090006 1 A Diameter tutup atas = 12 cm ⇒ DE= 00310000616373704d53465 × 12 40000000049454320735247 2 42000000000000000000000 ffd8ffe000104a4649460001020100c800c8 2= 6 cm 0000000f6d6000100000000 0000ffe20c584943435f50524f46494c45000 1 d32d4850202000000000000 s=10100000c484c696e6f021000006d6e74725 ffd8ffe000104a464 00000000000000000000000 9460001020100c800 Diameter alas = 24 cm ⇒ AB=2 × 24 247422058595a2007ce0002000900060031 c80000ffe20c584943 22 0000616373704d534654000000004945432 =00000000000000000000000 12 +5 435f50524f46494c4 = 12 cm Perhatikan bahwa ∆ABC sebangun 00000000000000000000000 07352474200000000000000000000000000 500010100000c484c dengan ∆DEC sehingga berlaku: 00000000000000116370727 696e6f021000006d6 00f6d6000100000000d32d4850202000000 40000015000000033646573 e747252474220585 00000000000000000000000000000000000 =63000001840000006c77747 169 = 13 cm 95a2007ce00020009 DE EC 00000000000000000000000000000000000 Lselimu 0006003100006163 = =074000001f000000014626b πrs AB BC 6 EC 73704d5346540000 t 00000000000000000001163707274000001 70740000020400000014725 = 3,14 × 5 × 0000494543207352 ⇔500000003364657363000001840000006c7 = 8595a000002180000001467 4742000000000000 12=BE204,1 + EC 1 13 7747074000001f000000014626b70740000 0000000000000000f 58595a0000022c000000146 EC2 =6d6000100000000d cm12 cm+ 3. ⇔0204000000147258595a000002180000001 d= ⇒r 258595a0000024000000014 2 8EC + 32d4850202000000 46758595a0000022c000000146258595a00 646d6e64000002540000007 = 6t8tabun cm = 16 cm + 0000000000000000 EC = 2EC 00024000000014646d6e640000025400000 g 0646d6464000002c4000000 EC tkeruc =0000000000000000 8 =cm 8 cm 070646d6464000002c40000008876756564 0000000000000000 ut 88767565640000034c00000 V = Vtabung + Vkerucut 1 0000000000000000 pasak 0000034c0000008676696577000003d4000 08676696577000003d40000 2 0000000000000000 = 000246c756d69000003f8000000146d6561 πr ttabung + 3 πr2tkerucut 00246c756d69000003f8000 0000000001163707 730000040c0000002474656368000004300 000146d6561730000040c00 2740000015000000 1 0336465736300000 000000c725452430000043c0000080c6754 00002474656368000004300 = 3,14 × 62 × 16 + × 3,14 × 3 1840000006c777470 52430000043c0000080c625452430000043 000000c725452430000043c 2301,44 = 2.110,08 cm3 74000001f00000001 = 1.808,64 + 6 ×8 c0000080c7465787400000000436f7079726 0000080c675452430000043 4626b70740000020 c0000080c62545243000004 4000000147258595 9676874202863292031393938204865776c 4. Misalkan jari-jari a000002180000001 bola: 6574742d5061636b61726420436f6d70616 r3c0000080c7465787400000 Volume balok = p 46758595a0000022c 000436f7079726967687420 = 4r× 2r × 2r= × A ×e7900006465736300000000000000127352 t000000146258595a 28632920313939382048657 47422049454336313936362d322e3100000 0000024000000014 16r3 48 Volume seluruh bola = 2 × πr3 = πr3 76c6574742d5061636b6172 646d6e6400000254 3300000000000000000127352474220494543 Volume balok : volume 00000070646d6464 6420436f6d70616e7900006 36313936362d322e3100000000000000000 seluruh bola 8 000002c4000000887 46573630000000000000012 00000000000000000000000000000000000 3 67565640000034c00 = 16r : πr3 73524742204945433631393 0000867669657700 00 243cm = 144 + 25 0003d4000000246c7 6362d322e31000000000000 = 16 : 8π 3 =00000000001273524742204 656d69000003f80000 : 00146d6561730000 9454336313936362d322e31 π 22 040c0000002474656 00000000000000000000000 3680000043000000 = 600000000000000000000000 : 00c72545243000004 7 3c0000080c6754524 00000000 = 21 : 11 Jadi, perbandingan volume balok 30000043c0000080c dan 625452430000043c0 volume seluruh bola 21 : 11. 5.

000080c7465787400 000000436f7079726 19676874202863292 r = 0313939382048657 d = 0,9 m 2 Luas bahan plastik yang 76c6574742d506163 6b61726420436f6d7 diperlukan = L. L = 4πr2 = 3,140616e79000064657 × (0,9)2 3630000000000000 = 12,56 × 0127352474220494 54336313936362d3 0,81 = 22e3100000000000 10,1736 m2 0000000000012735 2474220494543363 13936362d322e310 0000000000000000 0000000000000000 0000000000000000 00000

tk2 + r2

4×

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104 ffd8ffe000 ffd8ffe0001 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 a464946000102 104a464946 04a4649460 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 0100c800c8000 0001020100 001020100c 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 0ffe20c5849434 c800c80000 800c80000ff 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 35f50524f4649 ffe20c58494 e20c584943 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 4c45000101000 3435f50524f 435f50524f4 2 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 00c484c696e6f 46494c4500 6494c45000 ffd8ffe000104a46 9. Volume 1 keping gula = πr × 1 1 3 3 3 r = × 4 2 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 021000006d6e7 010100000c 10100000c4 49460001020100c t V = 3,14 × 3 × 2,5 8 2 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 472524742205 484c696e6f 84c696e6f0 r2 = 2800c80000ffe20c5 Jadi, jari-jari bola setelah diubah V = 28,26 × 2,5 V 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 8595a2007ce00 021000006d 21000006d6 84943435f50524f4 3 V = 70,65 cm volume total 2 dm. 0000000000000000000000000000000000000000000000 020009000600 6e74725247 e747252474 6494c4500010100 3. Jawaban: d 422058595a V122058595a2 : V2 = 310000616373 000c484c696e6f02 Banyak keping gula = V 704d53465400 2007ce0002 007ce00020 1000006d6e74725 785 : 1.177,5 7 70.650 000000494543 0009000600 0090006003 247422058595a20 = = 1.000 πr12t1 207352474200 : πr22t2 = = 785 : 1.177,5 70,65Uang yang diperoleh = 2.500 × 3100006163 1000061637 07ce00020009000 2 2 1.000 = 2.500.000 Jadi, Pak Tomo akan 000000000000 3704d53465 5 t1 : 600310000616373 5 t2 = = 785 :73704d5346 1.177,5 000000000000 5400000000 4000000004 704d53465400000 memperoleh uang sebanyak t1 : t2 = = 785 : 1.177,5 00f6d60001000 4945432073 9454320735 000494543207352 Rp2.500.000,00. 00000d32d485 2474200000 tinggi tabung t1 : t2 474200000000000 = 2 : 3 Jadi,5247420000 perbandingan 020200000000 0000000000 0000000000 000000000000000 tersebut 2 : 3. 000000000000 0000000000 4 2 00f6d6000100000 4. Jawaban: c d10000000000 = 10. Volume bola = πr 3 000000000000 0000f6d600 000f6d6000 3 4 10 12 cm ⇒000d32d48502020 r1ffd8ffe000104a464946 = 6 cm d2 = 000000000000 0100000000 100000000d 000000000000000 ⎛⎛ ⇔ V= × 3,14 × ⎛⎛ 0001020100c800c80000 10 cm ⇒000000000000000 r2 = 5 cm t d32d485020 =32d4850202 32⎛⎛ 000000000000 ffe20c584943435f50524 000000000000 2000000000 0000000000 ⇔ V= × 3.1401 21 cm 000000000000000 f46494c4500010100000 Perubahan volume = πt(r12 – 000000000000 0000000000 0000000000 6Volume seluruh 000000000000000 c484c696e6f021000006 000000000000 0000000000 0000000000 22 bola = 12 × V 2 d6e7472524742205859 r22) 000000000000000 = 000116370727 0000000000 0000000000 7 × 21(6 – 000000000000000 5a2007ce00020009000 2 5 )0000000000 400000150000 0000000000 000011637072740 = 726 3 600310000616373704d = 12 ×61 × 3.140 = 6.280 cm3 000336465736 0000000000 0000000000 000015000000033 cm 5346540000000049454 5. 300000184000 0000000000 0000000000 646573630000018 goaltrai Jadi, volume seluruh bola pada 3207352474200000000 d = 6 0006c77747074 dm = 60 cm ⇒ r1 = 30 cm 0000000000 0000000000 3 40000006c777470 n setiap cm . 0000000000000000000 6.280 r2 = 30 –b 3 = 27 cm Volume 000001f000000 0000000000 0000000000 Jawaban: 74000001f000000 0f6d6000100000000d32 014626b70740 0000001163 0000011637 014626b70740000 udarad485020200000000000 dalam bola yang tersisa 000020400000 7072740000 0727400000 4 020400000014725 223 AC= 0000000000000000000 = ) 0147258595a00 0150000000 1500000003 3 8595a0000021800 0000000000000000000 000218000000 3364657363 3646573630 22 4 = 0000146758595a0 3 (BE + EC) + 12 0000000000000000000 = 3π(30 π(r –000022c00000014 r2 – 146758595a000 0000018400 0000184000 A. Pilihan 1 3 =3 0000000000000000000 27 )= 6258595a0000024 0022c00000014 00006c7774 0006c77747 Ganda 3 0000000011637072740 9.756πcm 6258595a00000 7074000001 074000001f 1. Jawaban: d Selisih 000000014646d6e 0000150000000336465 6. 3 240000000146 f000000014 0000000146 22 640000025400000 volume = 244,92 cm r1 = 8 7363000001840000006 DC= EC + DE 46d6e64000002 626b707400 c77747074000001f0000 V2070646d64640000 = × V1 26b7074000 cm t = 6 cm 125 22 540000007064 0020400000 Jawaban: c 64 02c400000088767 = 00014626b7074000002 8 + 6 0002040000 64 2 4 6d6464000002c 0014725859 0147258595 3 565640000034c00 ⇔ 04000000147258595a0 × 3,14 × r2 = × 4.186 1 =31253 4 64 12.560 400000088767 5a00000218 a000002180 000086766965770 Selisih volume = πt(r12 – 2) 0000218000000146758 31 3 565640000034c 0000001467 3 00003d400000024 ⇔ × 3,14 × r=2 = × 0000014675 595a0000022c0000001 Misalkan: L luas selimut kerucut besar r2 1 3 12.560 2 2 31253 64 100 000000867669 58595a0000 8595a00000 6c756d69000003f 244,92 = × 3,14 × 6(8 – r2 ) 46258595a0000024000 dengan jari-jari AB, tinggi BC L2 = luas 3 6577000003d4 022c000000 22c0000001 38000000146d6561 ⇔ r = × × × 2 000014646d6e6400000 244,92 = 6,28(64 – r22) 12543314 16 000000246c756 146258595a 46258595a0 730000040c00000 selimut kerucut kecil dengan jari-jari DE, 25400000070646d6464 2 3 3 d69000003f800 0000024000 0000240000 024746563680000 tinggi CE 39 = 64 – r22 r2 = 000002c400000088767 × 40 × 100 125 0000146d6561 000014646d 00014646d6 04300000000c725 565640000034c000000 r22 = = 25 = 512 π452430000043c00 × AB × 730000040c000 6e64000002 e640000025 rL21 = = 8676696577000003d40 000247465636 5400000070 4000000706 r2 = 5 cm AC 3,14 × 12jari-jari 00000246c756d690000 r2 ==800080c675452430 cm Jadi, bola 2. Jawaban: c d1 800000430000 646d646400 46d6464000 000043c0000080c × 20 = 753,6 03f8000000146d656173 sekarang 8 cm. 2 π625452430000043 0000c72545243 0002c40000 002c400000 L2 = × DE × = 8 dm ⇒r1 = 4 dm cm 0000040c00000024746 0000043c00000 0088767565 0887675656 c0000080c746578 DC = 5636800000430000000 3,14 × 6 V1 : V2 = 8 : 1 44 80c6754524300 640000034c 40000034c0 7400000000436f7 0c725452430000043c00 × 10 = 188,4 00043c0000080 0000008676 079726967687420 3 3 Luas lampu = L0000086766 – L2 = 753,6 – 00080c6754524300000 ⇔πr1 : πr2 = cm2 kap 1 c625452430000 6965770000 9657700000 286329203139393 3 33 43c0000080c625452430 2 188.48204865776c6574 = 565,2 cm03d4000000 Jadi, luas fiber 8 3: 1 3 043c0000080c7 3d40000002 r1 : r2 = =8:1 000043c0000080c74657 yang diperlukan 565,2 cm2. 465787400000 246c756d69 46c756d690 742d5061636b617 87400000000436f70797 43 : r23 = 8 : 1 000436f707972 000003f800 00003f8000 26420436f6d7061 2696768742028632920 696768742028 0000146d65 000146d656 6e7900006465736 31393938204865776c6 632920313939 6173000004 1730000040 300000000000000 574742d5061636b6172 38204865776c6 0c00000024 c000000247 127352474220494 6420436f6d70616e7900 574742d50616 7465636800 4656368000 54336313936362d 0064657363000000000 36b617264204 0004300000 0043000000 322e31000000000 0000012735247422049 36f6d70616e79 000c725452 00c7254524 000000000000012 454336313936362d322 000064657363 430000043c 30000043c0 735247422049454 e310000000000000000 000000000000 0000080c67 000080c675 336313936362d32 0000001273524742204 001273524742 5452430000 4524300000 2e3100000000000 9454336313936362d32 204945433631 043c000008 43c0000080 000000000000000 2e31000000000000000 3936362d322e3 0c62545243 c625452430 000000000000000 0000000000000000000 100000000000 0000043c00 000043c000 000000000000016 0000000000000000000 2 000000000001 00080c7465 0080c74657 0BC + 144 + AB =

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104 ffd8ffe00 ffd8ffe0001 ffd8ffe00 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 a46494600010 0104a464 04a4649460 0104a464 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 20100c800c80 94600010 001020100c 94600010 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 000ffe20c5849 20100c80 800c80000ff 20100c80 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 43435f50524f4 0c80000ff e20c584943 0c80000ff 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 6494c4500010 e20c5849 435f50524f4 e20c5849 B. Uraian 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 100000c484c6 43435f50 6494c45000 43435f50 Jadi, perbandingan volume logam sebelum 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 96e6f02100000 524f4649 10100000c4 524f4649 dan sesudah dipanaskan 9 : 14. 1. Misalkan volume kayu yang dibuang = 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6d6e74725247 4c450001 84c696e6f0 4c450001 sV = V – V 1 2 V 8. Jawaban: b Volume 422058595a20 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0100000c 21000006d6 0100000c 2 s = πr t – 0000000000000000000000000000000000000000000000 07ce00020009 484c696e e747252474 484c696e 1 1 kerucut tetap, sehingga: 000600310000 6f021000 22058595a2 6f021000 V1 = V2 11 πr22t2 22 22 2 616373704d53 006d6e74 007ce00020 006d6e74 = × 10 × 14 – × 72 × 7 3 ⇔πr12t1 = πr22t2 465400000000 72524742 0090006003 72524742 77 = 4.400 – 1.078 = 3.322 cm Jadi, 3 33 494543207352 2058595a 1000061637 2058595a 2 2 volume kayu yang dibuang 3.322 cm3. r1 t1 = (2r1) t22 ffd8ffe00010 ffd8ffe00010 474200000000 2007ce00 3704d53465 2007ce00 4a4649460001 4a4649460001 r12 · 10 = 4r12 · t22 000000000000 02000900 4000000004 02000900 020100c800c8 020100c800c8 4 0000ffe20c584 0000ffe20c584 00000000f6d60 06003100 9454320735 06003100 2. Perubahan volume = π(r23 10 = 4t22 943435f50524f 943435f50524f 3 00100000000d 00616373 2474200000 46494c450001 46494c450001 – r13) 400616373 4 t2 = 2,5 cm Jadi, tinggi kerucut 252π = π(r230100000c484c – 0100000c484c r13)704d5346 32d485020200 0000000000 704d5346 3 696e6f021000 696e6f021000 000000000000 54000000 0000000000 54000000 sekarang 2,5 cm. 3 006d6e747252 006d6e747252 756 = 4(r2 – 27) 47422058595a 47422058595a 00494543 000f6d6000 00494543 9. Jawaban: b r2 = jari- 000000000000 2007ce000200 2007ce000200 189 = r23 – 27 000000000000 20735247 100000000d 20735247 090006003100 090006003100 jari tabung lubang sebelum 3 000000000000 42000000 32d4850202 42000000 00616373704d 00616373704d r2 = = 216 lubang diper 534654000000 534654000000 000000000000 00000000 0000000000 00000000 004945432073 004945432073 besar = jari-jari luar – ⇔ r2 = 000000000000 00000000 0000000000 00000000 524742000000 524742000000 ketebalan 000000000000 000000000000 Tebal lapisan =000000000000 r2000000f6 –0000000000 r000000f6 =6–3= 1 0000000000f6 0000000000f6 kayu = 10 – 3 000000000116 d6000100 0000000000 d6000100 d60001000000 d60001000000 3 cm Jadi, tebal lapisan tersebut = 7 cm 00d32d485020 00d32d485020 370727400000 000000d3 0000000000 000000d3 200000000000 200000000000 3. Misalkan V = volume mula3 cm. r1 = jari-jari tabung lubang 150000000336 0000000000 2d485020 1 2d485020 000000000000 000000000000 465736300000 20000000 0000000000 20000000 mula V2 = volume setelah diubah 000000000000 000000000000 setelah lubang diperbesar 10 cm 000000000000 000000000000 1840000006c7 00000000 0000000000 00000000 000000000000 000000000000 V1 = V2 11 = 10 – 2 = 8 cm V1 = volume 774707400000 00000000 0000000000 00000000 000000000000 000000000000 000000000000 2 2 000000000000 2 1f00000001462 00000000 0000011637 00000000 ⇔πr1 t1 = πr t2 tabung dengan jari-jari 8 cm = πr1 t 000000000000 2 000000000000 33 6b7074000002 00000000 0727400000 00000000 116370727400 116370727400 t000150000000 1 00000000 22 2 2 000150000000 040000001472 1500000003 00000000 ⇔ r1 t1= r2 336465736300 · 2336465736300 2 00000000 2 3646573630 00000000 = × 82 × 35 = 7.040 cm358595a000002 000184000000 000184000000 7V = volume tabung dengan jari2r12 = r22 6c7774707400 180000001467 00000000 0000184000 00000000 2 6c7774707400 0001f0000000 0001f0000000 58595a000002 00000000 0006c77747 00000000 jari 7 cm 22 ⇔ r2= r1 14626b707400 14626b707400 2c0000001462 00000000 074000001f 00000000 2 2 000204000000 000204000000 = πr2 t = × 7 × 35 = 5.390 147258595a00 147258595a00 r : r = r : r 58595a000002 00000000 0000000146 00000000 7Volume kayu yang terbuang = 1 2 1 000218000000 1 000218000000 3 400000001464 00000000 26b7074000 00000000 V1 – V2 = 7.040 –cm 5.390 = 1.650 146758595a00 146758595a00 00022c000000 00022c000000 6d6e64000002 11637072 0020400000 11637072 cm3 146258595a00 146258595a00 540000007064 74000001 0147258595 74000001 000240000000 000240000000 10. Jawaban: a 4 6d6464000002 50000000 a000002180 50000000 14646d6e6400 14646d6e6400 4. 277,2 liter = 277,2 dm3 = 000254000000 000254000000 c40000008876 33646573 0000014675 33646573 r1 : r2 = 4 : 3 ⇒ r1 = r2 70646d646400 70646d646400 3 3 277.200 cm Volume air mula-mula 756564000003 63000001 8595a00000 63000001 0002c4000000 0002c4000000 t = 10 cm V1 : V2 = = V 887675656400 887675656400 V = 4c0000008676 84000000 22c0000001 84000000 1 1 00034c000000 00034c000000 2 696577000003 6c777470 6c777470 πr12t : πr22t 22 867669657700 867669657700 πr t 22 2 46258595a0 1 × r × 50 ⇔ 277.200=0003d4000000 0003d4000000 74000001 0000240000 74000001 2 d4000000246c 2 7 12.560 : V2 = r1 : r2 ⎛ 4 ⎛ 22 246c756d6900 246c756d6900 756d69000003f f0000000 00014646d6 f0000000 2 2 2 0003f8000000 0003f8000000 5.5443= r146d65617300 8000000146d6 14626b70 e640000025 14626b70 12.560 : V2 = ⎛ r ⎛ : r2 146d65617300 200040c000000 2 ⎛ 00040c000000 r2 = = 561730000040 74000002 4000000706 74000002 247465636800 247465636800 2 2 2 2 2 3 ⎛ c00000024746 04000000 46d6464000 04000000 r = = 1.764 000430000000 000430000000 12.560 : V2= r2 :16r22 0c7254524300 0c7254524300 9 563680000043 14725859 002c400000 r= 42 cm Volume air14725859 yang 00043c000008 00043c000008 12.560 : V2 = 16 : 9 9 9 7. tersisa Jawaban: a Misalkan volume logam 00000000c725 5a000002 0887675656 5a000002 0c6754524300 = V 0c6754524300 00043c000008 00043c000008 sebelum dipanaskan = V , volume logam 452430000043 18000000 40000034c0 18000000 V2 = × 12.560 1 0c6254524300 0c6254524300 16 c0000080c675 14675859 0000086766 14675859 setelah dipanaskan = V2. 00043c000008 00043c000008 V2 = 7.065 cm3 Jadi, volume452430000043 benda setelah 5a000002 9657700000 5a000002 0c7465787400 2 0c7465787400 V : V = πr t : 1 2 1000000436f70 1 000000436f70 c0000080c625 2c000000 3d40000002 2c000000 dihaluskan 7.065 cm3. 797269676874 797269676874 2 πr22t2 = r12t202863292031 452430000043 : r t 14625859 46c756d690 = 14625859 202863292031 1 2 2 393938204865 393938204865 c0000080c746 5a000002 00003f8000 5a000002 36 × 776c6574742d 35 776c6574742d : 49 × 578740000000 40000000 000146d656 40000000 5061636b6172 5061636b6172 40 = 6420436f6d70 9 : 6420436f6d70 1414646d6e 0436f70797269 1730000040 14646d6e 616e79000064 616e79000064 676874202863 64000002 c000000247 64000002 657363000000 657363000000 292031393938 54000000 4656368000 54000000 000000001273 000000001273 524742204945 524742204945 204865776c65 70646d64 0043000000 70646d64 433631393636 433631393636 74742d506163 64000002 00c7254524 64000002 2d322e310000 2d322e310000 000000000000 000000000000 6b6172642043 c4000000 30000043c0 c4000000 000000127352 000000127352 6f6d70616e790 88767565 000080c675 88767565 474220494543 474220494543 36313936362d 36313936362d 000646573630 64000003 4524300000 64000003 322e31000000 322e31000000 000000000000 4c000000 43c0000080 4c000000 000000000000 000000000000 012735247422 86766965 c625452430 86766965 000000000000 000000000000 000000000000 000000000000 049454336313 77000003 000043c000 77000003 000000000000 000000000000 936362d322e3 d4000000 0080c74657 2 =21 : d4000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 ⇔22 × 6 × 42 × t = s110.880 7. a. Benda tersebut merupakan gabungan 62d322e31000000000000000000000000000000000000000000000000000000

setengah bola dan kerucut. 14 1 V= × πr3 + πr2t

110.880

⇔ t=

s 5.544

⇔ t = 20 cm

23 3 1222 122

s

⇔1.745

Jadi, ketinggian air sekarang 20

= × × 73 + × × 72 ×

t

3 5. Vcm. × 2× ( )3 = 4.851 cm3 1 = 4πr 1 =21 3422 37

5.236 22+ × 7 × t ⇔ = × 22 × 249 333 3 5.236 = 2 × 22 × 49 + 22 × 7 × t 5 5.236 = 2.156 + 154t 5 154t = 3.080 1 t = 20 cm Diperoleh tinggi kerucut 20 cm. y= r + t = 7 + 20

Perubahan volume = V1 – V2 1 1 4.671 = 4.851 – V2 3

33737

2

V2 = 179 3 42

⇔πr23 = 179 33 539 3

πr2=

3

×

3 3

r23 = = 42,875

4 4

r2 = 3,5 cm Jadi, jari-jari akhir bola 3,5 cm. 6.

a. V1 = volume kerucut kecil 1 ⇔ 550 = πr12t1 3

1.650= × r122 × 21

2 2

7

1.650 = 22 × r12 × 3 r12 = 25 ⇔ r1 = 5 cm r2 = jarijari kerucut besar = 2 × r1 = 10 cm V1 : V2 = 1 : 12 1 V2= 12 × V1 = 12 × 550 = 6.600 cm3 V2 = 1πr22t2 3 1 22

⇔ 6.600 = ×

× 102 × t 138.600 = 22 × 100 × t2222 37

7

138.600 2.200

t2 =

38.808 22

= 63 Jadi, tinggi kerucut besar 63 cm. b. L = luas alas kerucut besar = 2 πr2 2 =× 7 10 22

2.200

s

Vs

= V1 –

⇔πr2t s=22277.200 – 166.320 166.320

2

⇔ × 422 × t = 110.880 s

= =7314 cm2 7

7

2

2

Jadi, luas alas kerucut besar7314 cm .

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00010 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4a464946000 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 ffd8ffe000104a4 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 1020100c800 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 64946000102010 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 c80000ffe20c 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 0c800c80000ffe2 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 584943435f5 0000000000000000000000000000000000000000000

0c584943435f505 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 0524f46494c 24f46494c450001 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 45000101000 Jari-jari kapur barus yang tersisaVolume kayu mula-mula = V1 155 V1 = volume 3. ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 0100000c484c69 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 00c484c696e kerucut 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 = 224 × = cm mula-mula Vair = πrair2tair 6e6f021000006d6 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6f021000006 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001163707274000001500000 Jawaban: b 13 e7472524742205 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 d6e74725247 003364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c000 23 Volume kapur barus yang tersisa = V2 1 ⇔ 24π = × 82 2 000146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d400 0000000000000000000000000000000000000000000000 422058595a2 A. Pilihan4 =8595a2007ce000 πr t + πr t t 1 air 1 πr 3 3 0000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c00 VGanda = 21 × × 007ce000200 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 ⇔ 20009000600310 2 rai1222 = 22 3 00080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f 2 2 1. Jawaban: 09000600310 3 =000616373704d5 × × 6d70616e790000646573630000000000000012735247422049454336313936362d322e3100000000000000000000001273524742204945433631393 9 12 × 9 + × 12 × 31 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 πr2 r3 4 22 ⎛⎛ 5 ⇔= cm 377 r d air = 1 : 2 sehingga berlaku pula r : = 21 × × × ⎛⎛00061637370 374 34654000000004 6362d322e31000000000000000000000000000000000000000000000000000000 air ⎛⎛ 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 22 22 tair =3 : 94543207352474 twadah ffd8ffe000104a4649460001020100c800c8000 4d534654000 × × 144 + × 144 × 31 = 1 : 2 (aturan r 77 12 wadah 000000000000000000000000000000000000000000000000000000000000000000000000000000000 = 40ffe20c584943435f50524f46494c45000101000 × 22 × 20000000000000 00000494543 kesebangunan segitiga) = 2 × rair 64 00c484c696e6f021000006d6e74725247422058 9.504 98.208 5 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 00000000000000 20735247420 = + 595a2007ce0002000900060031000061637370 rwadah =77 2 × 3 = 6 cm = 11 ×12 0f6d60001000000 00000000000 4d534654000000004945432073524742000000 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 8 5 107.71 3 Jadi, jari-jari wadah 6 0000000000000000000000f6d6000100000000d 00d32d48502020 00000000000 = cm 2 7 1.375 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 3 32d48502020000000000000000000000000000 cm. = cm 00000000000000 00000f6d600 8 00000000000000000000000000000000000000 Tinggi tabung setelah diraut 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 00000000000000 0100000000d Volume00000000000000000000000000000116370727 kapur barus yang menyublim 4. = 31 Jawaban: b Luas – (16 – 9) 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 00000000000000 32d48502020 40000015000000033646573630000018400000 = V – V06c77747074000001f000000014626b70740000 permukaan bola = 4πr22 = 24 cm 1 2 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 00000000000000 00000000000 2 2 0204000000147258595a000002180000001467 576π =kayu 4πrsetelah 1.375 00000000000000 00000000000 Volume diraut = V 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 = 1.375 58595a0000022c000000146258595a00000240 – 2 15 cm 2 8 00000000000000 00000000000 r = = 144 00000014646d6e640000025400000070646d64 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 V = volume kerucut 2 + volume 27 cm Jadi, tinggi 11.000 −1.375 2 = 00000000001163 00000000000 64000002c400000088767565640000034c00000 r= 12 cm 4 = 00000000000000000000000000000000000000000000000 1 22 22 tabung 2 12 08676696577000003d4000000246c756d69000 benda 27 2 cm. 2 8 70727400000150 00000000000 = 377 ×× 16 3 + × 12 × 24 10 cm 10 cm 10 cm 10 cm Volume bola =×πr 9.62 003f8000000146d6561730000040c0000002474 00000033646573 00000000000 3 b. = 656368000004300000000c725452430000043c0 5 8 permukaan Luas = luas permukaan tabung 22 22 00000000000 4 =63000001840000 × 4 × 12 000080c675452430000043c0000080c62545243 2 = π××16 12+3 × 144 × 24 77 + 1.203,125 2 × 0000043c0000080c7465787400000000436f707 luas penampang berbentuk persegi 006c7774707400 = cm3 00000000000 3 1 16.896 76.032 panjang × 20 97269676874202863292031393938204865776 4 = volume2benda dengan t2==20001f0000000146 t==10 00000011637 + Jadi, volume kapur barus yang menyublim cm =V π2× 12 × 12 1 77 2c6574742d5061636b61726420436f6d70616e79 2 3 26b70740000020 07274000001 3 = 16π × = 2πr 00006465736300000000000000127352474220 + 2πrt 1.203,125 cm + . 2(2r)t = 2 × 3,14 × 10 + 92.928 tinggi kerucut 10 cm 14 1 3 40000001472585 50000000336 = cm 49454336313936362d322e3100000000000000 144 = 2 × 3,14 × 10 × 15 + 2 × 20 × 15 = 628 7 9. 95a00000218000 46573630000 0000000012735247422049454336313936362d 2.304π cm3 + 942322e3100000000000000000000000000000000 + 600 = 2.170 cm2 000146758595a0 01840000006 Volume bagian Volume tabung kayu yang = 0000000000000000000000 3 2 000022c0000001 c7774707400 hilang = V –V 3 = × × πr + πr tcm 1 2 volume bola = 2.304π 2 16 cm 233 222 122 46258595a00000 107.712 2. 0001f000000 5. = −92.928 24000000014646 014626b7074 Vtabung = =7× × 73 + × × 72 × 10 373 d6e64000002540 00000204000 14.784 Jawaban: c Tinggi bandul = t71 = 34 mm, tinggi kerucut 2t πr 2.156 1.540 ffd8ffe000104a4649460 2 = 311 cm 0000070646d646 22 00014725859 7 Jawaban: c+ = × 21 × 28 = t2 = = 24 mm Jari-jari kerucut = jari-jari 001020100c800c80000ffe 4000002c400000 7 = 22 × 4 × 5a000002180 = 2.112 333 20c584943435f50524f46 3.696 bola 494c4500010100000c484 00000146758 441 = cm 08876756564000 c696e6f021000006d6e74 = t1 0034c000000867 – t= = 10 3 mm 3595a0000022 2 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50 38.808 cm 725247422058595a2007c 3 1.232 cm Besar 6696577000003d e0002000900060031000 = 1,54 × (30 ×= 60 detik) Tinggi Panjang garis pelukis Vtabung = c0000001462 524f46494c4500010100000c484c696e6f021000006d6e74725247422 0616373704d534654000 = 1,54 × 1.800 = 4000000246c756 58595a00000 perubahan volume = kerucut = t1 00000494543207352474 = s 2.772 liter 3 mula-mula 058595a2007ce00020009000600310000616373704d5346540000000 Vbola 3 3 d69000003f80000 4 4 24000000014 V – V 38.808 = πr 20000000000000000000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 2 t = ? 3 s= Jadi, 1 000000000f6d600010000 049454320735247420000000000000000000000000000f6d60001000 00146d65617300 646d6e64000 volume tabung penampung air 9.261 = r 1 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 38.808 0000d32d485020200000 22 3 00040c00000024 = 1.745 – 1.232 00254000000 ××r 00000d32d48502020000000000000000000000000000000000000000 00000000000000000000 31 2 2 22 2.772 liter. = 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 72 = 74656368000004 24 + 10 = 00000000000000000000 15 −12 70646d64640 43 b. Diketahui r = 0,7 3m = 7 dm 00000000000000000000000000000000000000000000000000000001 00000000000000000000 = 513 cm 000000000000000000000000000000000000000000000000000000000000000000000000000000000 300000000c7254 00002c40000 × 2× r3 ? 22 * 4 * 441 = 00000000000000000000 225 −144 besar perubahan volume benda 1 = 3Jadi,576 + 100 163707274000001500000003364657363000001840000006c7774707 V = V 7 52430000043c00 00887675656 00000000001163707274 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 tabung = 81 = 00000150000000336465 1 r3 40000034c00 4000001f000000014626b707400000204000000147258595a0000021 43 ×3× 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 ⇔ 2.772 = πr2t==676 900080c67545243 cm = 263cm Luas 7363000001840000006c 7 0000043c000008 00008676696 513 cm . ? 77747074000001f000000 8000000146758595a0000022c000000146258595a000002400000001 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 3 selimut s kerucut×22 =72πrs 014626b7074000002040 0c625452430000 577000003d4 ⇔ 2.772= ×t 441 7 00000147258595a00000 4646d6e640000025400000070646d6464000002c4000000887675656 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 043c0000080c74 000000246c7 8. Volume kapur barus mula⇔ 218000000146758595a0 r= = = 22 × 7 × t 40000034c0000008676696577000003d4000000246c756d69000003f 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 65787400000000 000022c00000014625859 56d69000003 1 mula L1= =⇔ 3,14 V12.772 4 × 10 × 26 3 V =436f70797269676 2.772 21rL× π 816,4 = 215a000002400000001464 cm Jadi, jari-jari bola f8000000146 1 = ⇔1 3 ⇔ 8000000146d6561730000040c0000002474656368000004300000000 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 t= 6d6e6400000254000000 1 = 18 dm 3 87420286329203 154 Jadi, tinggi5 tabung 18 dm 70646d6464000002c4000 2 tersebut 21 cm. d6561730000 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 c725452430000043c0000080c675452430000043c0000080c6254524 ⎛⎛ mm 1 ×422 =atau 21 ×× ⎛⎛ 00088767565640000034 13939382048657 040c0000002 Luas setengah bola = 1,8 m. 2 125 37⎛⎛ c0000008676696577000 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 30000043c0000080c7465787400000000436f7079726967687420286 76c6574742d506 47465636800 × 4πr22 003d4000000246c756d69 2volume =1636b617264204 4× 22 × 00000000000000000000000000000000000000000000000 + tabung mula-mula 000003f8000000146d656 3292031393938204865776c6574742d5061636b61726420436f6d706 00043000000 8 = 11 2× 2 1730000040c0000002474 L = 2πr 36f6d70616e7900 00c72545243 2 16e79000064657363000000000000001273524742204945433631393 65636800000430000000 125 = 0c725452430000043c000 0000043c000 L = 00646573630000 2 × 3,143 × 1022 2 6362d322e31000000000000000000000012735247422049454336313 0080c675452430000043c 1.375 cm 00000000001273 0080c675452 0000080c6254524300000 L2 = 52474220494543 628 mm2 936362d322e310000000000000000000000000000000000000000000 430000043c0 43c0000080c7465787400 000000436f70797269676 000080c6254 0000000000010. a. V = debit 36313936362d32 × waktu 87420286329203139393 2e310000000000 52430000043 8204865776c6574742d50 61636b61726420436f6d7 00000000000012 c0000080c74 0616e790000646573630 73524742204945 65787400000 00000000000001273524 74220494543363139363 4336313936362d 000436f7079 62d322e3100000000000 322e3100000000 72696768742 00000000000127352474 00000000000000 02863292031 22049454336313936362 d322e310000000000000 00000000000000 39393820486 00000000000000000000 00000000000000 5776c657474 00000000000000000000 2 2 2 2 2d5061636b6 0000t2 + r 0 s −r

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104 ffd8ffe00010 ffd8ffe000 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 a464946000102 4a464946000 104a464946 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 0100c800c8000 1020100c800c 0001020100 ffd8ffe000104a4 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 0ffe20c5849434 80000ffe20c5 c800c80000 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 64946000102010 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 35f50524f4649 84943435f505 ffe20c58494 0c800c80000ffe2 0000000000000000000000000000000000000000000

4c45000101000 0c584943435f505 Luas00c484c696e6f permukaan bandul = L1 24f46494c450001 021000006d6e7 0100000c484c69 + L2 = 816,4 + 628 = 1.444,4 mm2 4725247422058 6e6f021000006d6 595a2007ce000 e7472524742205 2000900060031 6. 8595a2007ce000 0000616373704 20009000600310 d534654000000 2 000616373704d5 2 L1 : L0049454320735 = 4πr Jawaban: a 1 : 4πr 2 2 34654000000004 2 2 2 r1 := r2r=22474200000000 8 :6 : r = 8 : 6 = 94543207352474 1 2 0000000000000 64 : 36 = 16 : 920000000000000 L1= 0000000f6d600 00000000000000 240 0100000000d32 cm2 0f6d60001000000 9 d485020200000 00d32d48502020 L2= 1 × 240 0000000000000 00000000000000 6 = 0000000000000 135 2 00000000000000 cm 0000000000000 00000000000000 0000000000000 7. Jawaban: 00000000000000 a Luas 2 0000000000000 alas drum = πr = 00000000000000 3,14 × 0000000000000 2 00000000000000 20 = 3,14 × 400 0000000000001 00000000001163 1637072740000 70727400000150 =00000033646573 1.256 0150000000336 2 4657363000001 cm 3 Air menempati 63000001840000 bagian, sehingga udara 840000006c777 4 006c7774707400 1 47074000001f0 0001f0000000146 menempati 4 00000014626b7 26b70740000020 bagian. 0740000020400 Tinggi bagian 40000001472585 yang ditempati 0000147258595 udara 11 95a00000218000 = a000002180000 × t = × 100 = 25 cm 000146758595a0 44Volume 00146758595a0 2 000022c0000001 = πr × t udara udara = 1.256 × 25 = 000022c000000 46258595a00000 3 31.400 cm = 31,4 dm3 = 31,4 liter 146258595a000 24000000014646 Jadi,0024000000014 volume udara dalam drum d6e64000002540 646d6e6400000 31,4 liter. 0000070646d646 ffd8ffe000104a4649460 2540000007064 001020100c800c80000ff 4000002c400000 e20c584943435f50524f4 6d6464000002c 08876756564000 6494c4500010100000c4 4000000887675 84c696e6f021000006d6e 0034c000000867 8. Jawaban: d Diketahui r1 = 74725247422058595a20 65640000034c0 6696577000003d 07ce0002000900060031 12 cm, 0000086766965 s1 = 20 cm4000000246c756 Tinggi selimut 0000616373704d534654 kerucut77000003d4000 mula-mulad69000003f80000 = t100000000494543207352 47420000000000000000 000246c756d69 22 00146d65617300 000000000000f6d600010 t1 = 000003f800000 11 0000000d32d485020200 00040c00000024 00000000000000000000 0146d65617300 74656368000004 = 00000000000000000000 ffd8ffe000104a4649460001020100c8 00040c0000002 00000000000000000000 300000000c7254 00c80000ffe20c584943435f50524f464 00000000000000000000 4746563680000 = 256 94c4500010100000c484c696e6f02100 = 16 cm52430000043c00 Penampang selimut 00000000000001163707 04300000000c7 0006d6e74725247422058595a2007ce 27400000150000000336 00080c67545243 kerucut sebagai berikut. D 2545243000004 46573630000018400000 00020009000600310000616373704d5 0000043c000008 06c77747074000001f000 3465400000000494543207352474200 3c0000080c675 0c625452430000 000014626b7074000002 00000000000000000000000000f6d60 452430000043c 04000000147258595a00 043c0000080c74 00100000000d32d4850202000000000 00021800000014675859 0000080c62545 0000000000000000000000000000000 65787400000000 5a0000022c0000001462 2430000043c00 0000000000000000000000000000000 58595a00000240000000 436f70797269676 0000000000000000000000001163707 14646d6e640000025400 00080c7465787 87420286329203 000070646d6464000002 2740000015000000033646573630000 400000000436f c4000000887675656400 13939382048657 01840000006c77747074000001f0000 7079726967687 00034c00000086766965 00014626b7074000002040000001472 76c6574742d506 A 12 cm B 77000003d4000000246c 4202863292031 58595a00000218000000146758595a0 1636b617264204 756d69000003f80000001 000022c000000146258595a00000240 3939382048657 BD00000014646d6e64000002540000007 = t1 = 16 cm, AB46d6561730000040c000 = r1 = 12 cm, AD = s1 36f6d70616e7900 00024746563680000043 76c6574742d50 00646573630000 0646d6464000002c400000088767565 00000000c72545243000 = 20 cm 61636b6172642 640000034c000000867669657700000 0043c0000080c6754524 00000000001273 30000043c0000080c625 0436f6d70616e 3d4000000246c756d69000003f80000 52474220494543 452430000043c0000080 00146d6561730000040c00000024746 7900006465736 c7465787400000000436f 36313936362d32 56368000004300000000c7254524300 3000000000000 70797269676874202863 2e310000000000 00043c0000080c675452430000043c0 29203139393820486577 0012735247422 000080c625452430000043c0000080c 00000000000012 6c6574742d5061636b61 0494543363139 7465787400000000436f70797269676 726420436f6d70616e790 73524742204945 874202863292031393938204865776c 00064657363000000000 36362d322e310 4336313936362d 00000127352474220494 6574742d5061636b61726420436f6d7 0000000000000 54336313936362d322e3 322e3100000000 0616e79000064657363000000000000 10000000000000000000 0000000012735 0012735247422049454336313936362 00000000000000 00012735247422049454 2474220494543 d322e31000000000000000000000012 336313936362d322e310 00000000000000 735247422049454336313936362d322 36313936362d3 00000000000000000000 00000000000000 e310000000000000000000000000000 00000000000000000000 22e3100000000 0000400 − 144 00000000000000000000000000 20 cm 2 00000000000002

24f46494c450 3435f50524f 0010100000c4 46494c4500 ∆ABD sebangun dengan ∆ECD, 84c696e6f021 010100000c sehingga: EC CD = 000006d6e74 484c696e6f AB BD r2 72524742205 021000006d 4 = ⇔ 8595a2007ce0 6e74725247 12 16 r2 00200090006 422058595a 1 = ⇔ 00310000616 2007ce0002 12 4 0009000600 ⇔ r2373704d5346 = 3 cm 54000000004 3100006163 EC ED 94543207352 73704d5346 = 47420000000 5400000000 AB AD 3 00000000000 4945432073 ⇔ s2 = 0000000000f6 5247420000 12 20 1 d6000100000 0000000000 ⇔ s2 = 4 2 20 000d32d4850 0000000000 s2 =20200000000 5 cm Luas permukaan0000f6d600 selimut kerucut 00000000000 0100000000 tersisa = L22 00000000000 d32d485020 L2 =00000000000 πr2s2 = 3,14 × 3 × 5 = 47,1 cm2 2000000000 00000000000 0000000000 00000000000 0000000000 Jadi,00000000000 luas permukaan selimut kerucut yang 0000000000 2 00000000000 0000000000 tersisa 47,1 cm . 00000000001 0000000000 9. 16370727400 0000000000 Luas00015000000 sebuah parasut = 0000000000 2 Jawaban: 1b 2 03364657363 0000000000 × 4πr 1 00000184000 = 0000001163 ×4× 2 2= 25,12 m Luas plastik 0006c7774707 3,14 ×7072740000 22 4000001f0000 minimal = 15 × 25,12 = 0150000000 00014626b70 2 3364657363 376,8 m 74000002040 0000018400 ffd8ffe000104a464946 00000147258 00006c7774 0001020100c800c8000 0ffe20c584943435f505 10. Jawaban: a 1 122 595a0000021 7074000001 24f46494c4500010100 80000001467 f000000014 000c484c696e6f02100 2 V = πr t ⇔ 1.232= × × 72 × t 3 37 0006d6e747252474220 58595a00000 626b707400 58595a2007ce0002000 22c000000146 0002040000 9000600310000616373 ⇔ t = 24 cm Panjang 704d534654000000004 258595a0000 0014725859 garis pelukis: 9454320735247420000 02400000001 5a00000218 0000000000000000000 22 0000001467 00000f6d60001000000 s = 4646d6e6400 00d32d4850202000000 00025400000 58595a0000 0000000000000000000 070646d6464 022c000000 0000000000000000000 0000000000000000000 000002c40000 146258595a = × 7 × 25 = 550 cm2 0000000000000000000 7 00887675656 0000024000 0000000000000116370 7274000001500000003 40000034c000 000014646d 11. Jawaban: c Misalkan: kedalaman 3646573630000018400 00086766965 6e64000002 air pada tabung =00006c7774707400000 t1 77000003d40 5400000070 1f000000014626b7074 kedalaman air0000020400000014725 pada kerucut = t2 00000246c756 646d646400 8595a00000218000000 Volume air = V + V d69000003f80 0002c40000 146758595a0000022c0 1 2 2 2 00000146d65 0088767565 11.775 = πr2t100000146258595a0000 + πr t 1 1 2 024000000014646d6e6 61730000040c 640000034c 2 4000002540000007064 11.775=3,14 × 152 × t1 +0000008676 × 3,14 × 15 × 20 6d6464000002c400000 00000024746 3 3 088767565640000034c 56368000004 6965770000 0000008676696577000 11.775 = 706,5t + 4.710 1 300000000c72 03d4000000 003d4000000246c756d 69000003f8000000146 706,5t = 7.065 = 54524300000 246c756d69 1 d6561730000040c0000 43c0000080c6 000003f800 0024746563680000043 t1 = 10 cm 00000000c7254524300 75452430000 0000146d65 00043c0000080c67545 043c0000080c 6173000004 2430000043c0000080c 625452430000043c000 62545243000 0c00000024 0080c74657874000000 0043c0000080 7465636800 00436f7079726967687 c74657874000 0004300000 4202863292031393938 204865776c6574742d5 00000436f707 000c725452 061636b61726420436f 97269676874 430000043c 6d70616e79000064657 3630000000000000012 20286329203 0000080c67 7352474220494543363 13939382048 5452430000 13936362d322e310000 0000000000000000001 65776c657474 043c000008 2735247422049454336 2d5061636b6 0c62545243 313936362d322e31000 1726420436f6 0000043c00 0000000000000000000 0000000000000000000 d70616e7900 00080c7465 000000000000022

Kedalaman air = t1 + t2 = 10 + 20 = 30 cm 12. Jawaban: a Diketahui t = 9 cm, r1 = 11 cm, dan r2 = 15 cm Misalkan besar perubahan volume = Vp V =1 P

31

πt(r22

1.58 = 4 722 + 528 = 226 + 528 = 754 cm2 7

15. Jawaban: a Tinggi tabung = diameter bola sehingga: jari-jari bola =

1

× tinggi tabung =

2221 21

– r12) = × 3,14 × 9(152 – 3

2 11979,68 ) = cm3

cm tabung = Jari-jari bola = jari-jari Vtabung πr2t cm = 2

21⎛

=

13. Jawaban: a Misalkan V1 = volume tabung V2 = volume kerucut V1 =

2

22 ⎛

× ⎛⎛ × 21

7⎛ 2 ⎛441

= 22 × 4× 3 = 29.106 = 7.276,5 4 4 3 cm Vbola = 3 4 22πr⎛3 21⎛3

πr2t1 22 = × 72 × 12 7= 22 × 7 × 12 = 1.848 cm3 1 V2 =

7

= 37⎛ 2×⎛× ⎛⎛ 4 22 21 212 = ××× = 11 × 441 = 4.851 cm3 Volume ruangan – Vbola = Vtabung= 3742

3 1 22

= × × 72 × (18 πr2t2 3 7 – 12) 1 = × 22 × 7 × 6 3 = 22 × 7 × 2 = 308 cm3 V= V1 + V2 = 1.848 + 308 = 2.156 cm3 Jadi, volume benda tersebut 2.156 cm3. 14. Jawaban: b Volume 2 batang logam = volume air yang tumpah v 2 × V =6.336 2 2 × πr2t = 3.168 22 2 2 × × 62 × t=3.168 7

× 72 ×2t =3.168 27 2 1.58 ⇔ × t = 3.168 4 7

3.168 × 7 1.58 7 t4 t=

t = 14 cm Luas permu 22 22 2 = 2 × × 6 + 2 × kaan × 6 × 14 7744 ×36 setiap = 7 + 44 × 6 ×poton 2 gan logam =L L= 2πr2 + 2πrt

7.276,5 – 4,851 = 2.425,5 cm3 Jadi, volume ruangan di luar bola 2.425,5 cm3. 16. Jawaban: a Volume ruang kosong = V1 4

ffd8ffe000104a464946000102010 3 3 cm 0c800c80000ffe20c584943435f505 24f46494c4500010100000c484c69 6e6f021000006d6e7472524742205 42.390 3 = 3,14r1 8595a2007ce00020009000600310 4 000616373704d534654000000004 3 9454320735247420000000000000 10.597,5 = 3,14r1 000000000000000f6d60001000000 3 r1 = 3.375 00d32d4850202000000000000000 0000000000000000000000000000 r1 = 15 cm Jari-jari luar bola = r1 + 0000000000000000000000000000 0000000000000000000000001163 ketebalan dinding 7072740000015000000033646573 63000001840000006c7774707400 0001f000000014626b70740000020 4000000147258595a00000218000 000146758595a0000022c0000001 46258595a0000024000000014646 d6e640000025400000070646d646 4000002c40000008876756564000 0034c0000008676696577000003d 4000000246c756d69000003f80000 00146d6561730000040c00000024 74656368000004300000000c7254 52430000043c0000080c67545243 0000043c0000080c625452430000 043c0000080c7465787400000000 436f7079726967687420286329203 1393938204865776c6574742d506 1636b61726420436f6d70616e7900 0064657363000000000000001273 5247422049454336313936362d32 2e31000000000000000000000012 735247422049454336313936362d 322e310000000000000000000000 0000000000000000000000000000 0000

⇔ 14.130 =

1

πr1 3

3

=

0

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104a d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4649460001020 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 100c800c80000ff 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 e20c584943435f 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 50524f46494c45 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 00010100000c48 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 4c696e6f021000 18. Jawaban: c Diameter = 1 m 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 006d6e7472524 ⇒r1 = 50 cm Volume air dalam wadah 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 7422058595a20 mula-mula = V1 V1 = πr12t1 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 07ce0002000900 0000000000000000000000000000000000000000000000 0600310000616 22 373704d534654 2 = × 50 × 70 3 0000000049454 7= 550.000 cm Volume air 3207352474200 yang dipindahkan = V2 V2 = 5 0000000000000 × volume tabung kecil = 5 × 0000000000000f πr22t2 6d60001000000 00d32d4850202 22 2 0000000000000 =5× × 14 × 0000000000000 7= 30 5 × 18.480 = 92.400 3 0000000000000 cm Sisa air = V1 – V2 = 0000000000000 550.000 – 92.400 = 0000000000000 457.600 cm3 = 457,6 0000000000000 0000000000000 dm3 = 457,6 liter 0000116370727 4000001500000 0033646573630 0000184000000 19. Jawaban: d 6c777470740000 = rk = rb = r t= db ⇒ t= 01f00000001462 12 6b70740000020 * 2r 4000000147258 4 3 ⇔ 14 = 2r ?r3 + 595a000002180 ⇔ r= 7 cm V = volume 1 3 setengah bola 0000014675859 + volume kerucut ?r2t 5a0000022c0000 = 00146258595a0 23 0000240000000 * 14646d6e64000 22 7 0025400000070 * 73 + 646d646400000 13 2c400000088767 * 565640000034c0 22 7 * 72 * 14 0000086766965 = 77000003d4000 23 000246c756d690 * 22 * 49 + 00003f80000001 13 46d6561730000 * 22 * 7 * 14 040c0000002474 = 15 + 3 = 18 cm Jadi, jari-jari = 6563680000043 luar bola 18 cm. 23 00000000c72545 * 1.078 + 17. Jawaban: cd= 2430000043c000 13 0080c675452430 14 cm ⇒ r = 7 cm * 2.156 000043c0000080 2 2 s= c6254524300000 2 2 = 43c0000080c746 7 + 24 5787400000000 = 436f7079726967 49 + 576 6874202863292 0313939382048 = 625 = 25 cm L= πr2 + πrs = 65776c6574742d 2 π×7 + π × 7 × 25 = 49π + 5061636b61726 175π420436f6d70616 = 224π satuan luas e790000646573 6300000000000 0001273524742 2049454336313 936362d322e31 0000000000000 0000000001273 5247422049454 336313936362d 322e310000000 0000000000000 0000000000000

2

= =

2.156 +2.156 3 4.312 31

= 1.437 cm3 3

20. Luas bagian kubah yang akan dicat Jawaban: = L. 1 a 2 L= × 4πr =

2 1 227 × 4 27⎛⎛ 2 1 22 49

2

× ×⎛⎛ ⎛⎛

=

×4×× = 11 × 7 = 77 m2 77 274

Banyak cat = kaleng.

11

=7

Jadi, diperlukan 7 kaleng cat untuk ffd8ffe000104a46 mengecat49460001020100c kubah tersebut. C 21. Jawaban:800c80000ffe20c5 ffd8ffe000104a4649 84943435f50524f4 d 460001020100c800c 22 s = BC= 6494c4500010100 80000ffe20c5849434 35f50524f46494c450 000c484c696e6f02 22 0010100000c484c69 = 1000006d6e74725 10 +24 6e6f021000006d6e7 247422058595a20 = 100 +576 4725247422058595a 07ce00020009000 2007ce00020009000 60031000061637370 600310000616373 = 676 = 26 cm A 4d534654000000004 704d53465400000 94543207352474200 D 10 cm 000494543207352 00000000000000000 Luas selimut sebuah kerucut = πrs = 474200000000000 000000000f6d60001 00000000d32d48502 000000000000000 3,14 × 10 × 26 = 31,4 × 26 = 816,4 cm2 02000000000000000 00f6d6000100000 00000000000000000 000d32d48502020 00000000000000000 000000000000000 00000000000000000 Luas karton = 10 × 816,400000000000000000 = 8.164 cm2 000000000000000 00000000000011637 22. Jawaban:000000000000000 07274000001500000 000000000000000 c Paku tersebut terdiri atas00336465736300000 tabung I, tabung 000000000000000 1840000006c777470 II, dan kerucut. Volume tabung I = VI. VI= 74000001f00000001 000000000000000 4626b707400000204 2 000011637072740 πr1 t1 000000147258595a0 000015000000033 00002180000001467 58595a0000022c000 22 646573630000018 000146258595a0000 = ×40000006c777470 (1,4)2 7 024000000014646d6 ×1 74000001f000000 e6400000254000000 = 2 × 1,96 = 6,16 014626b70740000 7 70646d6464000002c 2 020400000014725 40000008876756564 cm3 tabung Volume II = 0000034c000000867 8595a0000021800 VII r=2 22 π2t2 0000146758595a0 6696577000003d400 0000246c756d69000 003f8000000146d65 = ×000022c00000014 (0,7)2 × 7 6258595a0000024 61730000040c00000 12 02474656368000004 000000014646d6e = 22 × 0,49 × 12 300000000c7254524 7129, 640000025400000 30000043c0000080c = 36 070646d64640000 = 18,48 675452430000043c0 7 02c400000088767 000080c6254524300 3 cm kerucut 00043c0000080c746 Volume = 565640000034c00 5787400000000436f VIII 1 000086766965770 2 70797269676874202 VIII= 3 πr3 t 00003d400000024 86329203139393820 22 1 3 6c756d69000003f 4865776c6574742d5 = × × (0,7)2 × 061636b6172642043 37228000000146d6561 3 6f6d70616e7900006 730000040c00000 46573630000000000 =× 0,49 7 024746563680000 00001273524742204 3 9454336313936362d = 1,5404300000000c725 322e3100000000000 452430000043c00 cm V= VI + VII + VIII = 00000000000127352 00080c675452430 47422049454336313 6,16 + 18,48 000043c0000080c + 1,54 936362d322e310000 3 00000000000000000 625452430000043 = 26,18 cm 00000000000000000 c0000080c746578 Jadi, volume paku tersebut 26,18 cm3. 0000000000000000B 7400000000436f7 079726967687420 23. Jawaban: a Misalkan volume 286329203139393 tabung di luar kerucut = V V= Vtabung8204865776c6574 – 742d5061636b617 Vkerucut21 26420436f6d7061 = πr t – 3πr2t 6e7900006465736 300000000000000 127352474220494 54336313936362d 322e31000000000 000000000000012 735247422049454 336313936362d32 2e3100000000000 000000000000000 000000000000000 0000000000000DB +DC

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 62d322e31000000000000000000000000000000000000000000000000000000

= =

πr2t

3 2 227 7

× × × × 12

372217

= 2 × 22 × × × 4 22= 2 × 22 × 7 = 44 × 7 = 308 cm3 24. 4

Volumec 6 bola besi= 6 × 3 Jawaban: 3 4 22 πrb 3 =6×××7 3 37= 8 × 22 × 49 = 8.624 cm Volume air = πr 2 t t air = 22 × 142 × 30 = 18.480 cm3 7V = Volume air + volume bola ⇔V = 27.104 cm3 Misalkan tinggi air sekarang =t 2 s V= πr t ts 22

27.104= × 142 ×s t

⇔

7

⇔ 27.104 = 616 ×s t ⇔ t = 44 cm s Jadi, tinggi air dalam tabung setelah dimasukkan enam bola besi 44 cm. 25. Jawaban: d = πr2t Vtabung t t ⇔ 6.280 = 3,14 × r2 ×t 20 ⇔ 6.280 = 62,8 2 t ×r 6.280 ⇔ r2= t 62,8 ⇔ r2 = 100 t ⇔ rt = 10 cm Alas kerucut berimpit dengan alas tabung sehingga rk = r = 10 cm t

Vk= 1 πrk2tk 31 3

2.512 = × 3,14 × 102 × tkk 3

7.536 = 314 × tkk tk =

314

= 24 cm 7.536

Jadi, tinggi kerucut yang berada di atas tabung (24 – 20) = 4 cm.

26. Jawaban: a d = 2,7 m ⇒r = 135 cm Luas selimut tabung = L L= 2πrt 22 594.000= 2 2 5 × × 135 × t 4.158.000 = 44 × 135 × t 4 4.158.000 = 5.940 t 4 t = 700 cm Panjang halte = t = 700 cm = 7 m

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104a ffd8ffe0001 ffd8ffe0001 ffd8ffe0001 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 4649460001020 04a4649460 04a4649460 04a4649460 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 100c800c80000f 001020100c 001020100c 001020100c 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c fe20c584943435 800c80000ff 800c80000ff 800c80000ff 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 f50524f46494c4 e20c584943 e20c584943 e20c584943 0000000000000000000000000000000000000000000

435f50524f4 500010100000c 435f50524f4 435f50524f4 484c696e6f0210 6494c45000 6494c45000 6494c45000 ffd8ffe000104a4649460001020100c800c80000ffe20c58 1 1 V2= × × πr2t2 00006d6e74725 10100000c4 10100000c4 10100000c4 ffd8ffe000104a46494600010 4943435f50524f46494c4500010100000c484c696e6f021 2 3 20100c800c80000ffe20c58494 000006d6e74725247422058595a2007ce0002000900060 247422058595a 84c696e6f0 84c696e6f0 84c696e6f0 1 3435f50524f46494c45000101 0310000616373704d5346540000000049454320735247 = 6 × 3,14 × 202 × 2007ce0002000 21000006d6 21000006d6 21000006d6 00000c484c696e6f021000006 420000000000000000000000000000f6d6000100000000 201 9000600310000 e747252474 e747252474 e747252474 d6e74725247422058595a2007 d32d48502020000000000000000000000000000000000 =6 × 3,14 × 400 × 20 616373704d534 22058595a2 22058595a2 22058595a2 ce00020009000600310000616 000000000000000000000000000000000000000000000 373704d53465400000000494 7 6540000000049 007ce00020 007ce00020 007ce00020 000000000000000011637072740000015000000033646 543207352474200000000000 25.120 3 4543207352474 0090006003 0090006003 0090006003 = cm 57363000001840000006c77747074000001f0000000146 00000000000000000f6d60001 3 cm 6V= V + 2000000000000 1000061637 1000061637 1 26b707400000204000000147258595a00000218000000 00000000d32d485020200000 61000061637 cm 000000000000000000000000 0000000000000 3704d53465 3704d53465 3704d53465 V2 146758595a0000022c000000146258595a000002400000 000000000000000000000000 25.120 0014646d6e640000025400000070646d6464000002c400 000f6d60001000 4000000004 4000000004 4000000004 = 6.280 + 6 000000000000000000000000 Volume balok9454320735 =p× A 000088767565640000034c0000008676696577000003d4 00000d32d4850 9454320735 9454320735 000000000000000000116370 ×000000246c756d69000003f8000000146d656173000004 t= 62474200000 × 3 ×2474200000 11 =727400000150000000336465 198 37.680 +25.120 2020000000000 2474200000 27. 0c0000002474656368000004300000000c725452430000 Jawaban: bd= = cm3 0000000000 7363000001840000006c77747 0000000000000 0000000000 0000000000 6 14 cm ⇔r = 7 cm 22 074000001f000000014626b70 043c0000080c675452430000043c0000080c6254524300 0000000000000 0000000000 0000000000 0000000000 Seluruh bagian balok dibentuk menjadi 62.800 00043c0000080c7465787400000000436f707972696768 740000020400000014725859 = s= 0000000000000 000f6d6000 000f6d6000 000f6d6000 tabung sehingga volume balok = volume 64 74202863292031393938204865776c6574742d5061636b 5a0000021800000014675859 2 100000000d 0000000000000 100000000d 100000000d 5a0000022c000000146258595 V 61726420436f6d70616e79000064657363000000000000 tabung. = πr t tabung 2 22 2 2 = 10.466 = 32d4850202 1432d4850202 +7 = 32d4850202 a0000024000000014646d6e64 0000000000000 0012735247422049454336313936362d322e310000000 6 2 0000025400000070646d6464 196 +49 198= 0000000000000 0000000000 × 0000000000 r ×7 0000000000 00000000000000012735247422049454336313936362d 7 000002c400000088767565640 = 10.4662 cm3 2 2 322e31000000000000000000000000000000000000000 0000000116370 198 0000000000 = 0000000000 22 × r 0000000000 000034c000000867669657700 3Jadi, volume gabungan belahan kerucut 000000000000000 11 cm 7 cm 7274000001500 0000000000 0003d4000000246c756d69000 r=2 = = 0000000000 9 0000000000 tersebut 2 003f8000000146d6561730000 0000033646573 0000000000 0000000000 0000000000 r= 30000000000 cm Luas 0000000000 selimut 3 040c000000247465636800000 6300000184000 0000000000 10.466 cm . tabung = L L=0000000000 2πrt4300000000c72545243000004 3 0006c77747074 0000000000 0000000000 3c0000080c675452430000043 =×7× 000001f0000000 0000000000 0000000000 30. Jawaban: c Perhatikan gambar 7=0000000000 22 c0000080c625452430000043c 0000080c74657874000000004 14626b7074000 0000000000 0000000000 0000000000 potongan bola berikut. 22 36f7079726967687420286329 0020400000014 0000000000 0000000000 0000000000 = 2 × × 3 × 7 2 1 1 2031393938204865776c65747 7= 132 cm Jadi, luas selimut 7258595a00000 0000011637 0000011637 0000011637 L= 84 × luas permukaan bola + 3 × luas 28. 42d5061636b61726420436f6d 11 2 tabung 132 cm . 2180000001467 0727400000 0727400000 lingkaran 70616e790000646573630000 Volume bak 0727400000 = × (4πr2) + 3 × × 000000000012735247422049 58595a0000022 1500000003 1500000003 1500000003 8 4 = V 1 b2 Jawaban: ffd8ffe000104a464946000 454336313936362d322e3100 V= × πr t c000000146258 3646573630 3646573630 3646573630 2 3. Diameter tabung 4,2 cm sehingga jari-jari 22 πr 1020100c800c80000ffe20c5 000000000000000000001273 = 13 πr2 + πr2 595a000002400 0000184000 0000184000 84943435f50524f46494c450 524742204945433631393636 tabung 0000184000 2,1 Perhatikan bahwa tabung kaca 24 1 cm. 22 7 ⎛⎛ 0010100000c484c696e6f02 0000014646d6e 0006c77747 = 0006c77747 × ⎛⎛ × 0006c77747 ⎛⎛ × 2d322e310000000000000000 15tabung dan setengah 2 27 merupakan gabungan dari 5 1000006d6e7472524742205 2 11 49 ffd8ffe000104a4649460001020100c800c = πr 000000000000000000000000 6400000254000 074000001f 074000001f 074000001f 8595a2007ce000200090006 4 bola. Jari-jari bola = jari-jari tabung 00000000000000 80000ffe20c584943435f50524f46494c450 00310000616373704d53465 00070646d6464 0000000146 0000000146 0000000146 =× × 15 40000000049454320735247 74= 288,75 = 5 × 3,14 × 000002c400000 26b7074000 26b7074000 26b7074000 0010100000c484c696e6f021000006d6e74 4 42000000000000000000000 = 2,1 cm Misalkan t = tinggi air dalam 3 0020400000 1 2 2 0887675656400 0020400000 0020400000 0000000f6d6000100000000 m Volume 725247422058595a2007ce0002000900060 10392,5 cm = d32d4850202000000000000 tabung Volume air = volume air dalam 00034c0000008 0147258595 0147258595 0147258595 00000000000000000000000 air = aV2 0310000616373704d53465400000000494 + volume 1 6766965770000 a000002180 a000002180 a000002180 00000000000000000000000 tabung B. 5432073524742000000000000000000000 00000000000000000000000 V= V a × 03d4000000246 0000014675 0000014675 0000014675 3 bola 00000000000000116370727 214 c756d69000003f 8595a00000 8595a00000 2 8595a00000 1. 0000000f6d6000100000000d32d48502020 Luas permukaan setengah 40000015000000033646573 =22c0000001 × 288,75 = Uraian 63000001840000006c77747 2 di dalam 8000000146d65 22c0000001 22c0000001 0000000000000000000000000000000000 bola = L1 1 2 3Jadi, volume air = t1 + × πr3 bak 3 πr 074000001f000000014626b ⇔158,004 192,5 m L = × 4πr 23 22 222 3 61730000040c0 46258595a0 46258595a0 46258595a0 2 1 0000000000000000000000000000000000 70740000020400000014725 192,5 m . 2= 2π· (3,5) 227 0000024746563 0000240000 0000240000 0000240000 8595a000002180000001467 ⇔158,004 = × (2,1)2 2× t + × × (2,1)3 0000000000000000000000000011637072 7 58595a0000022c000000146 7372221 2 1 29. 6800000430000 00014646d6 00014646d6 00014646d6 258595a0000024000000014 7400000150000000336465736300000184 ⎛⎛ 0000c72545243 e640000025 e640000025 e640000025 Misalkan V⎛⎛ volume 646d6e64000002540000007 ⇔ 158,004 =× t1 22 + ×belahan × =2×× × 1 =× 3 722 0000006c77747074000001f000000014626 0646d6464000002c4000000  710 37 0000043c00000 4000000706 4000000706 4000000706 Jawaban: a kerucut besar V = volume belahan 88767565640000034c00000 2 b707400000204000000147258595a00000 21 46d6464000 80c6754524300 46d6464000 46d6464000 08676696577000003d40000 ⎛⎛ 00246c756d69000003f8000 kerucut ⎛⎛ kecil 002c400000 10 002c400000 00043c0000080c 002c400000 218000000146758595a0000022c00000014 2 22321 t1 ×22221 ×× ×× 000146d6561730000040c00 t1 : t2= 6254524300000 0887675656 0887675656 0887675656 ⇔ 158,004= 2+ 3 00002474656368000004300 6258595a0000024000000014646d6e6400 2 40000034c0 10 ⇔:t22= ×10 3010=×1.386t 43c0000080c746 40000034c0 40000034c0 000000c725452430000043c 3 2 3 00025400000070646d6464000002c40000 =2×× × 2 = 44 cm0000080c675452430000043 + +19.404 1 1 5787400000000 0000086766 0000086766 0000086766 158,004 = 72 c0000080c62545243000004 20 cm 1 0088767565640000034c00000086766965 1 9657700000 436f7079726967 9657700000 9657700000 3c0000080c7465787400000 V1= × = 13.860t × Luas permukaan bagian lampu yang dibuat dari 158.004 + 19.404 2 3 1 000436f7079726967687420 77000003d4000000246c756d69000003f80 6874202863292 3d40000002 3d40000002 3d40000002 kaca 28632920313939382048657 πr2t 1 13.860t = 138.600 = 0313939382048 46c756d690 46c756d690 00000146d6561730000040c00000024746 76c6574742d5061636b6172 = L1 + 1 × 3,14 =146c756d690 × 6420436f6d70616e7900006 6 = 3,14 × 400 65776c6574742 00003f8000 00003f8000 2 00003f8000 2 56368000004300000000c7254524300000 t = 10 cm L277 + 44 = 121 cm = 46573630000000000000012 1 20 × 30 d5061636b6172 000146d656 ×000146d656 5000146d656 = 6.280 73524742204945433631393 43c0000080c675452430000043c0000080c 31730000040 6362d322e31000000000000 6420436f6d7061 1730000040 1730000040 cm 625452430000043c0000080c74657874000 00000000001273524742204 6e79000064657 c000000247 c000000247 c000000247 9454336313936362d322e31 00000436f70797269676874202863292031 3630000000000 4656368000 4656368000 4656368000 00000000000000000000000 00000000000000000000000 393938204865776c6574742d5061636b61 0000127352474 0043000000 0043000000 0043000000 00000000 2204945433631 00c7254524 00c7254524 00c7254524 726420436f6d70616e79000064657363000 3936362d322e3 30000043c0 30000043c0 30000043c0 0000000000012735247422049454336313 1000000000000 000080c675 000080c675 000080c675 936362d322e31000000000000000000000 0000000000127 4524300000 4524300000 4524300000 012735247422049454336313936362d322 3524742204945 43c0000080 43c0000080 43c0000080 e310000000000000000000000000000000 4336313936362 c625452430 c625452430 c625452430 d322e31000000 000043c000 000043c000 000043c000 000000000000000000000002 cm = 77 cm2 0000000000000 0080c74657 0080c74657 0080c74657 Luas selimut tabung = L2 L2= 2πrt 22 7 2.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe000104a 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 4649460001020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 100c800c80000f 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 fe20c584943435 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 f50524f46494c4 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 500010100000c 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 484c696e6f0210 Ketinggian air dalam tabung kaca Volume kerucut terpancung berlubang = V 62d322e31000000000000000000000000000000000000000000000000000000 00006d6e74725 1 V = volume kerucut besar – volume bola = tinggi air dalam tabung + tinggi 247422058595a 2 kerucut kecil – volume

air dalam = t1 + r = 10 + 2,1 = 12,1 2007ce0002000 9000600310000 cm 616373704d534 6540000000049 d 18 4. t = 12 cm, r =22 = = 9 cm 4543207352474 2 2 s= 2000000000000 0000000000000 000f6d60001000 = 00000d32d4850 144 + 2020000000000 225 81==15 cm Luas 0000000000000 selimut kerucut = L 0000000000000 0000000000000 s Ls 0000000000000 = πrs =0000000000000 3,14 × 9 × 15 0000000000000 = 423,9 cm2 Luas kertas yang0000000116370 tersedia = L L = 60 7274000001500 × 80 0000033646573 = 4.800 cm2 6300000184000 0006c77747074 L 4.800 =000001f0000000 Ls 423,9 ≈ 14626b7074000 11,32 0020400000014 Jadi,7258595a00000 banyak hiasan yang dapat dibuat 11 buah. (banyak hiasan tidak boleh bilangan 2180000001467 pecahan dan dibulatkan ke bawah) 58595a0000022 5. c000000146258 ffd8ffe000104a46494600010201 595a000002400 00c800c80000ffe20c584943435f5 0000014646d6e 0524f46494c4500010100000c484 6400000254000 c696e6f021000006d6e747252474 00070646d6464 22058595a2007ce000200090006 000002c400000 00310000616373704d534654000 0887675656400 000004945432073524742000000 00034c0000008 0000000000000000000000f6d600 6766965770000 0100000000d32d4850202000000 03d4000000246 000000000000000000000000000 c756d69000003f Misalkan t1 = tinggi kerucut besar t2 = tinggi 000000000000000000000000000 8000000146d65 000000000000000000000000000 kerucut kecil (kerucut pancungan) tb = 61730000040c0 000000001163707274000001500 0000024746563 tinggi tabung Dengan kesebangunan 000003364657363000001840000 6800000430000 segitiga diperoleh: 006c77747074000001f000000014 0000c72545243 626b70740000020400000014725 0000043c00000 tr22= 8595a0000021800000014675859 tr t3 80c6754524300 11 5a0000022c000000146258595a0 2 00043c0000080 ⇔ = 000024000000014646d6e640000 tbc625452430000 + t 12 t12 025400000070646d6464000002c 2 ⇔ 12043c0000080c7 + 2t = 400000088767565640000034c00 4 4 4657874000000 00008676696577000003d400000 4t2 =00436f70797269 12 + t22 0246c756d69000003f8000000146 3t2 = =6768742028632 12 d6561730000040c000000247465 9203139393820 6368000004300000000c7254524 t2 = 4865776c65747 4 cm t1 = tb + t2 = 12 + 4 30000043c0000080c6754524300 42d5061636b61 = 16 cm 00043c0000080c6254524300000 726420436f6d70 43c0000080c7465787400000000 616e790000646 436f707972696768742028632920 5736300000000 31393938204865776c6574742d5 0000001273524 061636b61726420436f6d70616e7 7422049454336 900006465736300000000000000 313936362d322 127352474220494543363139363 e310000000000 62d322e31000000000000000000 0000000000001 000012735247422049454336313 2735247422049 936362d322e3100000000000000 4543363139363 000000000000000000000000000 62d322e310000 0000000000000 0000000000000

tabung 112 = πr1 t1 – πr22t2 – 33 11

πr22tb = × 3,14 × 122 × 16 – 3

2

3

× 3,14 ×

– 3,14 × 3 × 12 32 × 4 1 = × 3,14 × 144 × 16 – 3,14 × 3 3 ×=4 – 3,14 × 9 × 12 2.411,52 – 37,68 – 339,12 = 2.034,72 cm3 Jadi, volume kerucut terpancung berlubang tersebut 2.034,72 cm3. 6. Dari gambar tersebut disimpulkan: 1 diameter bola = 3 × rusuk =1 3 kubus × 60 = = Jari-jari bola = 1 2 20 cm × 20 4 a. volume 1 bola 3 = 10?r cm 3

= 43

* 3,14 * 103 Volume seluruh bola== 27 × 3=9× 43 12.560 * 314 *=10 113.040 cm3 b. Volume ruang kosong = volume kubus – volume seluruh bola = 603 – 113.040 = 216.000 – 113.040 = 102.960 cm3 Jadi, volume seluruh bola 113.040 cm3, sedangkan volume ruang kosong dalam kubus 102.960 cm3. 7. Perhatikan segitiga penampang potongan kerucut besar berikut.

C ffd8ffe000104a4649460001020100c800c80000ffe2 0c584943435f50524f46494c4500010100000c484c6 96e6f021000006d6e74725247422058595a2007ce0 0020009000600310000616373704d5346540000000 049454320735247420000000000000000000000000 000f6d6000100000000d32d4850202000000000000 000000000000000000000000000000000000000000 000000000000000000000000000000000000000001 163707274000001500000003364657363000001840 AB 1.000 000006c77747074000001f000000014626b7074000 00204000000147258595a000002180000001467585 CF, FD, dan AD adalah tinggi bangun95a0000022c000000146258595a000002400000001 bangun ruang hasil pemotongan kerucut 4646d6e640000025400000070646d6464000002c40 0000088767565640000034c0000008676696577000 besar. CF = FD = AD sehingga 003d4000000246c756d69000003f8000000146d656 27 1730000040c0000002474656368000004300000000 CF = FD = AD =3= 9 cm c725452430000043c0000080c675452430000043c0 000080c625452430000043c0000080c74657874000 00000436f707972696768742028632920313939382 04865776c6574742d5061636b61726420436f6d706 16e790000646573630000000000000012735247422 049454336313936362d322e3100000000000000000 0000012735247422049454336313936362d322e310 000000000000000000000000000000000000000000 00000000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0000000000000000000000000000000000000000000

95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Perhatikan pula bahwa FG merupakan jariVolume tabung 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 jari bangun ruang I, DE jari-jari bangun II, = V2 V2 = πr22t2 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 AB jari-jari bangun ruang III. Dari 22 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 = × 352 × 20 kesebangunan segitiga diperoleh: 3 0000000000000000000000000000000000000000000000 7= 22 × 5 × 35 × 20 = 77.000 cm Volume CF FG = air dalam tabung mula-mula = V3 V3 = V2 – CA AB CF

V1 = 77.000 – 18.018 = 58.982 cm3 Jadi,

⇔ =FG

3CF 21 1

FG ⇔=

3

21

⇔ FG= 7 cm CF FG = CD DE CF

⇔ =7

2CF DE 17

⇔=

2

DE

⇔ DE= 14 cm Volume bangun ruang I = VI, volume bangun ruang II = VII, dan volume bangun ruang III = VIII. 1 VI = π × FG2 × CF 3 1 22

× × 72 × 9 22 × 7×3= 462 cm3 Misalkan volume kerucut berjari-jari DE dan tinggi CD adalah VDE. = VDE – VI VII =

37=

=1

π × DE2 × CD – 462

3 1 22

× × 142 × 18 – 462 6 × 22 × 2 × 14 – 462 = 3.696 – 462 = 3.234 cm3 =

37=

VIII = volume kerucut besar – VDE1 = π × AB2 × AC – 3.696 3 1 22

× × 212 × 27 – 3.696 9 × 22 × 3 × 21 – 3.696 = 12.474 – 3.696 = 8.778 cm3 =

37=

Jadi, volume bangun-bangun ruang tersebut berturut-turut 8.778 cm3, 3.234 cm3, dan 462 cm3. 8. Volume kerucut = V2 1 1 V1 = πr1 t1 3 1 22

= 37

× × 212 × 39

= 22 × 21 × 39 = 18.018 cm

3

volume air dalam tabung mula-mula ffd8ffe000104a464 9460001020100c800 c80000ffe20c584943 435f50524f46494c45 00010100000c484c6 96e6f021000006d6e 58.982 cm3. 74725247422058595 9. a2007ce0002000900 1 cm 06003100006163737 04d53465400000000 49454320735247420 40 cm 00000000000000000 0000000000f6d6000 100000000d32d4850 20200000000000000 00000000000000000 Misalkan luas permukaan bambu yang 00000000000000000 tertutup cairan pewarna = L. Jari-jari 00000000000000000 tabung luar = r = 21 cm 00000000000000000 1 00000000000001163 Tebal bambu = 1 cm. Jari-jari tabung 12.560 3 70727400000150000 dalam = r2 = r1 – 1 = 320=cm cmL = luas 00033646573630000 12.560 tabung selimut tabung luar + luas selimut 01840000006c77747 dalam + 2 × luas cincin lingkaran 074000001f0000000 2 = 2πr1t + 2πr2t + 2(πr1 – πr22) = 2πt(r1 + 14626b70740000020 4000000147258595a r2) + 2π(r12 – r22) =2 × 3,14 × 40 × (21 00000218000000146 2 + 20) + 2 × 3,14 × (21 – 202) = 6,28 × 758595a0000022c00 40 × 41 + 2(3,14 × 41) 0000146258595a000 = 10.299,2 + 257,48 0024000000014646d =6e640000025400000 10.556,68 cm2 Jadi, luas permukaan 070646d6464000002 bambu yang tertutup cairan pewarna c4000000887675656 adalah 10.556,68 cm2. 40000034c00000086 10. 76696577000003d40 Misalkan luas potongan tabung = L 00000246c756d6900 1 L 0003f8000000146d6 = 2(luas4 lingkaran) + 2 × luas persegi panjang 561730000040c0000 1 + × luas selimut 00247465636800000 4 4300000000c725452 tabung 1 1 430000043c0000080 = 2( × πr2) + 2 × r × t + × 4411 c675452430000043c 2πrt 0000080c625452430 = × 3,14 × 102 + 2 × 10 × 20 + × 3,14 2 2 000043c0000080c74 2 × 10 × 20 =65787400000000436 157 + 400 + 314 = 871 cm Jadi, luas f7079726967687420 permukaan potongan tabung tersebut 871 2 28632920313939382 cm . 04865776c6574742d 5061636b617264204 36f6d70616e790000 64657363000000000 00000127352474220 49454336313936362 d322e310000000000 00000000000012735 24742204945433631 3936362d322e31000 00000000000000000 00000000000000000 00000000000000000 21 cm

Latihan Ulangan Tengah Semester 1

A. Pilihan Ganda 1. Jawaban: b Syarat dua bangun datar sebangun adalah besar sudut-sudutnya sama dan sisi-sisi yang bersesuaian sebanding. Bangun datar yang sebangun dengan persegi panjang di atas ada pada pilihan b, karena besar keempat sudutnya sama 90° dan perbandingan sisi-sisi yang bersesuaian 1512 3

sama yaitu108 == . 2

2. Jawaban: c Kedua trapesium kongruen maka besar sudutsudut yang bersesuaian sama, yaitu: ∠A = ∠G, ∠B = ∠F, ∠C = ∠E, dan ∠D = ∠H ∠A + ∠B = 180° ⇒∠A + 84° = 180° ⇔∠A = 180° – 84° = 96° Jadi, besar ∠G = ∠A = 96°.

7. Jawaban: c Skala 1 : 500 Panjang pada gambar = 6,8 cm Panjang sebenarnya = 6,8 × 500 = 3.400 cm = 34 m Lebar pada gambar = 3,7 cm Lebar sebenarnya = 3,7 × 500 = 1.850 cm = 18,5 m Luas kapling sebenarnya L = 34 × 18,5 = 629 m2 8. Jawaban: a Foto A (4 cm × 8 cm) diperbesar menjadi foto B (6 cm × p cm). Oleh karena foto A dan foto B sebangun maka perbandingan sisi-sisi yang ⎛p A ⎛ bersesuaian sama. ⎛ A = A A Sehingga: p   BB  pAA 8 68 = A4⇒ pA BB p6 B

1Jawaban: c ABCD ≅ PQRS AB bersesuaian

=

4

⇔ pB =

×

= 12

Jadi, ukuran foto yang baru (foto B) adalah dengan QR, berari AB = QR. BC bersesuaian 6 cm × 12 cm. dengan PQ, berari BC = PQ. Sudut yang diapit dua sisi yang bersesuaian besarnya sama yaitu ∠ABC = 9. Jawaban: d ABC ≅ DEF maka sudut∠RQP. Jadi, pernyataan a, b, dan d benar. sudut yang bersesuaian sama besar. Oleh 2Jawaban: d karena ∠F bersesuaian dengan ∠A, maka besar ∠F = ∠A = 46°.

ABCD sebangun dengan

10.

EFGH. EFEH =⇒ = EF5 ABAD 93

Jawaban: b × 95

⇔ EF = = 15 cm Jadi, panjang EF = 15 cm. 3

1Jawaban: c Dua bangun datar dikatakan kongruen jika bentuk dan ukurannya sama. Bangun yang kongruen dengan bangun di atas ada pada pilihan c. 2Jawaban: b Panjang kebun pada denah = 12 cm Lebar kebun pada denah = 8,4 cm Panjang kebun sebenarnya = 20 m = 2.000 cm Kebun dan denahnya sebangun, maka:

Lebar kebun sebenarnya=Lebar kebun sebenarnya Lebar kebun pada denah Lebar kebun pada Lebar kebun sebenarnya 2.000 ⇒ denah= 8,

⇔ =3

4

12

8,4 ×

Lebar kebun sebenarnya 500

= 1.400 cm = 14 m Jadi, lebar kebun sebenarnya 14,0 m.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 12. Jawaban: b Dua segitiga kongruen 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 mempunyai besar sudutsudut yang sama dan 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 panjang sisi-sisinya juga sama. Sisi yang 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 panjangnya a cm berada di depan sudut 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 bertanda o dan panjangnya 8 cm. Jadi, nilai a = 0000000000000000000000000000000000000000000000 8. 13. ∆PQR dan ∆STU Jawaban: d ST TU ST2 sebangun = ⇒= PQQR 96 92 18

⇔ ST × = = = 3 cm 66

14. Jawaban: d Besar setiap sudut yang bersesuaian pada kedua segitiga sama sehingga kedua segitiga tersebut sebangun. y 12

=

15 ⇔ 15 ×12 y= 99

= 20 cm

15. Jawaban: d Panjang AC = DE maka ∠B = ∠F. Panjang AB = DF maka ∠C = ∠E. Panjang BC = EF maka ∠A = ∠D. Jadi, pernyataan yang benar adalah d. 16. Jawaban: b Sisi-sisi ∆ABC dan ∆DEF sama panjang, maka ∆ABC dan ∆DEF kongruen. Dua segitiga yang kongruen mempunyai sudutsudut yang sama besar. Jadi, sudut-sudut yang bersesuaian pada ∆ABC dan ∆DEF sama besar. A

17. ∆ABC sebangun dengan ∆ADE, Jawaban: c maka: ACBC= AC20 =⇒ AEDE

1512

15 = ×20 ⇔ AC= 25 cm CE = AC – AE = 25 – 15 = 10 cm Jadi, panjang CE = 10 cm. 12

18. Jawaban: a Segitiga ABC sebangun dengan segitiga DEC, sehingga: AB = BC DEEC AB 12

⇔= ⇔=

43 AB 4 4

AB · 1 = 4 · 4 A

1 1

AB = 16 m Jadi, lebar sungai 16 m.

sis R P ffd8ffe000104a4649460001020100 i c800c80000ffe20c584943435f50524f 46494c4500010100000c484c696e6f 021000006d6e74725247422058595 a2007ce0002000900060031000061 sisiC Q B 6373704d534654000000004945432 07352474200000000000000000000 Jadi, berdasarkan teorema SAS yaitu sisi: 00000000f6d6000100000000d32d48 AC = PQ, sudut: ∠C = ∠P, dan sisi: BC = 50202000000000000000000000000 PR, maka ∆ABC dan ∆PQR kongruen. 00000000000000000000000000000 11. 00000000000000000000000000000 Jawaban: b Pada belah ketupat ABCD dengan perpotongan diagonal di titik O tedapat 00000000000001163707274000001 4 segitiga kecil yang saling kongruen, yaitu 50000000336465736300000184000 ∆AOD ≅ ∆AOB ≅ ∆BOC ≅ ∆COD. Selain itu juga 0006c77747074000001f0000000146 ada dua pasang segitiga yang kongruen yaitu 26b70740000020400000014725859 ∆ABC5a00000218000000146758595a000 ≅ ∆ACD dan ∆ABD ≅ BCD. Adapun ∆ABD dan ∆ACD tidak kongruen karena tidak semua 0022c000000146258595a00000240 sisi yang 00000014646d6e640000025400000 bersesuaian sama panjang, yaitu 070646d6464000002c40000008876 panjang AC ≠ BD. 7565640000034c000000867669657 7000003d4000000246c756d690000 03f8000000146d6561730000040c00 00002474656368000004300000000 c725452430000043c0000080c67545 2430000043c0000080c62545243000 0043c0000080c74657874000000004 36f707972696768742028632920313 93938204865776c6574742d506163 6b61726420436f6d70616e79000064 65736300000000000000127352474 22049454336313936362d322e3100 00000000000000000000127352474 22049454336313936362d322e3100 00000000000000000000000000000 00000000000000000000000sisi sisi

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe000104a 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 4649460001020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 100c800c80000f 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 fe20c584943435 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 f50524f46494c4 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 500010100000c 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 484c696e6f0210 19. Jawaban: c PT : TS = 3 : 5 berarti PT = Volume tabung: 62d322e31000000000000000000000000000000000000000000000000000000 2 00006d6e74725 V= πr t = 3,14 × 2 247422058595a 10 × 12 = 3.768 cm32007ce0002000 9000600310000 23. Jawaban: 616373704d534 a Tinggi6540000000049 kerucut: ffd8ffe000104a464946 t= 4543207352474 0001020100c800c80000 2000000000000 ffe20c584943435f50524 f46494c4500010100000 = 0000000000000 169 − 25 c484c696e6f021000006 13 cm 000f6d60001000 d6e74725247422058595 00000d32d4850 a2007ce0002000900060 = 144 = 12 cm 2020000000000 0310000616373704d53 Volume kerucut: 4654000000004945432 0000000000000 0735247420000000000 2 1 0000000000000 V= πr t 000000000000000000f6 3 0000000000000 d6000100000000d32d48 5020200000000000000 0000000000000 1 = 0000000000000 × 3,14 × 5 × 5 × 12 0000000000000000000 3 0000000000000000000 3 = 0000000000000 314 0000000000000000000 0000000116370 0000000000000000000 cm 24. Jawaban: 7274000001500 0000011637072740000 d Kerucut 0000033646573 r = 12 cm dan 0150000000336465736 V = 720π 3000001840000006c777 6300000184000 3 47074000001f00000001 2 cm 11 t ⇒ 720π = × π × 122 × t V = 0006c77747074 πr 4626b707400000204000 3 3 3 000147258595a0000021 000001f0000000 720π = 48πt 8000000146758595a000 7 720π 14626b7074000 t= 0022c000000146258595 0020400000014 a0000024000000014646 48π 7258595a00000 d6e64000002540000007 ⇔ t= 15 Jadi, tinggi 0646d6464000002c4000 2180000001467 kerucut 15 cm. 58595a0000022 0008876756564000003 4c000000867669657700 c000000146258 0003d4000000246c756d 25. Jawaban: b 12 595a000002400 69000003f8000000146d 0000014646d6e Volume bola = 3718 cm36561730000040c000000 2 6400000254000 2474656368000004300 1 4 2.156 000000c7254524300000 ⇔ ×00070646d6464 πr3 = 43c0000080c675452430 23 3 2 22 2.156 000002c400000 000043c0000080c62545 3 2430000043c0000080c7 0887675656400 ⇔× r = 465787400000000436f7 3 73 00034c0000008 07972696768742028632 2.156 3 7 7 9203139393820486577 r3= = 6766965770000 ×× 6c6574742d5061636b61 3 03d4000000246 r = =c756d69000003f 343 726420436f6d70616e79 r=7 cm Jadi, jari-jari bola 7 8000000146d65 0000646573630000000 0000000127352474220 cm. 61730000040c0 49454336313936362d32 0000024746563 2e310000000000000000 6800000430000 0000001273524742204 9454336313936362d322 0000c72545243 e3100000000000000000 0000043c00000 0000000000000000000 80c6754524300 000000000000000000 00043c0000080c 6254524300000 43c0000080c746 5787400000000 436f7079726967 6874202863292 0313939382048 65776c6574742 d5061636b6172 6420436f6d7061 6e79000064657 3630000000000 0000127352474 2204945433631 3936362d322e3 1000000000000 0000000000127 3524742204945 4336313936362 d322e31000000 0000000000000

3x dan TS = 5x untuk suatu nilai x.

S 24 cm R ffd8ffe000104a4649460001020 100c800c80000ffe20c584943435 f50524f46494c4500010100000c4 TU 3x 84c696e6f021000006d6e747252 47422058595a2007ce000200090 00600310000616373704d53465 PW Q 40000000049454320735247420 00000000000000000000000000 Pada gambar di atas SW sejajar RQ, 0f6d6000100000000d32d485020 20000000000000000000000000 sehingga: WQ = VU = SR = 24 cm PW = PQ00000000000000000000000000 – WQ = 40 – 24 = 16 cm Segitiga STV 00000000000000000000000000 sebangun dengan segitiga SPW. 00000000000000000011637072 74000001500000003364657363 TVST= ⇒TV =5x 000001840000006c77747074000 PWSP 165x + 3x TV 5 001f000000014626b7074000002 04000000147258595a00000218 ⇔= 000000146758595a0000022c000 16 8 000146258595a0000024000000 16 ×5 014646d6e64000002540000007 ⇔ TV= = 10 cm 8 TU = TV + VU = 10 + 24 = 0646d6464000002c40000008876 347565640000034c0000008676696 cm Jadi, panjang TU = 34 577000003d4000000246c756d69 cm. 000003f8000000146d656173000 0040c0000002474656368000004 20. 300000000c725452430000043c0 000080c675452430000043c0000 ffd8ffe000104a4649460 080c625452430000043c0000080 001020100c800c80000ffe Jawaban: cC c7465787400000000436f707972 20c584943435f50524f464 69676874202863292031393938 94c4500010100000c484c 204865776c6574742d5061636b6 696e6f021000006d6e747 1726420436f6d70616e79000064 25247422058595a2007ce 65736300000000000000127352 A000200090006003100006 E B 47422049454336313936362d32 16373704d534654000000 2e310000000000000000000000 Panjang bayangan pohon: AB = 12 m 004945432073524742000 12735247422049454336313936 Panjang bayangan Budi: EB = 2 m Tinggi 000000000000000000000 362d322e310000000000000000 0000f6d6000100000000d Budi: DE = 150 cm = 1,5 m Segitiga ABC 00000000000000000000000000 32d485020200000000000 0000000000005x sebangun dengan segitiga EBD, maka 000000000000000000000 berlaku: 000000000000000000000 000000000000000000000 CA AB = ⇔CA12= 000000000000000000000 DEEB 1,5 2 12 ×1,5 116370727400000150000 000336465736300000184 ⇔ CA= = 9 m 0000006c7774707400000 2 Jadi, tinggi pohon 9 m. 1f000000014626b707400 000204000000147258595 21. a00000218000000146758 Jawaban: 595a0000022c000000146 d 4 Volume 258595a00000240000000 bola = πr3 14646d6e6400000254000 3 422 212121 00070646d6464000002c4 000000887675656400000 = ×××× 37222 = 11 × 21 × 21 34c000000867669657700 0003d4000000246c756d6 = 4.851 cm3 9000003f8000000146d65 61730000040c000000247 Jadi, volume bola adalah 4.851 cm3. 465636800000430000000 22. 0c725452430000043c000 0080c675452430000043c Tabung: t = 12 0000080c6254524300000 43c0000080c7465787400 cm 11 b Jawaban: r = d = × 20 = 10 cm 000000436f70797269676 874202863292031393938 2 2 204865776c6574742d506 1636b61726420436f6d70 616e79000064657363000 000000000001273524742 2049454336313936362d3 22e310000000000000000 000000127352474220494 54336313936362d322e31 000000000000000000000 000000000000000000000 000000000000

D

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104 ffd8ffe00 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 a464946000102 0104a4649 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 0100c800c8000 460001020 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 0ffe20c5849434 100c800c8 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 35f50524f4649 0000ffe20c 0000000000000000000000000000000000000000000

Lselimu = πrs = 3,14 × 3 × t 12,36 = 2 116,4312 cm(L Luas bahan + =L4selimut × = alas 4) × (28,26 + 116,4312) = 578,7648 cm2 31. d : t = 4 : 3 berarti d = 4x dan t = 3x Jawaban: d untuk suatu nilai x. Luas permukaan tabung: L = 2πr2 + 2πrt = 720π

32.

r2 + rt = 360 (2x)2 + (2x)(3x) = 360 4x2 + 6x2 = = 360 2 10x = = 360 x2 = 36 x= 6 Diameter tabung: d = 4x = 4 × 6 = 24 cm

Volume = volume tabung + Jawaban: volumeckerucut 1 2 = πr t1 + 3 22 122

πr2t2 2 = × 7 × 10 + × × 72 × 6 737= 1.540 + 308 = 1.848 cm3

33. r : t = 3 : 4 berarti r = 3n dan t = 4n, maka: rb2 + t2 = s22 Jawaban: (3n)2 + (4n)2 = (15)22 9n2 + 16n2 = = 225 25n2 = = 225 n2 = =9 n= 3 Jari-jari kerucut: r = 3n = 3 × 3 = 9 cm Tinggi kerucut: t = 4n = 4 × 3 = 12 cm Volume 3222 kerucut:

26. Jawaban: d Tabung dengan diameter d dan tinggi t. Luas permukaan tabung: L = 2 × luas alas + luas selimut 1

= 2 × π( d)2 + 21 πdt = πd2 + πdt 2

=

2 1

2t)

πd(d +

4c45000101000 584943435 00c484c696e6f f50524f464 27. Jawaban: c Bola 94c450001 dengan r021000006d6e7 = 3,5 cm Luas 4725247422058 0100000c4 permukaan bola: 595a2007ce000 84c696e6f 2 2000900060031 021000006 L= 4πr 0000616373704 d6e747252 22= 4 × d534654000000 474220585 7 0049454320735 95a2007ce × 23,52 = 154 cm 2474200000000 000200090 0000000000000 006003100 28. Jawaban: d Panjang busur pada juring 006163737 lingkaran0000000f6d600 = keliling alas kerucut 0100000000d32 04d534654 d485020200000 000000004 α × 2πrjuring = o 0000000000000 360 o 945432073 0000000000000 524742000 2πr kerucut 270 0000000000000 000000000 o ⇔ 360 × 20 = 0000000000000 000000000 r⇔ = 15 cm r kerucut keruc 0000000000000 0000000f6 ut 0000000000000 d60001000 29. Jawaban: a 0000000000001 00000d32d Tabung tanpa tutup: 485020200 t = 1,2 m1637072740000 000000000 10150000000336 r= d 4657363000001 000000000 2840000006c777 000000000 000000000 = 47074000001f0 × 70 2=00000014626b7 35 1 000000000 cm0740000020400 = 000000000 0,35 m 0000147258595 000000000 Luasa000002180000 000000000 alas: L= a 00146758595a0 000000000 22 000000000 πr2 000022c000000 = 146258595a000 × 000116370 7 2 0,35 0024000000014 727400000 = 0,385 m 150000000 Luas646d6e6400000 2540000007064 336465736 selimut: L= 300000184 s 22 6d6464000002c 2πrt 4000000887675 0000006c7 = 2 × × 0,35 × 1,2 65640000034c0 774707400 7 0000086766965 0001f0000 = 2,64 m2 00014626b Luas77000003d4000 tangki 000246c756d69 707400000 air: L= a L + Ls 000003f800000 204000000 = 0146d65617300 0,385 + 147258595 2,64 = 00040c0000002 a00000218 4746563680000 000000146 30. 3,025 m2 04300000000c7 758595a00 00022c000 r= =2545243000004 cm 23 22d Jawaban: d 3c0000080c675 000146258 s= 452430000043c 595a00000 2 2 0000080c62545 240000000 = 12 +3 2430000043c00 14646d6e6 = 00080c7465787 153 = 400000254 000000706 3 400000000436f 46d646400 = 7079726967687 12,36 4202863292031 0002c4000 Lalas cm = πr2 3939382048657 =000887675 3,14 × 2 76c6574742d50 3656400000 = 28,26 61636b6172642 34c000000 2 cm 0436f6d70616e 867669657 7900006465736 7000003d4 3000000000000 000000246 0012735247422 c756d6900 0494543363139 0003f8000 36362d322e310 000146d65 0000000000000 617300000 0000000012735 40c000000 2474220494543 247465636 36313936362d3 800000430 22e3100000000 0000000c7

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0000000000000000000000000000000000000000000

95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 35. Papan dan lukisan sebangun, sehingga diperoleh: 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 39. Jawaban: b Luas permukaan bandul ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f02100000 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 terdiri atas selimut kerucut dan permukaan 6d6e74725247422058595a2007ce00020009000600310000616373704d53465400000000494543207352474200000000 3 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 00000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000 setenga bola. Luas permukaan bandul: Jawaban: a cangkir = volume 4 volume 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 mangkuk 00000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c 31 2 4 0000000000000000000000000000000000000000000000 3 77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000014 ⇒× πr t = × πr2 L= L + L 42 1 13 1 selimut kerucut permukaan 6258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086 2 bola 76696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c7 23 2 3 πr 25452430000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f707972 1 ⇔ t = πr 3 2 1 1 = πrs + × 4πr2 4 69676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000 2= πr(s + 2r) = 3,14 3 000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e 8r × 5(20 + 2 × 5) = 2 = 2 31000000000000000000000000000000000000000000000000000000 p1 A1= pA 37,5 = 45 30 37,5 × 30⇔ A1= 9r1 A1⇔ 15,7 × 30 = 471 3 2 × L= L – L 87 cmyang tidak tertutupi lukisan: = 25 Luas papan papan lukisan = 45 × 30 = – 37,52 × 25 = 1.350 – 937,5 = 412,5 cm2 3. 45 × 95 a. Perhatikan ∆BHF dan ∆EGD.bandul 1) ∠ EGD =2.∠FHG karena=sehadap. ∠ FHG = ∠BFH karena merupakan Jadi, luas permukaan 471 cm 12,2 cm Jadi, kedalaman atau tinggi pasangan sudut dalam berseberangan. Berarti ∠EGD = ∠BFH. 2) ∠DEG = ∠GCH karena merupakan pasangan cangkir 12,2 40. sudut dalam berseberangan. ∠GCH = ∠BHF karena sehadap. Berarti ∠DEG = cm. ∠BHF. 3) BHDE merupakan 36. Jawaban: c sudut) maka ∆BHF dan ∆EGD jajargenjang, maka = BH. Oleh (sudut, sisi, LuasED setengah bola:karena 1 2277 memenuhi syarat kongruen Luas adalah: setengah bola: kongruen. b. Pasangan kongruen selain2∆BHF dan ∆EGD ∆ABF ≅∆CDG ∆ABH ≅∆CDE ∆AFE ≅∆CGH ∆ABE Jawaban: d segitiga L1Perhatikan = × 4πr2∆ADF =2× ×× = 771)cm ≅∆CDH 4. a. dan ∆CEF. ∠AFD = ∠CFE karena mertolak belakang. 2) ∠ADF = ∠FCE karena 2722Luas lingkaran: 1 2 L= ×24πr keduanya sikusiku. 3) ∠DAF = ∠CEF karena merupakan pasangan sudut dalam berseberangan. Oleh karena besar ketiga sudutnya sama,2 maka dan ∆CEF sebangun. b. ∆ADF dan sebangun, maka: CECF CE15 − 12 = ⇒ = 227 7∆ADF 77 2 ∆CEF 2 L2CE = πr == × 3,75 ×== 38,5 cm =2× ×Jadi, 7022panjang BE = 18,75 cm. ADDF 1512 15 ×3⇔ = cm BE = BC + CE = 15 + 3,75 = 18,75 cm 2 7222Luas 12permukaan setengah bola 7= 30.800 dm Jadi, luas permukaan atap pejal = luas setengah bola + luas tersebut 30.800 dm2. lingkaran = 77 + 38,5 = 115,5 cm2 Jadi, luas permukaan besi cor 115,5 37. Jawaban: b Bola: r1 = 10 cm cm2. dan biaya Rp10.000,00 L1= 4πr12 = 4 B. 1. a. Pada trapesium siku-siku besar dua Uraian sudutnya siku-siku. Sudut yang lain yaitu sudut lancip yang besarnya 64° dan sudut tumpul yang besarnya 180° – 64° = 116°. 2p° = 116° ⇔ p = 58 (3q – 8)° = 64° ⇒ 3q = 64 + 8 ⇔

72

q=

3

=

× π × 102 = 400π cm2 Tempat penyimpanan berbentuk bola L2= 4πr22 = 4 ×

π × 1002 = 40.000π cm2 Biaya pengecatan tempat penyimpanan berbentuk bola 40.000π 400π

× Rp10.000,00 = Rp1.000.000,00

Jadi, p = 58 dan 24 q = 24. 38. b. 1 Pada bangun-bangun yang sebangun 2 11 V= πr t sisi-sisi yang bersesuaian panjang Jari-jari:c r = d = × 14 3 Jawaban: 2 2 sebanding. r8 × 68 48 = ⇔ r= = = 4,8 cm 1 = 7 cmtabung: t = 32 – 7 = 25 cm Tinggi 10 10s8 102×880 1 = 610 3× 3,14 × 9 × Permukaan bangun ruang terdiri atas = ⇔ s = = = 13 cm 12 = 106 113,04 6631 3 lingkaran, selimut tabung, dan setengah cm permukaan bola. Luas permukaan bangun . Jadi, r = 4,8 dan s = 13 34. 3 ruang: L= Llingkaran + Lselimut tabung + 1 2 2. Panjang43 lukisan p1 = panjang papan – 2 × 3,75 Jawaban: r2 Lbola 3 13d 111 Vπr 1r33 r r1 = 45 – 7,5 = 37,5 cm Lebar lukisan 1 = = 433 = = = 33 = 2r= V2πr r(2r)8r8 = πr2 +2 2πrt + × 4πr2 2211 1 A Panjang papan = p Lebar 1 3 2= 3πr + Jadi, perbandingan volumenya 1 : 8. papan =A 2πrt = πr(3r + 2t) 22 = × 7 × (3 × 7 + 2 × 7

25) = 22 × 71 = 1.562 cm2

8. Permukaan potongan kayu terdiri atas persegi panjang, setengah lingkaran, dan setengah selimut tabung. Luas persegi panjang: L1 = 14 × 20 = 280 cm2 Luas

5. Permasalahan tersebut dapat digambarkan sebagai berikut. E

6.

ffd8ffe000104a4649460001020100 c800c80000ffe20c584943435f50524 Af46494c4500010100000c484c696e6 f021000006d6e7472524742205859 5a2007ce000200090006003100006 16373704d53465400000000494543 20735247420000000000000000000 000000000f6d6000100000000d32d 48502020000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000011637072740000 01500000003364657363000001840 000006c77747074000001f0000000 14626b70740000020400000014725 8595a00000218000000146758595a Perhatikan bahwa ∆ABE dan ∆CDB 0000022c000000146258595a00000 24000000014646d6e640000025400 sebangun. AB AE = 000070646d6464000002c40000008 CDBC 8767565640000034c000000867669 6577000003d4000000246c756d690 3,2 1,6 ⇔ 00003f8000000146d656173000004 = 9,6 BC 0c000000247465636800000430000 0000c725452430000043c0000080c 9,6 × ⇔ 675452430000043c0000080c62545 BC1,6 = 3,22430000043c0000080c7465787400 = 4,8 m Ketinggian air kolam = 000000436f7079726967687420286 kedalaman kolam – BC = 10 – 4,8 = 3292031393938204865776c657474 m 2d5061636b61726420436f6d70616 Jadi, ketinggian air kolam 5,2 m. e7900006465736300000000000000 12735247422049454336313936362 d322e310000000000000000000000 12735247422049454336313936362 d322e310000000000000000000000 Selisih volume tabung dan kerucut 00000000000000000000000000000 000D = volume tabung – volume

setengah lingkaran: 1

L2= × πr2 2 1

510 1

kerucut = πr2t – 31πr2t =

2 3

πr2t

2 22 21 21

= × × × × 15 3 3722= 11 × 21 × 15 = 3.465 cm Jadi, volume ruang di antara tabung dan kerucut 3 3.465 cm . 7.

Tabung: V =

6.280 cm3 11 a.

b.

r= d = × 20 = 10 cm

22

V = πr2t ⇒ 6.280 = 3,14 × 102 × t 314t = 6.280 3 t= = 20 cm6.280 314Jadi,

tinggi tabung 20 cm. Luas permukaan tabung: L= 2πr2 + 2πrt = 2πr(r + t) = 2 × 3,14 × 10 × (10 + 20) = 62,8 × 30 = 1.884 cm2

22

= × × 72 = 77 cm2 setengah selimut ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f 9. = tabung: 50524f46494c4500010100000c484c696e6f021000006d6e7472524 1 L3= 2× πdt 7422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d 1 22 6000100000000d32d485020200000000000000000000000000000 = × × 14 × 20 = 440 cm2 27Luas permukaan 00000000000000000000000000000000000000000000000000000 potongan kayu: L= L1 + 2 × 00000000000001163707274000001500000003364657363000001 L2 + L3 840000006c77747074000001f000000014626b7074000002040000 = 280 + 2 × 77 + 440 = 874 00147258595a00000218000000146758595a0000022c0000001462 cm2 Jadi, luas daerah yang dicat 58595a0000024000000014646d6e640000025400000070646d6464 874 cm2. 000002c400000088767565640000034c0000008676696577000003 d4000000246c756d69000003f8000000146d6561730000040c0000 + x12 x 002474656368000004300000000c725452430000043c0000080c67 5452430000043c0000080c625452430000043c0000080c74657874 10x = 5x + 60 5 5 5,2 00000000436f70797269676874202863292031393938204865776c 6574742d5061636b61726420436f6d70616e790000646573630000 5x = 60 1 12 000000000012735247422049454336313936362d322e310000000 x= 12 00000000000000012735247422049454336313936362d322e3100 0000000000000000000000000000000000000000000000000000x 27Luas

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c450 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 00310000616373704d53465400000000494543207352474200000000000000000000 penuh: 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 Debit = 00000000f6d6000100000000d32d4850202000000000000000000000000000000000 Vt 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 Waktu yang diperlukan untuk mengisi 00000000000000000000000000000000000000000000000000000000000000116370 ? t= drum sampai PJawaban:cd 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 7. Jawaban: V debit Banyak sampel 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 Jumlah balita tahun 7274000001500000003364657363000001840000006c77747074000001f000000014 penelitian =500 75 × 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 2010 =5× =banyak 2.500 626b707400000204000000147258595a00000218000000146758595a0000022c0000 desa = balita 75 × 13 = 975 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 Jumlah tahun 297 orang 0000000000000000000000000000000000000000000000 A. Pilihan 00146258595a0000024000000014646d6e640000025400000070646d6464000002c4 2009 1 = 3 × 500 + × 500 Ganda ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 2 = 1.500 + 250d=Sampel 1.750 Persentase 00000088767565640000034c0000008676696577000003d4000000246c756d690000 1. Jawaban: Berat Badan (kg) 8. Jawaban: penelitiannya 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 Frekuensi kenaikan banyak balita tahun 2009 c adalah 55 siswa kelas IX dari setiap SMP di 03f8000000146d6561730000040c0000002474656368000004300000000c72545243 35 = 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 Kabupaten Rahayumulya. 6 Jadi, waktu yang diperoleh untuk mengisi

0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465 2.500 −1.750 000000000000000000000000000000000000000000000000000000000000000000000000000000000 100% drum sampai penuh3633 menit. 9. = × Jawaban: c Populasi penelitiannya 6 1.750 750 6 787400000000436f70797269676874202863292031393938204865776c6574742d50 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 adalah siswa kelas IX di Kabupaten 37 9 61636b61726420436f6d70616e790000646573630000000000000012735247422049 = × 100% = 42 % 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 2 Rahayumulya. 1.7507 38 Bab III Statistika dan 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 454336313936362d322e310000000000000000000000127352474220494543363139 10. Jawaban: 3 8. Jawaban:caBanyak Kenaikan jumlah 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 Peluang 39 siswa yang wanita diobservasi 23 × 2010 = 17.500 – penduduk pada=tahun 36362d322e31000000000000000000000000000000000000000000000000000000 = 33 menit 5 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 55 + 28 = × 500. 55 = Kenaikan 2.805 orang 17.000 jumlah penduduk laki40 laki pada tahun 2010 = 16.000 – 15.000 = 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 4 Banyak siswa yang 41 memiliki berat badan 1.000. Selisih kenaikan jumlah penduduk laki4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 kurang dari 38 adalah 4 6 + 6 + 2 = 14. laki dan wanita = 1.000 – 500 = 500. B. 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 42 A. Pilihan 2. Jawaban: d Jangkauan = nilai datum 6 00000000000000000000000000000000000000000000000 Ganda terbesar – nilai datum terkecil 1. Data kualitatif adalah data yang Uraian 1Jawaban: d Mengerjakan 20 soal bukan kegiatan 9. Jawaban: a Kenaikan jumlah

pengamatan suatu–objek. mengumpulkan data karena objek data tidak dapatdiperoleh pendudukdari pada tahun 2011 sifat = 34.000 = 42 – 35 = 7 Hasil kali jangkauan dan Data kuantitatif adalah data yang diolah. Mengukur tinggi badan Andi dan 33.500 = 500 Persentase kenaikandiperoleh jumlah nilai datum terbesar = 7 × 42 = 294 dari hasil pengukuran menimbang berat sebuah semangka bukan penduduk pada tahun atau 2011perhitungan. Data kualitatif kegiatan mengumpulkan data karena objek data Data kualitatif tunggal. Menghitung luas permukaan buku di 3. Tahun perpustakaan merupakan kegiatan mengumpulkan Data kuantitatif 33.500 Penurunan Hasil Padi Data kuantitatif data karena data Jawaban: b objek data tidak tunggal dan2005–2006 10. 10 Bangun Jawaban: d Jumlah produksi gula dari tersebut terdiri atasdi2seluruh dapat diolah. 375 – 350 = 25 ton 1Populasinya adalah lingkungan daerah tahun 2008 yaitu: sampai dengan 2010 =(atas) 350 + 500 bagian, 1) Kerucut kecil 2007–2008 2Jawaban: c Rata-rata tinggi badan sekelompok di Sentono yang terkena wabah penyakit flu 450 = 1.300 ton. 425 – 350 = 75+ton anak dapat diketahui dengan cara mengukur lingkungan di beberapa rkk = 5 cm,adalah skk = 12 2009–2010 burung.⇒Sampelnya tinggi berat badan sekelompok anak450 tersebut, tempat di daerah Sentono yang terkena wabah – 400 = 5011. ton 2) cm, Kerucut Penurunan terbesar rata-ratanya. hasil padi sebanyak 75 kemudian menghitung penyakit flu burungbesar yang diperiksa petugas ⇒ r = 10 cm, skb = 24 cm Luas bahan ton terjadi pada tahun 2007–2008. kb a secara acak. puskesmas 3Jawaban: c Populasinya adalah pelanggan salon. 500 Jawaban: = * 100% ? 1,49% kap lampu seluruh = luas selimut kerucut kecelakaan lalubesar lintas.– 4 adalah pelanggan 1Jawaban: ca Sampelnya Banyak siswa wanita selama 5salon tahun2Populasinya Sampelnya beberapa kecelakaan lalu lintas yang luas selimut kerucut yang mengisi kuesioner. = 50 + 100 + 200 + 200 + 200 = 750 orang diteliti penyebabnya, yang diambil dari beberapa kecil = – 5Jawaban: cd Waktu Populasi adalah keseluruhan objek 2 yang kurang dari 74 detik tempat secara πrkbskb πrkkskkacak. = yang diteliti. ini, objek yang Tahun yaitu 73, 72, Dalam dan 71.hal Jumlah murid = 2 diteliti +4 + 5 + 83Populasinya adalah Jumlah Kenaikan π(rkbskb – rkk skk) gelas hasil produksi pabrik adalah kota + 12 + warga 5+6+ 3 =A yang mengalami kecelakaan Persentaseyang dikemas dalam 100 kotak. Banyak populasi = di jalan X. 2007 2009 2010 = – 5 gelas · 12) = 3,14(240 – 60) = 6 ×2011 63,14(10 × 100 =· 24 3.600 – 10.000 = 2.000 12.000 – 8.000 = 4.000218.000 – 12.000 = 6.000 18.000 – 16.000 = 2.0 6Jawaban: b Sampel adalah bagian dari 12.000 populasi 3,14(180) 565,2 cm Jadi, luas bahan 2.000 10.00012. * 100% =Jawaban: 20% 4.000= 8.000 * 100% = 50% 6.000 12.000 * 100% = 50% 2.000 16.000 * 10 d Persentase yang yang akan dicek. Dalam hal ini, jumlah tas yang yang digunakan untuk membuat kap berpendidikan SMP = 100% 3.600– (45% + 8% dicek tas. murid yang mencapai garis Banyak sampel × 1tua gelas = 300 45.400 Banyak lampu 565,2 cm2. =orang + 12%) = 35% Banyak siswa 12 finish kurang dari 74 detik = 5 + 4 + 2 = berpendidikan SMP sawah gelas seluas 0,5 ha = 5. a. Populasi: 11 orang. Persentase banyak murid yang 2 5.000 m Sampel: lahan seluas 10 m2 mencapai garis fin Dari sampel setiap 10 m2 diperoleh 10. b. Volume 11 ish kurang dari 74 detik45 = × 100% = 8 kg drum: V=gabah πr2t basah. Kira-kira hasil panen 24,44%. yang diperoleh = (5.000 : 10) × 8 = 4.000 6. Jawaban: c Jumlah persediaan 22 2 kg gabah basah = × 30 × 105 gandum = 20 × 35 + 25 × 20 + 30 × 25 + 50 7= 22 × 900 × 40 + 100 × × 15 = = 700 + 500 + 750 + 2.000 + 3.000 = 6.950 kg = 6,95 ton

297.000 cm3 = 297 liter

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 14. Jawaban: c Sudut pusat 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 penggemar musik pop = 360 – (60° 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 + 40° + 120°) = 360° – 220° = 140° 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 Jumlah anggota perhimpunan 0000000000000000000000000000000000000000000000 pemuda =

360° 140°

× 35 = 90

15. orang. Jawaban: c Persentase banyak anak yang gemar Bahasa Indonesia = 100% – (40% + 12,5% + 25%) = 22,5% Persentase banyak anak gemar Matematika Persentase banyak anak gemar Bahasa Indonesia = Banyak anak gemar Matematika Banyak anak gemar Bahasa Indonesia 12,5% 10 5 10 ×

910

⇔ = ⇔ = ⇔ n = = 18 22,5%n 9n 5 Jadi, banyak anak yang gemar Bahasa Indonesia 18 orang. B. 1.

a. Panjang batang tanaman di tempat

Uraian bersuhu 25°C = 5 cm. Panjang batang tanaman di tempat bersuhu 30°C = 4 cm. Panjang batang tanaman di tempat bersuhu 35°C = 1 cm. Jadi, panjang batang tanaman jika tanaman ditanam di tempat yang bersuhu 25°C–35°C adalah 1 cm–5 cm.

b. Panjang batang tanaman 4 cm, di tempat bersuhu 30°C. Panjang batang tanaman 6 cm di tempat bersuhu 15°C. Panjang batang tanaman 7 cm di tempat bersuhu 20°C. Jadi, tanaman tersebut cocok ditanam di tempat yang bersuhu 15°C–30°C. 2. a.

Tabel

Panjang Potongan Kabel (cm) Frekuensi 8 9 10 11 12 13 14 15 56455542

=

35 100

× 900 = 315

13. orang Jawaban: a Besar sudut pada daerah basket = 360° – (90° + 36° + 108° + 72°) = 54° Banyak siswa yang mengikuti ekstrakurikuler 108°

basket =360 × 60 = 9 orang °

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 000000000000000000000000000000000000000000000000000000000000000000000000000000000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 = 15 × 100% 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 f021000006d6e74725247422058595a2007ce00020009000600310000616373704d534654000000004945432 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 362 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000 ×+× 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 = % AA BB x 41= 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 000000000000000000000000000000000000000000000000000000000000000000000001163707274000001 gabn +n 3 AB 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 0000000000000000000000000000000000000000000000 500000003364657363000001840000006c77747074000001f000000014626b70740000020400000014725859 A. Pilihan 3. a. Pendapatan terjadi pada 72 ×+× n 80terendah 15 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 5a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e6400000254000000 A Ganda ⇔kamis. 75 Banyak hari wisatawan dewasa pada 1. Jawaban: nA+15 70646d6464000002c400000088767565640000034c0000008676696577000003d4000000246c756d69000003 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 = hari Kamis = 475. Banyak wisatawan anakd Banyak data = 7=(ganjil) f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452 ⇔ 75nA + 1.125 72 nA + 1.200 Median = nilai datum ke-4. 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 anak = 450. Pendapatan pada 430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031 ⇔ pada 3 7. nAJawaban: =hari 75 Kamis 00000000000000000000000000000000000000000000000 c Rata-rata + 450 × hari Kamis = 475 × Rp12.000,00 19 = x – 3 393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735247 ⇔ nA = 25 Rp10.000,00 = Rp5.700.000,00 + x = 22 422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e310 Jadi, banyak siswa kelas 9A adalah 25 orang. 2. Jawaban: d Rp4.500.000,00 00000000000000000000000000000000000000000000000000000xn xn = Rp10.200.000,00 Jadi, 8. Jawaban: cterendah 9. pendapatan objek Rata-rata banyak pasien per hari = 19 Misal rata-rata nilai siswa laki-laki = xwisata 118 19 tersebut Jawaban: Rp10.200.000,00. +20 + +++ + n 21 17 15 b Misal = 19 rata-rata 4 rata-rata nilai siswa perempuan = x2 Banyak siswa perempuan = 17. 7 110 + n = 133 ⇔n = 23 Banyak bilangan yang siswa laki-laki = 13. lain =x. b. Pendapatan pada hari Minggu = 3. Jawaban: 900 × Rp20.000,00 x1 – x2 = 0,6 ⇔x2 = + x11.000 – 0,6 × d Banyak data = 40 Median terletak di Rp15.000,00 = Rp18.000.000,00 + 13x1 antara datum ke-20 dan datum 2 Rp15.000.000,00 = Rp33.000.000,00 = 6,76 +17x ke-21. 30 Pendapatan pada hari Sabtu = 850 × Media Rp15.000,00 ⇔13x1 + 17(x1 – 0,6)= 202,8 1 + 900 × Rp12.000,00 = n Rp12.750.000,00 + Rp10.800.000,00 = = (nilai datum ke-20 + nilai datum 30 x1 – 10,2 = 202,8 2 Rp23.550.000,00 Kenaikan pendapatan ke-21) k 1 30x 213 pada hari Minggu = 1 =wisata objek = (8 + 8) = 8 2 ⇔ x1 = 7,1 Jadi, rata-rata nilai siswa laki-laki Rp33.000.000,00 – Rp23.550.000,00 = 7,1. 4. Jawaban: Rp9.450.000,00 Berat Ikan a 15 16 4. a. Pertumbuhan panjang kecambah hari 17 ke-7 dalam 18 kondisi gelap = 4 cm. Pertambahan panjang19kecambah hari ke-7 dalam kondisi 201,3 cm. Selisih panjang kecambah hari terang = Frekuensi ke-78 = 4 – 1,3 = 2,7 cm Jadi, selisih panjang kecambah 4 antara kondisi gelap dan terang 2,7 5 pada hari ke-7. cm terjadi 5 6 4

b. Panjang kecambah hari ke-5 dalam kondisi gelap = 1,2 + 1,5 + 1,9 + 2,6 + 4,3 + 5 = 16,5 cm Panjang kecambah hari ke-5 dalam kondisi terang = 1,2 + 2 + 2,2 + 1,5 + 1 + 1 = 8,9 cm 5. Persentase banyak siswa yang ikut basket dan sepak bola = 100% – (25% + 12% + 23%) = 40% Persentase banyak siswa yang ikut basket b. Jumlah potongan kabel = 36 Banyak potongan kabel yang panjangnya antara 10 cm dan 14 cm = 5 + 5 + 5 = 15. Persentase potongan kabel yang mempunyai panjang antara 10 cm dan 14 cm

=2 × 40% = 16% 5Banyak siswa yang ikut basket = 16% × 200 = 32 anak Jadi, banyak siswa yang ikut basket ada 32 anak.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 12. Jawaban: c p6 p ++++ B. 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 −+16 ++20p 7p 5 x 1. Tabel berat badan sekelompok balita. = Berat Badan (kg) 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6 4p 4 +42 Uraian Frekuensi 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 9 19 = 1 6 0000000000000000000000000000000000000000000000 1 114 = 4p + 42 1 10 4p = 72 4 2 p= 18 11 13. Jawaban: c Rata-rata = 70,5 ⇔

⇔

((n −3) × 90)+(5 × 80)+(4 × 70)+(n × 60)+(2 × 50) n −3+5+4+n+2

= 70,5

510 +150n = + 82n 8

70,5 5 510 + 150n = 564 + 141n 150n – 141n = 564 – 510 1 9n = 54 9 n= = 6 54 9Jadi,

banyak anak yang nilainya 60 ada 6. 14. Jawaban: d Misal banyak siswa pria = x Maka, banyak siswa wanita = 40 – x Nilai rata-rata kelas = 72 Jumlah nilai siswa pria dan = 72 Jumlah siswa pria dan wanita wanita 69x +74(40 −x) 6 = 72 4048,8. Nilai datum Rata-rata 15 bilangan =

⇔

4 12 4 13 3 14 4 15 5 b. Modus data berat badan = 15 16 Banyak data = 30 Median terletak di antara 4 datum ke-15 dan datum ke-16. Median 17 2 18 1 1

=

2 1

(nilai datum ke-15 + nilai datum

ke-16) = (14 + 14) = 14 Jadi, modus data 2

berat badan balita 15 dan mediannya 14. 2. a. Rata-rata berat badan 30 balita 9(2 10) (4 11) (4 12) (3 13) (4 14) (5 15) (4 16) (2 17) 18 +× +×+× +× +× +× +× +× + = +++++++++ 12 44345421 407

= 30sering ≈13,57 yang paling muncul 6 69x + 2.960 – 74x = 2.880 kg adalah 15. 5x = 80 5 552 ×+ +×4x 648 b. Usia balita paling banyak adalah Jadi, modus data 15. ⇔ = 48,8 x= 16 Jadi, banyak siswa pria 16 orang. 15 2 tahun Data terurut: 11, 12, 12, 13, 5. Jawaban: ⇔ 260 + 4x + 288 732 Banyak data 14,=15 Banyak data == 6 Median data c 2+4+5+5+9+3+4 balita usia 2 tahun ⇔ 4x = 184 15. = 32 ⇔ = 1x(nilai = 46datum ke-3 + nilai datum Median Jadi,(genap) rata-rata empat jumlah data bilangan yang lain 46. 2 Rata-rata = Jawaban: d banyak data 1 32 32 ke-4) 1 = (nilai datum ke- + nilai datum ke-( + = (12 + 13) = 12,5 Rata-rata berat 10. Jawaban: 2 2 2 2 jumlah data j 1))1 b 7,5= 80)+ (8balita ×90) +×(20 100)2 × (2 badan = (nilai 1 datum ke-16 + nilai datum ke-17) 12 Rata-rata nilai ujian =usia 2 tahun 11 30 1++ Jumlah data = 12 × 7,5 = 90 Misal +×212131415 + = = (7 + 8) = 7,5 2.520 6 77 = = 84 x= 2 jumlah nilai 3 siswa baru Rata= ≈12,83 kg 30Nilai rata baru = 7,8 6. Jawaban: d Jadi, rata-rata nilai ujian6 di kelas tersebut 84. 5 Rata-rata 90 +x 6 ⇔ = 12 7,8 3. a. Tabel b 11. Jawaban: 7 5) ×+ (8 7) (108)(11 9) (2 (6 10) ×+×+× Jumlah nilai 4 siswa = 4 × 60 = 8240 +x6) +3 = (3+×+×+ 90 9 =117 +++2 3 ++ 810116 Jumlah nilai 5 siswa = 240 + 70 = 310 295 9 x= 27 == 7,375 40 x 27 Rata-rata nilai 3 siswa = = = 9. 3di atas 37,375 Siswa yang mendapat nilai baru sebanyak 11 + 6 + 2 = 19 anak.

310

10

Nilai rata-rata 5 siswa Frekuensi = = 62 5 9 8 11 5 4 3

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 b. Banyak data = 40 Median data 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 Rata-rata berat badan 8 anak = x – terletak di antara datum ke-20 dan datum 0,25 9x 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 9 −44 = x – 0,25 = ke-21. Median 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 8 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 9x – 44 = 8 x – 2 9 1 = (nilai datum ke-20 + nilai datum 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 x = 42 Jadi, rata-rata berat badan sembilan 2 anak 42 kg. 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 ke-21) 1 = (7 + 7) = 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 7. Misal tinggi Adi = tinggi Dodi = tinggi Rita 2 Jadi, median data 7. 7 =x 00000000000000000000000000000000000000000000000 (12 ×152) 4.

Rataan = +3x12 +3

a. Rata-rata tinggi bibit tanaman cabai =

(16 ×10) +(15 ×12) +(18 ×14) +(18 ×16) +(20 (n ×18) +× 20) 16 +15 +18 +18 +20 +n

1.240 +20n 87 +n 8

⇔ 15 =

1.305 + 15n = 1.240 + 20n 1

1 1.824 +3x 1 153 = 15 2.295 = 1.824 + 3x 2 3x = 2.295 – 1.824 = 471 3 x= = 157 471

5n = 65 ⇔n = 13 Jadi, banyak bibit tanaman 3Jadi, tinggi Adi 157 cabai yang tingginya 20 cm adalah 13. cm. 8. Rata-rata 25 bilangan = 51 b. Banyak data = 100 Median terletak Jumlah 25 bilangan = 25 × 51 = di antara datum dan data ke-50 dan datum 1.275 Misal p = rata-rata 15 ke-51. Median bilangan Jumlah15 bilangan = 15 × p = 15p Rata-rata 40 bilangan = 1 52,5jumlah 40 bilangan = (nilai datum ke-50 + nilai datum = 52,5 2 40 ke-51) 1 (1.275 +15p) = (16 + 16) = = 52,5 2 40 Modus data = 18. 16 15p = 40 × 52,5 – 1.275 1 5. Mean ×+× +(2n 21) (3 22) +× (3 24) +× (2n 19) (4 15p = 825 1 20) × +× (3 23) +× (n 25)

=

2n ++ ++++ 42n 333 n

825

p= = 55 15Jadi, rata-rata 15 bilangan tersebut 55.

287 +105n ⇔ 21,5 = 9. Misal n1 = jumlah siswa ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e747252474220585 13 +5n 1 95a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32 n2 = jumlah siswa laki-laki wanita 279,5 + 107,5n = 287 + 105 n 2 d485020200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000116370727400 jumlah nilai 2,5n = 7,5 2 0001500000003364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a0000021800000014675859 Rata-rata = banyak siswa 5a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676 n= 3 Nilai datum yang paling tinggi ⇔ Jumlah nilai = rata-rata × banyak 696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c000008 frekuensinya adalah 19 dan 21 sebanyak 6. Jadi, 0c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574 siswa Jumlah nilai siswa wanita = n1 × modus data 19 dan 21. Banyak data = 28. 742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336313936362d322e3100000000000000000000 73 = 73n1 Jumlah nilai siswa laki-laki = Median terletak di antara datum ke-14 dan 0012735247422049454336313936362d322e31000000000000000000000000000000000000000000000000000000

datum ke-15. Median

=

1

(nilai datum ke-15 + nilai datum

2 ke-16) 1 = 2 (21

n2 × 71 = 71n2 Nilai rata-rata kelas = 73n1 2 +71n ⇔ 71,8 = n1+n 2 2 71,8 73n1 + 71n2 = 71,8n1 + 71,8n22 1,2n1 = 0,8n2

+ 21) = 21 Jadi, modus data 19 dan 21, sedangkan median data 21. 6.

Misal x = rata-rata berat badan sembilan anak, maka rata-rata berat badan delapan anak = x x – 0,25. = 9 jumlah data ⇔ jumlah data =9x

Rata-rata nilai Matematika kelas IX B (9 ×5) +× 6) +(11 ×7) +× 8) (4 9) (3 10) (8 (5

+×+× =

40 276

= = 6,9 40

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454 320735247420000000000000000000000000000f6d6000100000000d32d485020200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000116370727400 0001500000003364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e6400000254 00000070646d6464000002c400000088767565640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c000008 0c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000 012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e31000000000000000000000000000000000000000000000000000000

10. Misal kelima bilangan bulat tersebut adalah a – 10, a – 5, a, a + 5, dan a + 10. Ratarata = 14 a10 − + a − 5 + a + a + 5 + a + 10 − = 14 = a= 14 Nilai datum terbesar = a + 10 = 14 + 10 = 24. Nilai datum terkecil = a – 10 = 14 – 10 = 4. Jangkauan = 24 – 4 = 20. Jadi, jangkauan data 20.

n1 0,8 2 n2 1,23

⇔==

n1 : n2 = 2 : Persentase banyak siswa wanita 3 n2

1 = n +n 23 × 100% = × 100% = 40% + 1

2

Persentase banyak siswa laki-laki n3

2 × 100% = × + = n1 +n 23 100% = 60% 2

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 7. Jawaban: a Data setelah 3 62d322e31000000000000000000000000000000000000000000000000000000 Nilai Q1 = X12 +1 (X13 – 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 diurutkan: 4556667788 3 26 27 4 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 3 X26 ) 1 == 78 ++(8(9 – – 8) = 81 = 12 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 4 44 Jadi, nilai kuartil atas 21 = Letak median = (10 + 1) 0000000000000000000000000000000000000000000000 54 7) 73 = 7,758,25 8,25. = 4 1 5 10. Jawaban: 2 Nilai median = X5 + 1 (X6 – 5. Jawaban: a Data setelah a 2 = 6 + (6 – 6) X ) diurutkan: 58911 1922242 293136

=6 21 Letak Q1 = datum ke- (10

Letak Q2 = datum ke- (34 41 + 1) 1 Letak Q1 = = datum datum ke-17 ke- (10

+ 1) 3 = datum ke-2 4 Nilai Q = X + 1 2

+ 1) 1 ke-2 = datum Nilai Q2= X17 2+(X18 – 1X17) Nilai Q1= X2 +

5

4

3

4 (X3 – X2) A. Pilihan = 5 +3 (5 – 5) = 4 Ganda 1. Letak Jawaban: b Data 3 datum Q5 = ke- setelah (10 4 1517 diurutkan: 323 45911 1314 + 1) 1 = datum ke-8 41

3

Letak Q3 = datum ke- (10 Nilai Q3 = X8 + (X94– X8) 4 + 1) 1 = datum ke-8 11 = 7 + (8 – 7) = 7 = 7,25 4 Jadi, 4 4 Nilai Q3 = X8 + pernyataan yang 4 benar (1) dan 1 8. (2). Jawaban: (X9 – X18d) Data setelah diurutkan: 1417 + 19 (15 – 2457911 13=15 4 20 23 24 3) 1 = 14 4 Letak Q1 = datum ke- 1(13 2. Jawaban: b Data4setelah diurutkan: 566 7991112 + 1) 1 34 = datum ke-3 13 21

Nilai + datum (X – 1 3 Nilai Q Q11= =Xnilai 2 4 ke-4 (11 + 1) = nilai X3) 1 =5+ (7 – 5) =3 datum ke-3 = 5. 3. 2 Letak Jawaban: Q3 = datum d Data ke-setelah (13 diurutkan: 6 12 34 579 1113 14 1821 41 + 1) = datum ke-10 2 2 Letak Q2 = datum ke- (12 1 4 Nilai + 1) Q3 = X10 + (X11 – 2 11 1 = datum ke-6 X10) 2 1 – 19) = 19 1 = 19 + (20 2 2 Nilai Q2= X6 + Nilai 113 (Q3 (X – Q–1)X=)(19 – 6) = 6 = 6,75 16 2 2 7 224 = 7 + (9 – 2=8 9. 7) Banyak data = 34 3 4. Jawaban: c Letak Q3data = datum (34 Banyak = 16 ke3 4 Jawaban: d + 1) 1 Q1= Letak = datum datum ke-26 ke- (16 4

+ 1) 3

= datum ke-12 4

4

4

32 4

4

3 =

7 + (7 – 7) = (X3 – X 2 32) = 7 84+3 (9 – 1 Letak Q8) = datum ke- (34 1 8 = 8,75 = 4 4 + 1) 3 = datum ke-8 3 (10 Letak Q = datum ke3

43

4

+ 1) Nilai Q1==8datum + (X9ke-8 –1 44 1

X8) 3 = 6 + (6 – 6) = 6 1 111 Nilai Q3= X8 + 4 (X9 – X8) 1 29Q+ +(31 Nilai (Q4=1 + Q3)–= (6 + 7 + 8 ) = 2 4 3 3 412 7 29) = 291 = 29,5 B. Uraian 2Nilai (Q – 2Q ) = 29,5 – 2 × 3 1 8,75 = 12 6. Jawaban: c Data 1. Banyak data = 33 setelah diurutkan: 1 245891112 13 1417 1921 Letak Q1 = datum ke- 4(33 + 1 = datum ke-8 1 1) Letak Q = datum 2 1 ke- (12 1 4 Nilai Q1= x8 + + 1) 1 = datum ke-3 (X – X ) 14 1 2 = 89 + 8(9 – 8) = 2 Nilai Q1=1X3 + 22 Letak Q = datum ke8(X2 – X 1 ) 4 4 + 3(8 – 4 = datum (33 + 1)= 54ke-17 3 Nilai Q25) = x9 = 9 3 = 5 = 5,75 43

Letak Q3 = datum ke- (33 Letak Q3 = datum ke-4 (12 4 + 1) 1 = datum ke-25 + 1) 3 1 2 Nilai Q = =datum x25 4+3 ke-9 (X26 – 3 1 X ) Nilai + 25 Q32==X13 9 + (15 – 13) – X9) =(X14 10 2 3 4 = 14 + (17 – 1 8 1 Jadi, nilai , Q2 = 9, dan Q3 14)Q14 = 2 = 16 = = 16,25 14.diurutkan: 140, 141, 411 2. a. Data setelah 142, 143,(Q 145, 150, 151, + 152, 153, Nilai + 148, Q1 ) = (16,25 5,75) = 153 2 3 = 12 2 Banyak data 11

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e7472524 1 Jadi, peluang terambilnya kartu bernomor 62d322e31000000000000000000000000000000000000000000000000000000 Letak Q1 = datum keNilai(12 Q1 = X6 + (X7 – X6) 1 11 7422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d 4 4 prima ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 6000100000000d32d485020200000000000000000000000000000000000000000000000000000000000000000000000000000000000 adalah2 . + 1) 13 0000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b707400000204000000 = datum ke-3 = 10 + (11 – 10) = 10 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 4 4 4 147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d64640 7. Jawaban: d S = {(A, A, A), (A, A, G), 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 1 2 00002c400000088767565640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c00000 A. Pilihan (A, G, A), (A, G), ke- (24 + 1) Nilai Q = X + (X4 – X ) Letak Q G, = datum 02474656368000004300000000c725452430000043c0000080c675452430000043c0000080c625452430000043c0000080c746578740 1 3 4 3 2 000000000000000000000000000000000000000000000000000000000000000000000000000000000 41 Ganda 1. Jawaban: a S 1= {1, 2, 3, 4, 5, 6}, 0000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e7900006465736300000 (G, A, A), (G, A, G), (G, G, A), (G, G, G)} 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 = datum ke-12 00000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e310000 = 142 + muncul (143 – 142) n(S) = 6 A = kejadian mata dadu n(S) = 8 A = kejadian muncul angka dan 4 21 000000000000000000000000000000000000000000000000001 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 kurang dari 3 1 gambar = 142 Nilai Q = X + (X 2 – X ) 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 4 2 (G, 12 A, G), 13(G,12 = {1, = {(A, G, G), G, A), (A, A, G), 2 1 (A, G, A), (G, A, A)}1 n(A) 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 2} n(A) Letak Q = datum ke- (12 + 1) = 14 + (15 – 14) = 14 2 4 22 1 = 6 Peluang muncul satu =2 n(A) 2 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 P(A) =Jadi, peluang == datum = muncul gambar:Letak Q = datum ke-3 (24 + 1) ke-6 n(S)63 mata dadu 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 3 1 2 413 kurang dari 3 1 n(A) 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 P(A) = = datum ke-18 Nilai Q = X + (X – X ) n(S) adalah . 2 6 2 7 6 413 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 3 =)63 = 00000000000000000000000000000000000000000000000 = 148 + (150 Nilai – Q3 = X18 + 4 (X19 – X18 8= 149 4 2. 2 8. Jawaban: c A– 33 = kejadian keluarga 148) muncul mata dadu Frekuensi relatif = 17 + (18 17) = 17 Jadi, 4 4 = 12 n(S) = Jawaban: c Amir memenangkan undian n(A) 1 prima banyak muncul mata dadu prima = b. banyak Nilai (2Q – Q2yang ) = dilakukan 2 × 42 142 – 149 = 2.526 nilai kuartilnya adalah percobaan 1 11 3 1 Q1 = 10 , Q2 = 14 , dan Q3 = ++28 9910 14 135 3. a. terendah diperoleh pada hari = Keuntungan == 422 n(A) 12 50 5025 . = P(A)17 = 4= ke-6 sebesar 1,5 juta. n(S)2.526421 peluang keluarga Amir 1 1131 Keuntungan tertinggi diperoleh pada hari b. Nilai (Q1 + QJadi, 3) = 10 + × 17 = 16 3. Jawaban: d Ruang memenangkan 2 3 undian 4346

sebesar sampel:ke-12 S = {M, A, T, 2,3 E, I,juta. K} n(S) data = 12 vokal = = 6b.VBanyak = himpunan huruf Data terurut: {A, E, I} n(V) = 3

421 1 + 1) 5. Letak Q1 = datum ke- (30 4 . 9. Jawaban: b Jika3kotak II ditambah 1 1,5 1,6 1,6 1,7 1,8 2,0 2,0 2,0 2,1 2,2 2,2 2,3 = datum ke-7 bola hitam maka jumlah4bola dalam kotak II 1 n(V) Q 3 1 = datum ke- (12 Letak ada 9 bola, yaitu 33 bola hitam dan 6 bola 4 P(V) = n(S = = 1 62 Nilai Q = X7 + (X8 – 4 putih. n(S)1= 9 A = kejadian terambil bola + 1) 1 ) 1 X ) 3 hitam n(A)==44 3 Peluang datum ke-3 4 huruf vokal= 44 + (44 Jadi, peluang=terambilnya –7 44) Nilai Q1 terambil = X3 + bola hitam: 4 adalah 1 2. 2 4. Jawaban: b Banyak 1 4 (X4 – X3) kelereng hijau: n(H) = 8 Letak Q2 = datum ke-4 (30 + 1) = 1,6 + (1,7 – 1,6) = 1 4 Banyak kelereng seluruhnya: n(A) 1 = datum =Q3datum ke-15 1,625 2 Letak P(A) = n(S)93 ==3 n(S) = 8 + 12 + 10 = 30 1 3 Nilai Q = X15 + (X16 – 4 84 2 2 ke(12 + 1) Peluang terambil kelereng 10. Jawaban: c A = kejadian siswa laki3 hijau: P(H) = 30 X15) 1 = datum ke-9 laki menjadi ketua kelas n(A) = 16 = 45 + 4 A = kejadian 22 (46 – 45) = 5. = 15Jawaban: a n(S) = 25 3 1 Nilai Q = X + (X – X ) 3 terambilnya 3kartu 9 bernomor 9 faktor 45 4 10 Letak Q = datum ke- (30 + 1) 4 3 n(A) 16 2 3 P(A) = = = = 2,1 + (2,2 – 2,1) = n(S)405 Peluang siswa perempuan menjadi 1 dari 16 = {1, 2, 4 = datum ke-23 ketua kelas: 23 4 kuartil atas 2,175 dan nilai 4, 8,Jadi, 16} nilai n(A) = 2,175 1 kuartil bawah 1,625. P(A ′) Q =3 1 5 n(A) 5 Nilai = –XP(A) +4= –1X–235) = 23 4. Data setelah diurutkan: 1 (X24 1 P(A) = n(S)25 = 11. 5 Jawaban: A –=46) = 46 = 46 + c (47 44 4 6 7 8 810111212131414 kejadian muncul angka n(A) 6. 151515 Jawaban: a n(S) = 52 A = kejadian 1(Q + 2Q + Q ) 161717182021212325 2 3 = 90Nilai – 40 3 =1 50 Peluang terambilnya kartu merah bernomor a. Banyak data = 24 muncul angka: 115 = 31(44 + 2 × 45 + 46 ) = 60 24 12 1 n(A) 50 5 Letak Q15,=7,datum (24 ganjil = {3, 9 dari ke2 jenis P(A) = n(S)909 == 4 kartu} n(A) = 4 × 2 = 8 n(A) 8 + 1) 1 = datum ke-6 2 4

P(A) = n(S)5213 ==

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 12. Jawaban: d Banyak sampel yang 2. Banyak percobaan yang dilakukan: 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 diambil di setiap kota sama, yaitu 40 anak. n(S) = 200 Banyak muncul angka: n(A) = 85 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 Banyak muncul gambar: n(B) = 200 – 85 = 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 115a. banyak anak buta warna Frekuensi relatif Peluang anak buta warna = 0000000000000000000000000000000000000000000000 muncul angka: n(A) 85 17 banyak == n(S)20040 sampel 63 P(anak buta warna di kota A) = 40 b. Frekuensi relatif muncul = 21 gambar: n(B) 115 23 20 P(anak buta warna di kota B) = == 40

= 4 20di kota C) = = P(anak buta warna 4010 P(anak buta warna 1 di ketiga kota) 624 1 ++ 12

===

Jadi, pernyataan yang benar pada pilihan d. 13. Jawaban: d A = kejadian terambil bola lampu rusak n(A) = 12 n(S) = 40 ffd8ffe000104a464946000102 0100c800c80000ffe20c5849434 n(A) 12 3 35f50524f46494c45000101000 == = P(A) n(S) 4010 Peluang terambil bola lampu 00c484c696e6f021000006d6e7 yang tidak rusak: 4725247422058595a2007ce000 3 7 P(A′) =2000900060031000061637370 1 – P(A) = 1 –10 = 10

n(S)20040

3. Kartu pos yang jawabannya benar = 1.861 – 475 = 1.386 A = kejadian kartu pos Roni terambil menjadi pemenang n(A) = 9

40 ++4040 12010

4d53465400000000494543207 14. Jawaban: c A = kejadian mata dadu 3524742000000000000000000 yang muncul berjumlah lebih dari 9 = {(6, 4), (5, 5), (4, 6), (6, 5), (5, 6), (6, 6)} 0000000000f6d6000100000000 d32d485020200000000000000 n(A) = 6 61 0000000000000000000000000 P(A) = 0000000000000000000000000 = Peluang mata dadu yang 36

6

muncul0000000000000000000000000 berjumlah lebih 1 0000001163707274000001500 dari 9 adalah. 6 0000033646573630000018400 15. ffd8ffe000104a4649460001020100c8 Jawaban: a Misal: pebulu tangkis 00006c77747074000001f00000 pria00c80000ffe20c584943435f50524f464 = A, B, C, D 0014626b70740000020400000 pebulu tangkis wanita = x, 94c4500010100000c484c696e6f02100 0147258595a00000218000000 0006d6e74725247422058595a2007ce y, z x →(A, x) 00020009000600310000616373704d5 146758595a0000022c00000014 y →(A, y) C 3465400000000494543207352474200 6258595a00000240000000146 z →(C, z) x →(B, x) 00000000000000000000000000f6d60 46d6e64000002540000007064 00100000000d32d4850202000000000 x →(D, x) B 6d6464000002c4000000887675 0000000000000000000000000000000 y →(D, z →(B, z) (B, y) D 0000000000000000000000000000000 65640000034c00000086766965 z →(D, z) 0000000000000000000000001163707 77000003d4000000246c756d69 2740000015000000033646573630000 Jadi, ada 12 pasangan ganda campuran 000003f8000000146d65617300 01840000006c77747074000001f00000 yang terbentuk. 00040c00000024746563680000 0014626b70740000020400000014725 B. 8595a00000218000000146758595a00 04300000000c72545243000004 00022c000000146258595a000002400 1. Titik sampel yang muncul dari satu 3c0000080c675452430000043c 0000014646d6e640000025400000070 Uraian angka dan0000080c625452430000043c00 mata dadu genap: B = {(AGG, 2), 646d6464000002c4000000887675656 (AGG,4), (AGG, 6), (GAG, 2), 40000034c0000008676696577000003 00080c7465787400000000436f (GAG, 4), (GAG, 6), (GGA, 2), (GGA, 4), d4000000246c756d69000003f8000000 7079726967687420286329203 (GGA, 6)} Jadi, banyak titik sampel = 9. 146d6561730000040c0000002474656 1393938204865776c6574742d5 368000004300000000c725452430000 061636b61726420436f6d70616 043c0000080c675452430000043c000 0080c625452430000043c0000080c74 e790000646573630000000000 65787400000000436f70797269676874 0000127352474220494543363 202863292031393938204865776c657 13936362d322e310000000000 4742d5061636b61726420436f6d70616 0000000000001273524742204 e790000646573630000000000000012 735247422049454336313936362d322 9454336313936362d322e3100 e310000000000000000000000127352 0000000000000000000000000 47422049454336313936362d322e310 0000000000000000000000000 0000000000000000000000000000000 0000000000000000000000x 00y →(C, y) z →(A, z) →(C, x) A

P(A) =

9

=

1

1.386

154

Jadi, peluang Roni menjadi

. 1 154 pemenang 4. a. A = kejadian jarum menunjukkan daerah putih n(A) = besar sudut putih = 360° – (90° + 75° + 75°) = 120° n(S) = 360° 120° 1 P(A)= = 360°3 Jadi,

peluang lempengan menunjukkan daerah 1 putih

3

.

b. B = kejadian jarum menunjukkan daerah merah B′= kejadian jarum menunjukkan daerah selain merah 75° 5 P(B) = = 360°24 5 19

P(B′) = 1 – P(B) = 1 –

24

=

24

Jadi,

peluang jarum menunjukkan daerah 19 selain merah . 24

5. a. A = kejadian muncul mata dadu berjumlah 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} n(A) = 6 n(S) = 36 P(A) =

6 36

=

6

1 Jadi, peluang muncul

mata dadu berjumlah 7 1 adalah . 6

b. B = kejadian muncul mata dadu berjumlah kurang dari atau sama dengan 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} n(B) = 10 10 5

P(B) =

36

=

18

Peluang kejadian muncul mata dadu berjumlah lebih dari 5:

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 4. Jawaban: a Misal: dadu pertama 5 13 P(B′) = 1 – P(B) = 1 – = 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 merah dan dadu kedua hitam 18 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 Jadi, peluang muncul mata dadu Dadu Kedua (B) 18 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 berjumlah 13 lebih dari 5 adalah . 0000000000000000000000000000000000000000000000 18 D 2 c. Misal dadu pertama adalah dadu merah. ad 3 u C = kejadian muncul mata dadu pertama 5 Pe bilangan ganjil = {(1, 1), (1, 2), (1, 3), (1, 2 rt (2, 2) 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), a (2, 3) m (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, (2, 5) a 5), (5, 6)} n(C) = 18 Peluang muncul mata 5 (A 5, 2) dadu merah bilangan ganjil: ) (5, 3) (5, 5)

P(C) = n(S)362

n(C)

18 1

== Jadi, peluang muncul mata dadu

merah 1 bilangan prima adalah . D = kejadian muncul kedua mata

d. 2 dadu prima = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)} n(D) = 9 Peluang muncul kedua mata dadu prima: P(D) =

n(D)

91

== Jadi, peluang muncul mata dadu keduanya 1 n(S)364

prima adalah . 4

A. Pilihan Ganda 1Jawaban: b Kejadian yang mungkin terjadi mempunyai peluang antara 0 dan 1. 2Jawaban: a Kejadian yang pasti terjadi adalah setiap makhluk hidup akan mati. 3Jawaban: d 1

Semua kejadian A mempunyai peluang P(A) = :, kecuali 2

A = kejadian muncul mata dadu 3 1 → P(A) = {3} =

6

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 10. Jawaban: d Misal: A = {anak gemar 3 P(B) = 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 Matematika} B = {anak gemar Fisika} x = 36 A ∩ B = {(4, 6), (5, 5), 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 banyak anak yang tidak gemar Fisika dan (6, 4)} n(A ∩ B) = 3 3 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0000000000000000000000000000000000000000000000 P(A ∩ B) = Matematika 36 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 183 3181 Diagram =+ –== Venn: S AB 35 – 15 363636362 ffd8ffe000104a46494 60001020100c800c80 000ffe20c584943435f A ∩ B = {(2,50524f46494c450001 2), (2, 3), (2, 5), (5, 2), (5, 3), 0100000c484c696e6f0 21000006d6e7472524 (5, 5)} n(A ∩ B) = 6 61x 7422058595a2007ce0 P(A ∩ B) = 002000900060031000 = 6 n(S) = 60 36 n 0616373704d5346540 n(A U B) + x000000049454320735 n =60 Jadi, peluang muncul dadu merah faktor ((35 – 15) + 247420000000000000 15 + (30 – 15) + x = 60 prima dari 1 10+dan hitam x 50 = 60000000000000000f6d ⇔prima x=10 adalah . 6000100000000d32d4 850202000000000000 6 5. Jawaban: b Banyak anggota ruang 000000000000000000 sampel: n(S) =000000000000000000 12 A = kejadian muncul gambar = {(G,000000000000000000 1), (G, 2), (G, 3), (G, 4), (G, 5), 4c696e6f021000006 000000000000000000 d6e7472524742205 muncul mata dadu faktor (3,8595a2007ce000200 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (G, 6)} B = kejadian 000000000001163707 274000001500000003 prima dari 12 = {(A, 2), (A, 3), (G, 2), (G, 3)} (6, 1), (6, 5)} n(A ∩ B) = 14 0900060031000061 364657363000001840 6373704d53465400 P(A ∪ B) = P(A) + P(B) – P(A ∩ 000006c77747074000 0000004945432073 B) 001f000000014626b7 5247420000000000 074000002040000001 18 151419 0000000000000000 A ∩ B = {(G, 2), (G, 3)} → n(A ∩ =+–= 47258595a000002180 00f6d600010000000 36363636 Jadi, peluang jumlah mata dadu 00000146758595a000 B) = 2 ∩ B) 2 n(A 1 0d32d48502020000 yang muncul merupakan bilangan ganjil P(A ∩ B) = 0022c0000001462585 = = n(S) 126 0000000000000000 atau jumlah mata dadu 19 95a000002400000001 0000000000000000 6. Jawaban: b Misal P = perempuan dan L 4646d6e64000002540 0000000000000000 bilangan prima adalah . 0000070646d6464000 = laki-laki. Kemungkinan anak lahir: PP, PL, LP, 36 0000000000000000 002c40000008876756 LL n(S) = 4 A = kejadian lahir laki-laki dan 0000000000000000 9. 5640000034c0000008 0000000000011637 perempuan A =676696577000003d40 {(PL, LP)} n(A) = 2 Peluang 0727400000150000 kejadian lahir anak lahir laki-laki dan 00000246c756d69000 Jawaban: a 0003364657363000 003f8000000146d656 S BK001840000006c777 Diagram16Venn: 1730000040c0000002 – x 47074000001f00000 474656368000004300 0014626b70740000 000000c72545243000 0204000000147258 595a000002180000 perempuan0043c0000080c67545 n(A) 2 1 2430000043c0000080 00146758595a0000 c625452430000043c0 15 022c000000146258 P(A) = = = 000080c74657874000 n(S)42 595a000002400000 00000436f707972696 0014646d6e640000 7. Jawaban: d 768742028632920313 Banyak anggota ruang sampel: 025400000070646d B = {siswa memelihara 93938204865776c657 n(S) = 36 A = kejadian muncul mata dadu 6464000002c40000 4742d5061636b61726 burung} K0088767565640000 = {siswa berjumlah genap = {(1, 1), (1, 3), (1, 5), (2, 2), 420436f6d70616e790 034c000000867669 memelihara kucing} (2, 4), (2, 6), (3,000646573630000000 1), (3, 3), (3, 5), (4, 2), (4, 4), ((16 – x) siswa + 6577000003d40000 x += (21 Banyak B 40– x) + 15 = 40 (4, 6), (5, 1), (5, 000000012735247422 3), (5, 5), (6, 2), (6, 4), (6, 6)} 52 – x =4000246c756d690000 5 049454336313936362 03f8000000146d656 d322e3100000000000 x = 52 – 40 = 12 x = banyak siswa memelihara 1730000040c00000 n(A) = 000000000001273524 burung dan kucing = 12 siswa 0247465636800000 742204945433631393 18 18 4300000000c72545 P(A) = 6362d322e310000000 2430000043c00000 36 000000000000000000 n(S) = 40 x 123 80c6754524300000 B = kejadian muncul mata dadu 000000000000000000 43c0000080c625452 P(A)= = =430000043c000008 00000000000 berjumlah 10 ={(4, 6), 15 (5, 5), (6, 4)} n(S)4010 Jadi, peluang terpilih siswa 0c74657874000000 n(B) = 3 30 – 15 memelihara burung dan 3 00436f70797269676 8742028632920313 kucing .93938204865776c6 10 574742d5061636b6 1726420436f6d7061 6e79000064657363 0000000000000012 7352474220494543 36313936362d322e 3100000000000000 0000000012735247 4220494543363139 36362d322e310000 0000000000000000 0000000000000000 0000000000000000 00x 21 – x

8. Jawaban: d Banyak anggota ruang sampel: n(S) = 36 Kejadian muncul jumlah mata dadu bilangan ganjil: A={(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, (5, 6), 2), (6, (5, 4), 1), (6, 3), (6, 5)} n(A) = 18 Kejadian muncul jumlah mata dadu bilangan prima: B = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), ffd8ffe000104a464 (3, 2),9460001020100c80 (3, 4), (4, 1), (4, 3), (5, 2), (5, 0c80000ffe20c58494 (6, 5)} 6), (6, n(B) 1), = 15 A ∩ B = {(1, 2), (1, 4), (1, 3435f50524f46494c 6), (2, 1), (2, 3), (2, 5), 4500010100000c48

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 14. Jawaban: c Peluang turun hujan: P(H) 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 B = {(A, 2), (G, 2), (A, 5), = 0,4 Peluang tidak turun hujan: P(T) = 1 – 0,4 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 (G, 5)} n(B) = 4 n(B) 4 = 0,6 Frekuensi harapan tidak turun hujan 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 selama bulan April (n = 30): Fh(T) = P(T) × n = 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 P(B) = = n(S)12 A ∩ B = {(A, 0000000000000000000000000000000000000000000000 0,6 × 30 = 18 2), (A, 5)} n(A ∩ B) =2 Jadi, di kota tersebut tidak turun hujan selama 18 hari. P(A ∩ B) = 2 15. Jawaban: b A = kejadian lolos tes masuk akademi P(A) = 1 – P(A′) = 1 – 0,57 = 0,43 Fh(A) = P(A) × n = 0,43 × 3.000 = 1.290 Jadi, banyak peserta yang diterima 1.290 orang. B.

12

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 642 =+– 121212 82

== Jadi, peluang muncul angka pada mata uang atau muncul mata dadu faktor prima 40 2 123

1. a. Kemustahilan adalah . Uraian 3 Kemustahilan Faktor prima dari 330 adalah 2, 3, 5, dan 11. b. = kejadian muncul matatidak dadu PeluangCdipanggilnya anak yang Kepastian bilangan genap C= {(A, 2), (A, 4), (A, 6), gemar 10 1 Kemustahilan Bilangan 36 mempunyai 9 faktor, (G, 2), (G, = 4), (G, 6)} keduanya = . 606 yaitu 1, 2, 3, 4, 6, 9, 12, 18, dan 36 11. Jawaban: a Banyak anggota ruang 61 sampel: n(C) n(S) = = 68 n(C) Kejadian muncul paling 2. Jumlah kartu = 52 ⇒ n(S) = sedikit dua gambar: A = {(A, G, G), (G, A, G), P(C) = = = 52 A = kejadian terambil kartu hitam n(S)122 = n(A) kejadian (G, G, A), (G, G,DG) = 4 muncul mata dadu ganjil n(A) = 8 bilangan prima D= {(A, 2), (A, 3), (A, n(A) 8 5), n(A) (G, 42), 1 (G, 3), (G, 5)} n(D) 6 1 P(A) = n(S)52 = P(A) = = = n(S)82 Frekuensi harapan muncul B = kejadian terambil kartu Queen dua n(D) gambar: = P(A) × n 1 =n( = F=h(A) n(B) = 4 122 n(B) 4

P(B) = = n(S)52 A ∩ B = kejadian terambil kartu hitam ganjil dan Queen n(A ∩ B) = 0 P(A ∩ B) = 0 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 8 4 123

S)

2

C ∩ D = {(A, 2), (G, 2)} → P(C ∩ D) = × 400 = 200 = 2 12∪ D) = P(C) + P(D) – P(C ∩ P(C 12. Jawaban: c Banyak anggota ruang 662 sampel: D) n(S) = 12 =K+=– kejadian muncul angka 121212 10 5 dan mata dadu faktor dari 30 = {(A, = =1), (A, 2), (A, 3), peluang muncul mata dadu (A,126 5) Jadi, (A, 6)} n(A) 5 bilangan genap atau muncul mata n(K)dadu = = bilangan prima 5 n(S)12 Frekuensi harapan muncul angka

=+–0== Jadi, peluang terambil kartu hitam ganjil atau 3 . dan adalah mata dadu genap: Fh(K) = P(K) × n 5 6 ffd8ffe000104a4649460001020100c800c8000 Queen adalah . 13 0ffe20c584943435f50524f46494c45000101000 1M = kemungkinan menang di kandang lawan 3. 00c484c696e6f021000006d6e74725247422058P(M) = 0,76 Banyak pertandingan seluruhnya: n = 595a2007ce000200090006003100006163737025 kali. F=(M) × 60 = 25 × n = 0,76 × 25 = 19 Tim = P(M) h 12 4d534654000000004945432073524742000000 tersebut diharapkan dapat menang Jawaban: b Banyak anggota 19 kali di 0000000000000000000000f6d600010000000013. kandang lawan. ruang sampel: n(S) = 6 A = kejadian d32d4850202000000000000000000000000000 2a. n(S)mata = 12dadu + 13lebih + 15dari = 40 muncul 4 00000000000000000000000000000000000000 n(S) = 12 = {5, 6} n(A) = 2 00000000000000000000000000000011637072 a. Kejadian muncul angka pada mata n(A) 2 1 74000001500000003364657363000001840000 uang A = {(A, 1), (A, 2), (A, 3), (A, 4), (A, 006c77747074000001f000000014626b707400 5), (A, 6)} n(A) = 6 P(A) = n(S = = 63 000204000000147258595a0000021800000014 n(A) 6 ) 1 6758595a0000022c000000146258595a000002 P(A) = = (A) = P(A) × n ⇒ 120= × n ⇔ n = 360 F n(S)12 B = kejadian muncul mata dadu h 3 Jadi, banyak percobaan yang dilakukan 4000000014646d6e640000025400000070646d faktor prima 40 360 kali. 6464000002c400000088767565640000034c00 00008676696577000003d4000000246c756d69 000003f8000000146d6561730000040c000000 2474656368000004300000000c725452430000 043c0000080c675452430000043c0000080c62 5452430000043c0000080c7465787400000000 436f7079726967687420286329203139393820 4865776c6574742d5061636b61726420436f6d 70616e79000064657363000000000000001273 5247422049454336313936362d322e31000000 00000000000000001273524742204945433631 3936362d322e31000000000000000000000000 000000000000000000000000000000 52525213

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Banyak siswa perempuan yang b. B = kejadian terambil sebuah 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 memperoleh nilai antara 60 dan 80 = 4 + bolaP(B) kuning 15 = 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 40 8 + 6 = 18. Perbandingnan siswa laki-laki 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 dan perempuan yang memperoleh nilai C = kejadian terambil sebuah 0000000000000000000000000000000000000000000000 antara 60ddan 80 = 22 adalah : 18 = 11 : 9. beberapa 1Jawaban: Sampelnya siswa bola biru 13 P(C) = SMP di provinsi A. 40 2Jawaban: b Kenaikan jumlah pengunjung B ∩ C = { } maka P(B ∩ C) terendah terjadi pada hari VII Banyak pengunjung = 0 P(B ∪ C) = P(B) + P(C) hari VI = 475. Banyak pengunjung hari VII = 500. 15 13 Persentase kenaikan banyak pengunjung hari VII =+ 4040 28

=7 =

4010 7

F (B ∪ C)= × 120 h10= 84 kali Jadi, frekuensi harapan terambil bola kuning atau biru 84 kali.

c. P(C′) = 1 – P(C)

=

500 −475 5 475× 19100%

= 5 %.

6. Jawaban: d Misal banyak buku sekolah yang terjual sebulan = n Persentase banyak buku umum = 100% – (25% + 20% + 18% + 15%) = 22%

= 1 – 13 = 27 4040

Fh(C′) = P(C′) × n = 27 × 120 = 81 kali 40Jadi, frekuensi harapan terambil bukan bola biru 81 kali.

22% 132 =⇔ 25%n

n = 150

Jadi, banyak buku sekolah yang terjual dalam sebulan 150 eksemplar.

1Jawaban: d Banyak siswa yang memperoleh nilai 6 = 5 orang Banyak siswa yang memperoleh nilai 9 = 9 orang Selisih banyak siswa yang memperoleh nilai 6 dan 9 = 9 – 5 = 4 orang 2Jawaban: c Sudut pusat kegiatan sosial = 360° – (70° + 60° + 60° + 50° + 40°) = 360° – 280° = 80° Banyak responden yang melakukan kegiatan

da 80°7 adalah 8 dan 9. Data setelah diurutkan: 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 9 Nilai yang lebih dari sosial =360 × 180 = 40 orang. 9.

°

Berat Badan (kg) Frekuensi 35 Jawaban: a 3 36 5 7 = 20 × 100% = 25%. 4 37 3 1Jawaban: c Objek yang akan diteliti adalah hasil 38 Ujian Nasional SMP di DKI Jakarta. Jadi, populasi 3 penelitian tersebut adalah hasil ujian nasional siswa 39 2 SMP di DKI Jakarta. 40 2Jawaban: d Banyak siswa laki-laki yang 3 A = kejadian terambil sebuah bola memperoleh nilai antara 60 dan 80 = 6 + 7 + 9 = 41 hijau P(A) =12 2 22. 40 42 12 3 F (A) = ×43120 = h40Jadi, frekuensi harapan terambil 36 kali 3 bola hijau 36 kali. 44 1 45 1

Banyak siswa yang memperoleh nilai lebih dari 7 adalah 3 + 2 = 5 orang. Persentase siswa yang memperoleh nilai lebih dari

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 15. Jawaban: a Jumlah nilai 36 siswa = 36 10. Jawaban: c Misal: x = banyak 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 × 7,2 = 259,2 Jumlah nilai 38 siswa = 38 × 7,1 penjualan bensin hari Rabu 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 = 269,8 Jumlah nilai 2 siswa baru = 269,8 – y = banyak penjualan bensin hari Jumat 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 259,2 = 10,6 Rata-rata penjualan bensin 1 minggu = 0000000000000000000000000000000000000000000000 10,6 3.000 liter 4.000 +2.000 xy ++ 3.000 +5.000 ++ 2.000 ⇔ = 3.000 Rata-rata nilai 2 siswa =2 = 5,3 7

16.000 + +xy 7 +

= 3.000 16.000 + x+y= = 5.000 liter Jadi, jumlah penjualan pada 2 21.000 hari Rabu x + ydan = Jumat 5.00021.000 liter. – 16.000 11. Mean jumlah data Jawaban: b = = =

16. Jawaban: a Misal: x = nilai yang telah diperoleh Ali n = banyak ujian yang diikuti Ali Pada ujian yang ke-n Ali memperoleh + x94 nilai 94. = 89 n

⇔x – 89n= –94 . . . (1) Pada ujian yang ke-n Ali memperoleh nilai 79. + x79 n

= 86 ⇔x – 86n= –79 . . . (2) Eliminasi x dari (1) dan (2): x – 89n = –94 x – 86n = –79 –––––––––––– – –3n = – 15 −15 ⇔ n= = 5 −3 Jadi, Ali akan mengikuti ujian yang ke-5.

banyak data ×+3 70 ×+7 80 1060 85×+2 65 ×+6 75 ×+×+5 90 ×+2 95 ×+3 100 × 2 ++523 2236++7 10 ++++ 3.260

= 81,5 40

12. Banyak data = 5 + 7 + 6 + 4 + 2 = 24 Jawaban: bdata ke-12 + data ke-13 (genap) Median = 2 67

= 6,5 2

13.

+

=

++ +++ n912 6 53

= 150,75 4.530 +300n = 150,75 30 +2n 3

1,5 n = 7,5 1 n= 5 Banyak anak yang tingginya lebih dari 155 cm = 5 + 3 = 8 14. Jawaban: b Banyak data = 40 Median terletak di antara datum ke-20 dan datum ke21. Median

=

1

2

=

2 1

Jawaban: a

(nilai datum ke-20 + nilai datum ke-21) (150 +

berat badan

keseluruhan

Rata-rata = 34,2 =

Rata-rata tinggi anak = 150,75 n ×140 +× 145 Jawaban: c 155 6+× n 3160 +× 165 +× ⇔ 150 +× 912

⇔

17.

banyak siswa berat badan keseluruhan b

1 10 berat badan keseluruhan = 34,2 × 10 = 342 Misal x = berat badan siswa baru Rata-rata baru = 35 342 +x ⇔= 3511 ⇔342 + x = 385 ⇔x = 43 Jadi, berat badan siswa baru 43 kg. 18. Jawaban: b Misal ketiga bilangan tersebut 2n – 2, 2n, 2n + 2. Rata-rata = 18. 2n −+ + + 2 2n 2n 2 2 = 18 6n =54 n =9 Bilangan terbesar = 2n + 2 = 2 × 9 + 2=20 Bilangan terkecil = 2n – 2 = 2 × 9 – 2=16 –––––––––––––––––––––––––––––––––––– Selisih = 4 Jadi, selisih bilangan terbesar dan bilangan terkecil 4.

150) = 150 Nilai datum yang paling sering muncul adalah 36. Jadi, modus data 36.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 19. Jawaban: b xy z 62d322e31000000000000000000000000000000000000000000000000000000 n(A)= 6 n(A) 6ffd8ffe000104a46 1 = ++

9

3

⇔ x+y+z ++++ =27 xyzab 5 5 27 + a + b =55 2 a + b =28 ab 28

= 11

+

Rata-rata a dan b =22 = = 14. 20. Jawaban: b Misal tinggi tanaman A = 2n tinggi tanaman B = 3n tinggi tanaman C = 5n 2n ++ 3n 5n = 3 3

60

10n = 180 1

n= 18 Tinggi tanaman A = 2n = 2 × 18 = 36 cm. 21. Jawaban: c Faktor prima dari 30 adalah 2, 3, dan 5. A = kejadian jumlah mata dadu yang muncul lebih 2, 3, dan 5 = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (4, 1)} n(A) = 7 7 P(A) =

36

P(A′) = 1 – P(A) =1–

7

36 29

= 36 Jadi, peluang muncul jumlah mata dadu yang 2 muncul faktor prima dari 30 adalah 9 . 36

22. P(A) = peluang terambil buah Jawaban: d 25 25 rambutan = 25 = ++1718 60 P(B) = peluang terambil buah jeruk 17 17 == 25 ++1718 60

Oleh karena A ∩B = { } maka P(A ∩B) = 0. Peluang terambil rambutan atau jeruk: P(A ∪ B) = P(A) + P(B) 25 17 6060

=+

42 7 6010

==

23. Jawaban: c n(S) = 12 A = kejadian muncul mata dadu genap = {(2, A), (2, G), (4, A), (4, G), (6, A), (6, G)}

49460001020100c P(A) = = = 800c80000ffe20c5 n(S)122 B = kejadian muncul mata dadu 84943435f50524f4 faktor prima6494c4500010100 dari 6 000c484c696e6f02 = {(2, A), (2, G), (3, A), (3, 1000006d6e74725 G)} n(B)= 4247422058595a20 (A ∩B) = {(2, A), (2, G)} n(A ∩ B) = 2 P(A ∪B) = 07ce00020009000 P(A) + P(B) 600310000616373 – P(A ∩B) 704d53465400000 000494543207352 642 =+ – 474200000000000 121212 82 000000000000000 = =00f6d6000100000 123 000d32d48502020 24. 000000000000000 Misal: A = {pengendara motor tidak 000000000000000 Jawaban: c SIM} 000000000000000 membawa B = {pengendara motor 000000000000000 tidak membawa STNK} 000000000000000 x = banyak pengendara motor yang 000000000000000 tidak membawa SIM dan STNK Diagram 000011637072740 Venn: 000015000000033

646573630000018 AB 3540000006c777470 –x 74000001f000000 014626b70740000 020400000014725 8595a0000021800 0000146758595a0 000022c00000014 6258595a0000024 Jumlah pengendara motor yang kena operasi = 50 000000014646d6e ((35 – x) + x + (25 – x) = 50 640000025400000 60 – x =50 6 070646d64640000 x = 60 – 50 = 10 Banyak pengendara yang 02c400000088767 hanya membawa SIM: n(A) = 35 – x = 35 – 10 = 565640000034c00 25 Peluang terpilih pengendara motor yang 000086766965770 hanya 00003d400000024 6c756d69000003f 8000000146d6561 730000040c00000 024746563680000 04300000000c725 3 452430000043c00 00080c675452430 000043c0000080c 625452430000043 c0000080c746578 7400000000436f7 079726967687420 286329203139393 8204865776c6574 742d5061636b617 26420436f6d7061 6e7900006465736 300000000000000 127352474220494 54336313936362d 322e31000000000 000000000000012 735247422049454 336313936362d32 2e3100000000000 000000000000000 000000000000000 0000000000000 x S

25 – x

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 ffd8ffe000104a46 Banyak siswa = 36 B 3 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 3. x:y=3 :2⇔x 49460001020100c 14 1 – x + x + 10 – x + 5 =36 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 = Rata-rata y 800c80000ffe20c5 nilai x= 7 2 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 ×+×+ (43) + 6)(64) 5x (11 ×+ +×+× 7y (5 8) (4 9) 84943435f50524f4 Peluang terpilih siswa gemar Matematika = 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 46 178 x11 ++ y 545x 6494c4500010100 +++ +++ dan n(x) 7 Fisika: P(x) =n(S = . 7y 000c484c696e6f02 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 36 ⇔ 5,88 = 30 ++xy 1000006d6e74725 ) 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 26. Jawaban: a P(bayi terkena 178+× 35 y +7y 247422058595a20 2 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 campak) = 0,05 Fh(bayi terkena campak) = ⇔ 5,88= 3 07ce00020009000 30 +y +y 2 00000000000000000000000000000000000000000000000 n × P(bayi terkena campak) = 1.500 × 600310000616373 0,05 = 75 Jadi, terdapat 75 bayi yang terkena campak. 27. Jawaban: c Kemungkinan banyak bibit yang tumbuh = 0,85 × 6.000 = 5.100 28. Jawaban: d A = kejadian muncul mata dadu faktor dari 10 = {1, 2, 5} n(A) = 3 n(A) 3 1 P(A) = n(S) = = 6

29

178 + y 704d53465400000 ⇔ 5,88 = 5 000494543207352 30 +y 2 474200000000000 5 000000000000000 ⇔ 5,88(30 + y)=22 2178 29 00f6d6000100000 176,4 1 = 178 + 14,5y + y + 14,7y 000d32d48502020 0,2y = 178 –000000000000000 176,4 1,6 1 y= = 8 000000000000000 000000000000000 0,2 000000000000000 33 Diperoleh x 000000000000000 = y= ×8= 2

2

2 1

Fh(A) = P(A) × n ⇒8 = × n ⇔n = 2Jadi, percobaan tersebut dilakukan 16 sebanyak 16 kali. 29. P(berawan) = 1 – (P(cerah) + P(hujan)) Jawaban: a = 1 – (0,6 + 0,3) = 0,1 n = banyak hari pada bulan Mei = 31 Fh(berawan) = 0,1 × 31 = 3,1 ≈3 hari Jadi, pada bulan Mei diperkirakan berawan selama 3 hari. 30. Jawaban: b P(bola tepat masuk ring) = 0,65 n = banyak pelemparan Fh(bola ttepat masuk ring) = 52 n × P(bola tepat masuk ring) = 52 n n × 0,65 = 52 5 52 n= 0,65

⇔ n = 80 kali Jadi, pemain basket telah melakukan pelemparan sebanyak 80 kali. B. 1. a. Data kualitatif Uraian Data kuantitatif Data kuantitatif Data kualitatif 2. Populasinya adalah seluruh air di sepanjang Sungai Ciliwung. Sampelnya adalah air yang diambil dari beberapa tempat di Sungai Ciliwung.

2

000000000000000 Nilai yang paling banyak diperoleh siswa 12 000011637072740 adalah 5. Dengan demikian modus data 5. Banyak data000015000000033 = 50. 1 646573630000018 Median = 40000006c777470 (nilai datum ke-25 + nilai 2 1 74000001f000000 datum ke-26) = (6 + 6) = 2 Jadi, modus data 5 dan 014626b70740000 6. 020400000014725 mediannya 6. 8595a0000021800 4. Misal: Jumlah pelajar di kelurahan 0000146758595a0 tersebut = N Banyak pelajar SMP, SD, dan 000022c00000014 SMK berturutturut: 6n, 4n, dan n. 6258595a0000024 Banyak pelajar SMA = 45% 000000014646d6e × N 45 640000025400000 ⇔ 90= × N 100100 070646d64640000 02c400000088767 ⇔ N= 90 × = 200 45Banyak pelajar SMP + SD + SMA = 565640000034c00 200 – 90 2 000086766965770 6n + 4n + n=110 6 00003d400000024 n(A) 25 1 11n = 110 6c756d69000003f 1 . membawa SIM adalah == 2 8000000146d6561 n= 10 Diperoleh 6n =n(S)50 6 × 10 = 60 730000040c00000 25. Jawaban: a Misal: A = {siswa gemar 4n = 4 ×B10 = 40 Jadi, banyak pelajar Matematika} =024746563680000 {siswa gemar Fisika} x= 04300000000c725 SMP, SD, tidak dan SMK berturutturut banyak siswa gemar Matematika60, dan40, 452430000043c00 dan 10. 00080c675452430 5. Misal: p = banyak siswa yang 000043c0000080c memperoleh nilai 8 Rata-rata = 6 Fisika 625452430000043 4) (40 ×+×+ 5) (706) ×+×+ (p 8) (10 ×10) Diagram (20c0000080c746578 ⇔ =6 20 + + ++40 70p10 800 Venn: S +8p AB7400000000436f7 ⇔=6 079726967687420 –x 114 +p 140 286329203139393 800 + 8p = 840 + 6p 8 8204865776c6574 8p – 6p = 840 – 800 8 742d5061636b617 2p = 40 40 4 26420436f6d7061 p= = 20 6e7900006465736 2 5 300000000000000 127352474220494 54336313936362d 322e31000000000 000000000000012 735247422049454 336313936362d32 2e3100000000000 000000000000000 000000000000000 0000000000000 x

10 – x

B ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 C d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 P 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 Q A 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 (A, A) 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 (A, B) 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 (A, C) Wakil Presiden 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 (A, P) Banyak anak yang memperoleh nilai di (A, Q) 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 atas nilai rata-rata = 20 + 10 = 30. B Skor 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 (B, A) 6. Frekuensi 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 Pr (B, B) 3 es 0000000000000000000000000000000000000000000000 (B, C) 1

Skor terbanyak Skor rata-rata

id en

4 4 5 8 6 6 7 3 yang 8 3

diperoleh peserta adalah 5.

(1 3) (4×+×+×+×+×+× 4) (85) (66) (3 7) (3 8) +++++ 1486 3 3 140

== 5,6 25 c.

Skor rata-rata setelah diskualifikasi (8×+×+×+× 5) (6 6) = +++ 8533 121 == 6,05 20

(3 7) (3 8)

7. a. A = kejadian anak yang lahir paling sedikit dua anak laki-laki = lahir 2, 3, atau 4 anak laki-laki A′= kejadian anak yang lahir paling banyak 1 anak laki-laki = {(L, P, P, P), (P, L, P, P), (P, P, L, P), (P, P, P, L)}

(B, P) (B, Q) C (C, A) (C, B) (C, C) (C, P)pasangan Ruang sampelnya adalah (C, Q) selain yang diarsir. P (P, A) n(S) = 25 – 5 = 20 (P, B) perempuan K = kejadian terpilih keduanya (P, C) = {(P, Q), (Q, P)} n(K)(P, = P) 2 (P, Q) Peluang terpilih keduanya Q perempuan (Q, A) n(K) (Q, B) = P(K) = (Q, C) n(S) 21 (Q, P) == (Q, Q) 2010 Peluang terpilih keduanya laki-laki = 1 – P(k) 19

= 1 – 1010 = d. L = kejadian terpilih presiden perempuan dan wakil presiden laki-laki = {(P, A), (P, B), (P, C), (Q, A), (Q, B), (Q, C)} n(L) = 6 63

n(A′) = 4 n(A ) ′41

P(L) =

P(A′) = n(S

Jadi, peluang presidennya perempuan dan 10 wakil presidennya laki.

)

==

164

13 4

P(A) = 1 – P(A′) = 1 –

=

4

Jadi,

peluang memiliki paling sedikit dua anak 3 adalah . laki-laki 4

b. B = kejadian lahir dua anak perempuan = {(L, L, P, P), (L, P, L, P), (P, L, P, L),(P, L, L, P), (L, P, P, L), (P, P, L, L} n(B) = 6 n(S) = 16 n(B) 6

3

P(B) = = = n(S)168 Jadi, peluang memiliki dua anak perempuan 3 adalah . 8

8. a. Calon laki-laki = A, B, C Calon perempuan = P, Q

20

3

=

10

laki 9. Misal banyak bola putih mula-mula = x Peluang terambil bola putih setelah dimasukkan 6 2

bola putih = . x6 2+ 3= ⇔ 21+63 2 2

x + 6 = 27 × x + 6 = 18 x = 18 – 6 = 12 Jadi, banyak bola putih mula-mula 12.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 62d322e31000000000000000000000000000000000000000000000000000000

4.

Latihan Ulangan Akhir Semester 1

Jawaban: b Perhatikan bahwa segitiga kecil sebangun dengan segitiga besar sehingga diperoleh: 6x 1x = ⇔ = 1816

A. Pilihan316 +8 Persegi Panjang Ganda 5. 1. Jawaban: c 11× AC × BD = × 6 × 9 = 27 cm2LABCD = 22L= Panjang 16 + x= 3x = LABCD 279 Oleh Lebar karena PQRS dan ABCD sebangun maka QS PR ABCD . LABCD 42k2 = 2x = 16 LPQRS Jawaban: c = = k dan AB = k=2 18 sehingga diperoleh BDAC ⇔ k = 3 9 QS2 QS2 = ⇔ = ⇔QS cm x = 8 cm AD = 9 cm ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f0210000 Jadi, x = 8 FGHI 06d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000 cm. GH = 6 cm 000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000000000000 HI = 4 cm 00000000000000000000000000000000000000000000000116370727400000150000000336465736300000184000000 AEFD AD = 9 cm 6c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c000000 AE = 6 cm 146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c000000 BCFE 8676696577000003d4000000246c756d69000003f8000000146d6561730000040c000000247465636800000430000000 BE = 12 cm 0c725452430000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f707 BC = 9 cm 97269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000 000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d3 22e31000000000000000000000000000000000000000000000000000000 = 6 cmBD3 93 PR2 PR2 = ⇔ = ⇔PR = 4 cmAC3 63 Jadi, panjang PR dan QS berturut-turut 4 cm dan 6 cm. 6. Jawaban: d Dua lembar kertas berbentuk persegi panjang sebangun sehingga: A1 p1363 = = = Ap2 2274 Perbandingan luas kedua kertas: 2L13 9 ⎛⎛ = ⎛⎛ = L24 16 ⎛⎛Jadi, perbandingan luas kedua kertas tersebut 16 : 9 atau 9 : 16. +x

1Jawaban: c ∠A = ∠E = 60°, ∠B= ∠F = 50°, dan

∠C = ∠D = 70°. Oleh karena ∆ABC dan ∆DEF 3 sebangun maka sisisisi di hadapan sudut yang bersesuaian sama panjang sehingga BC = DF, AC = DE, dan AB = EF. Jadi, pernyataan yang benar pilihan c. 10. Banyak percobaan: n = 60. Banyak 2Jawaban: c ∆TQP dan ∆SPQ kongruen dengan anggota ruang sampel: n(S) = 20. A = ∠PTQ = ∠QSP, ∠SPQ = ∠TQP, dan ∠TPQ = kejadian terambil kartu bernomor prima ∠SQP. Jadi, pernyataan yang benar pilihan c. = {31, 37, 41, 43, 47} n(A) = 5 Frekuensi harapan terambil kartu bernomor prima: F(A) = P(A) × n nn(A)

= n(S) × 60 5

= 20 × 60 = 15

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe0001 ffd8ffe000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 04a4649460 104a464946 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 001020100c 0001020100 ffd8ffe000104a4 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 800c80000ff c800c80000 64946000102010 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 e20c584943 ffe20c58494 0c800c80000ffe2 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 435f50524f4 3435f50524f 0c584943435f505 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce00 9. 11. 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 6494c45000 46494c4500 24f46494c450001 020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d48502020000000000000 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 10100000c4 010100000c 1 ∆ABC dan ∆ADE sebangun dengan 000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000033646573630000018400 0100000c484c696 L∆PSR = ×bPS × SR 00006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c000000146258595a00000240000000146 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 84c696e6f0 484c696e6f Jawaban: d Jawaban: 2 21 perbandingan BC ACAB e6f021000006d6e 46d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d4000000246c756d69000003f8000000146d656173 sisi-sisi yang bersesuaian == 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 21000006d6 021000006d 0000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000080c625452430000043c0000080c746578740000000 74725247422058 DEAEAD 93 125 = × PS × 25 1 0000000000000000000000000000000000000000000000 e747252474 6e74725247 0436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e79000064657363000000000000001273524742204945 ffd8ffe000104a464 2 PS = 10 cm Perhatikan bahwa ∆PSQ sebangun 4336313936362d322e31000000000000000000000012735247422049454336313936362d322e31000000000000000000000000000000000000000000000000 = 595a2007ce0002 = . 22058595a2 422058595a 9460001020100c80 ffd8ffe000104a464 2 6 00090006003100 000000 dengan ∆RSP 007ce00020 0c80000ffe20c58494 9460001020100c800 2 2007ce0002 00616373704d53 ⎛⎛ 3 0090006003 0009000600 Luasc80000ffe20c584943 3435f50524f46494c ∆ABC = ⎛⎛ luas ∆ADE. 46540000000049 ⎛⎛ 2 1000061637 3100006163 4500010100000c48 435f50524f46494c4 sehingga diperoleh: SQ PS 45432073524742 3704d53465 73704d5346 4c696e6f021000006 500010100000c484c 4 00000000000000 ⇔Luas ∆ADE = luas ∆ABC PS2 102 = ⇔SQ = = = 4 cm 4000000004 5400000000 d6e7472524742205 696e6f021000006d6 9 00000000000000f PSSR SR25Jadi, panjang SQ 9454320735 4945432073 8595a2007ce000200 e747252474220585 6d600010000000 = 4 cm. 4 5247420000 2474200000 = × 27 0900060031000061 95a2007ce00020009 0d32d485020200 9 0000000000 12. Jawaban: b ∆ABC sebangun dengan 6373704d53465400 0006003100006163 = 1200000000000000 cm Luas ∆ADE maka berlaku perbandingan: 0000000000 0000004945432073 73704d5346540000 00000000000000 trapesium BCED =0000f6d600 000f6d6000 5247420000000000 0000494543207352 luas00000000000000 ∆ABC – luas ∆ADE 100000000d 0100000000 0000000000000000 4742000000000000 2 x4 6 AE DE + = 2700000000000000 – 12 = 15 cmd32d485020 32d4850202 00f6d600010000000 =⇔= 0000000000000000f 00000000000000 x8 8 ACBC + 0000000000 2000000000 0d32d48502020000 6d6000100000000d 00000000000000 8x + 32 = 6x + 48 0000000000 10. Jawaban: a Misalkan 0000000000000000 32d4850202000000 00000000011637 2x = 16 ∆ABC = 0000000000000000 segitiga besar0000000000 07274000001500 x= 8 Jadi, nilai 0000000000 ∆DEB = segitiga kecil Perhatikan bahwa 0000000000000000 00000336465736 x = 8. 0000000000 ∆ABC dan ∆DEB sebangun sehingga 0000000000000000 30000018400000 0000000000 diperoleh: 0000000000000000 06c77747074000 13. 40 cm 0000000000 0000000000011637 0000000000000000 001f00000001462 C= DEDB A ffd8ffe000104a46494600010 1 0000000000 0727400000150000 0000000001163707 6b707400000204 20100c800c80000ffe20c58494Jawaban: Misal tinggi tabung = t2maka d = t atau r AC BC d tabung 0000000000 0003364657363000 24 Volume = πr t 2740000015000000 1 000000147258593435f50524f46494c45000101 DE 20 = t. 0000011637 0000001163 00000c484c696e6f021000006 001840000006c777 ⇔= 0336465736300000 5a000002180000 40 25 B 20 cm d6e74725247422058595a200 0727400000 7072740000 47074000001f00000 1840000006c777470 2 00146758595a00 22 1 7ce000200090006003100006 ⎛⎛ ⇒ 539 × ⎛⎛ × t =t 1500000003 0150000000 0014626b70740000 74000001f00000001 16373704d534654000000004 74 ⎛ ⎛⎛ ⇔DE00022c00000014 = 800 3646573630 3364657363 0204000000147258 4626b70740000020 6258595a000002 25 E = 32 cm945432073524742000000000 3 ∆ABC dan ∆DEB t = 2.744 D 0000184000 0000018400 0000000000000000000f6d600 595a000002180000 4000000147258595 4000000014646d merupakan segitiga sama kaki sehingga t = 14 cm 0100000000d32d4850202000 0006c77747 00006c7774 ffd8ffe000104a464946 00146758595a0000 a000002180000001 6e640000025400 000000000000000000000000 0001020100c800c80000 diperoleh: 074000001f 7074000001 022c000000146258 46758595a0000022c Jawaban: c Luas 000070646d6464 ffe20c584943435f50524 000000000000000000000000 14. 0000000146 f000000014 595a000002400000 000000146258595a f46494c4500010100000 000000000000000000000000 permukaan tabung tanpa tutup = ffd8ffe000104a46494600 000002c4000000 2 c484c696e6f021000006 t1= 0000024000000014 26b7074000 626b707400 01020100c800c80000ffe2 000000000000000000001163 0014646d6e640000 2 1⎛ 4 luas alas + luas selimut tabung = 88767565640000 d6e74725247422058595 0c584943435f50524f4649   0002040000 ⎛a2007ce0002000900060 2707274000001500000003364 20 0020400000 025400000070646d 646d6e6400000254 πr(r + 2t) = πr(r + 2 × 3r) = πr × 7r 034c0000008676 4c4500010100000c484c6 657363000001840000006c77 0310000616373704d53 0147258595 0014725859 2 6464000002c40000 00000070646d6464 96e6f021000006d6e7472 696577000003d4 = 7πr 747074000001f000000014626 = 4654000000004945432 5247422058595a2007ce0 a000002180 5a00000218 0088767565640000 000002c4000000887 b70740000020400000014725 0735247420000000000 000000246c756d 002000900060031000061 0000014675 0000001467 8595a0000021800000014675 000000000000000000f6 034c000000867669 67565640000034c00 6373704d5346540000000 2 69000003f800000 t2= 1 ⎛d6000100000000d32d48 2 58595a0000 8595a0000022c00000014625 049454320735247420000 8595a00000 6577000003d40000 0000867669657700 0146d656173000 5020200000000000000 ⎛2 ⎛ 000000000000000000000 8595a0000024000000014646 22c0000001 022c000000 00246c756d690000 15. Jawaban: b Oleh karena digulung 0003d4000000246c7 0000000000000000000 000f6d6000100000000d3 0040c000000247 d6e640000025400000070646 0000000000000000000 46258595a0 146258595a 2d4850202000000000000 03f8000000146d656 menjadi kerucut, jari-jari lingkaran berubah d6464000002c400000088767 = 56d69000003f80000 46563680000043 0000000000000000000 000000000000000000000 0000240000 0000024000 565640000034c00000086766 1730000040c00000 menjadi garis pelukis kerucut. Misalkan α= 00146d6561730000 0000000000000000000 00000000c72545 000000000000000000000 1 1 0000011637072740000 96577000003d4000000246c7 120° dan r = jari-jari kerucut. 00014646d6 000014646d 0247465636800000 040c0000002474656 000000000000000000000 L1 =2430000043c000 × AC × t1= ×0150000000336465736 40 × k 56d69000003f8000000146d65 000000000000000000001 22= 20 × 15 e640000025 6e64000002 4300000000c72545 3680000043000000 r 3000001840000006c777 61730000040c000000247465 163707274000001500000 15 0080c675452430 α= ×k 360° 4000000706 5400000070 47074000001f00000001 2430000043c00000 00c72545243000004 6368000004300000000c7254 003364657363000001840 000043c0000080c s 4626b70740000020400 000006c77747074000001 46d6464000 646d646400 52430000043c0000080c67545 80c6754524300000 3c0000080c6754524 r 2 62545243000004 0000147258595a000002 = 300 cm f000000014626b7074000 120° 1 = × 360° k k 2430000043c0000080c625452 002c400000 0002c40000 43c0000080c625452 30000043c0000080c 18000000146758595a00 00204000000147258595a 3c0000080c74657 430000043c0000080c7465787 00022c00000014625859 120° = rk × 24° 0887675656 0088767565 000002180000001467585 430000043c000008 625452430000043c0 11 400000000436f707972696768 = × DE × t2 = × 32 × 12 = 192 L 87400000000436f 5a000002400000001464 95a0000022c0000001462 222L – L2 = 300 –640000034c 192 = 40000034c0 0c74657874000000 000080c7465787400 742028632920313939382048 6d6e6400000254000000 rk = 5 cm 1 70797269676874 58595a000002400000001 2 cm 70646d6464000002c400 0000086766 0000008676 65776c6574742d5061636b61 2 00436f70797269676 4646d6e64000002540000 000000436f7079726 108 20286329203139 cm 0000887675656400000 726420436f6d70616e7900006 0070646d6464000002c40 9657700000 6965770000 8742028632920313 9676874202863292 34c00000086766965770 3938204865776c 000008876756564000003 465736300000000000000127 3d40000002 03d4000000 00003d4000000246c756 93938204865776c6 0313939382048657 4c0000008676696577000 6574742d506163 352474220494543363139363 d69000003f8000000146 003d4000000246c756d69 46c756d690 246c756d69 574742d5061636b6 76c6574742d506163 62d322e31000000000000000 6b61726420436f6 d6561730000040c00000 000003f8000000146d656 00003f8000 000003f800 1726420436f6d7061 6b61726420436f6d7 0247465636800000430 1730000040c0000002474 d70616e7900006000000012735247422049454 336313936362d322e3100000 0000000c725452430000 000146d656 0000146d65 656368000004300000000 6e79000064657363 0616e79000064657 46573630000000 043c0000080c67545243 000000000000000000000000 c725452430000043c0000 1730000040 6173000004 0000000000000012 3630000000000000 0000043c0000080c6254 00000001273524 000000000000000000000000 080c675452430000043c0 52430000043c0000080c c000000247 0c00000024 7352474220494543 000080c62545243000004 0127352474220494 74220494543363 0 25 cm 7465787400000000436f 3c0000080c74657874000 4656368000 7465636800 36313936362d322e 54336313936362d3 7079726967687420286 13936362d322e3 00000436f707972696768 0043000000 0004300000 3292031393938204865 3100000000000000 22e3100000000000 742028632920313939382 10000000000000 776c6574742d5061636b 04865776c6574742d5061 00c7254524 000c725452 0000000012735247 0000000000012735 00000000012735 61726420436f6d70616e 636b61726420436f6d706 30000043c0 430000043c 4220494543363139 2474220494543363 7900006465736300000 16e790000646573630000 24742204945433 0000000001273524742 000080c675 0000080c67 000000000012735247422 36362d322e310000 13936362d322e310 6313936362d322 2049454336313936362 049454336313936362d32 4524300000 5452430000 0000000000000000 d322e310000000000000 e3100000000000 2e3100000000000000000 0000000001273524742 43c0000080 043c000008 000001273524742204945 0000000000000000 00000000000000 2049454336313936362 4336313936362d322e310 c625452430 0c62545243 0000000000000000 d322e310000000000000 00000000000000 000000000000000000000 ⎛ ⎛0000043c00 000043c000 0000000000000000000 000000000000000000000 00000000000000 00000 00DB −BC⎛−⎛ DE = AC20  = 0000000000000000000 2 0080c74657 0625 0400 −400 = 00080c7465 25 −16 −−256  000000000002 ⎛ 000 22

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104a ffd8ffe000104 ffd8ffe00 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4649460001020 a46494600010 0104a464 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 100c800c80000f 20100c800c800 94600010 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 fe20c584943435 00ffe20c58494 20100c80 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 f50524f46494c4 3435f50524f46 0c80000ff 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 500010100000c 494c45000101 e20c5849 16. 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 484c696e6f0210 00000c484c696 43435f50 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 00006d6e74725 e6f021000006d 524f4649 t= 247422058595a 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6e7472524742 4c450001 Jawaban: a 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 2007ce0002000 2058595a2007 0100000c 2 0000000000000000000000000000000000000000000000 ce0002000900 484c696e = 9000600310000 25 2 616373704d534 060031000061 6f021000 − 7 = 6540000000049 576 = 24 cm 1 6373704d5346 006d6e74 4543207352474 540000000049 72524742 2000000000000 454320735247 2058595a 2 Volume kerucut = πr t 3 1 22 0000000000000 420000000000 2007ce00 2 000f6d60001000 000000000000 02000900 = × × 7 × 24 37 00000d32d4850 000000f6d6000 06003100 = 1.232 cm3 2020000000000 100000000d32 00616373 0000000000000 d48502020000 704d5346 17. Jawaban: b Volume 0000000000000 000000000000 54000000 bola = 20000000000000 b · 56,52 000000000000 00494543 4 30000000000000 000000000000 20735247 πr = = 113,04 3 0000000000000 000000000000 42000000 4 × 0000000000000 3,14 × r3 = = 113,04 000000000000 00000000 3 1 3 000000000000 00000000 3 0000000116370 r = 7274000001500 113,04 × × 000000000000 000000f6 4 3 3,14 ffd8ffe000104a464 0000033646573 000000011637 d6000100 9460001020100c800 r3 = =6300000184000 27 072740000015 000000d3 c80000ffe20c584943 0006c77747074 435f50524f46494c45 000000033646 2d485020 r = 3 cm Jadi, d = 2 × r = 2 × 3 00010100000c484c6 000001f0000000 573630000018 20000000 = 6 cm. 96e6f021000006d6e 14626b7074000 40000006c777 00000000 74725247422058595 18. Jawaban: d Luas alas 0020400000014 47074000001f0 00000000 a2007ce0002000900 kerucut 7258595a00000 = 78,5 cm22 06003100006163737 00000014626b 00000000 04d53465400000000 πr 2 2180000001467 = 78,5 = 707400000204 00000000 49454320735247420 000000147258 00000000 3,1458595a0000022 × r2 = 78,5 78,5 7 00000000000000000 595a00000218 00000000 2 c000000146258 0000000000f6d6000 r = 3 100000000d32d4850 595a000002400 3,14 000000146758 00000000 ffd8ffe000104a464946 20200000000000000 r2 =0000014646d6e 25 595a0000022c 00000000 0001020100c800c8000 00000000000000000 0ffe20c584943435f505 r=5 cm 6400000254000 000000146258 00000000 00000000000000000 24f46494c45000101000 00070646d6464 595a00000240 00000000 00c484c696e6f0210000 00000000000000000 06d6e74725247422058 00000000000000000 000002c400000 00000014646d 00000000 595a2007ce000200090 00000000000001163 0887675656400 6e6400000254 11637072 0060031000061637370 70727400000150000 4d53465400000000494 00034c0000008 00000070646d 74000001 00033646573630000 5432073524742000000 6766965770000 6464000002c4 50000000 01840000006c77747 0000000000000000000 074000001f0000000 03d4000000246 000000887675 33646573 000f6d6000100000000 d32d485020200000000 14626b70740000020 c756d69000003f 65640000034c 63000001 0000000000000000000 4000000147258595a 8000000146d65 000000867669 84000000 0000000000000000000 00000218000000146 0000000000000000000 61730000040c0 6577000003d4 6c777470 758595a0000022c00 0000000000000000000 0000146258595a000 0000024746563 000000246c75 74000001 0000000000011637072 0024000000014646d 7400000150000000336 6800000430000 6d69000003f80 f0000000 6e640000025400000 4657363000001840000 0000c72545243 00000146d656 14626b70 006c77747074000001f0 070646d6464000002 0000043c00000 1730000040c0 74000002 00000014626b7074000 c4000000887675656 0020400000014725859 40000034c00000086 80c6754524300 000002474656 04000000 5a00000218000000146 76696577000003d40 00043c0000080c 368000004300 14725859 758595a0000022c0000 15 00000246c756d6900 00146258595a0000024 6254524300000 000000c72545 5a000002 0003f8000000146d6 000000014646d6e6400 43c0000080c746 2430000043c0 18000000 561730000040c0000 00025400000070646d6 464000002c400000088 00247465636800000 5787400000000 000080c67545 14675859 Tinggi kerucut = 767565640000034c000 4300000000c725452 k 436f7079726967 2430000043c0 5a000002 0008676696577000003 430000043c0000080 6874202863292 000080c62545 = c675452430000043c 225 −25 = d4000000246c756d690 200 = 10 2c000000 00003f8000000146d65 0313939382048 2430000043c0 14625859 0000080c625452430 61730000040c0000002 000043c0000080c74 65776c6574742 000080c74657 5a000002 4746563680000043000 65787400000000436 00000c7254524300000 d5061636b6172 874000000004 40000000 f7079726967687420 43c0000080c67545243 6420436f6d7061 36f7079726967 14646d6e 0000043c0000080c625 28632920313939382 452430000043c000008 6e79000064657 687420286329 64000002 04865776c6574742d 0c74657874000000004 5061636b617264204 3630000000000 203139393820 54000000 36f7079726967687420 36f6d70616e790000 0000127352474 4865776c6574 70646d64 2863292031393938204 64657363000000000 865776c6574742d5061 2204945433631 742d5061636b 64000002 00000127352474220 636b61726420436f6d7 3936362d322e3 61726420436f6 c4000000 49454336313936362 0616e79000064657363 0000000000000012735 d322e310000000000 1000000000000 d70616e79000 88767565 2474220494543363139 00000000000012735 0000000000127 064657363000 64000003 36362d322e310000000 24742204945433631 0000000000000001273 3524742204945 000000000001 4c000000 3936362d322e31000 5247422049454336313 4336313936362 273524742204 86766965 00000000000000000 936362d322e31000000 00000000000000000 d322e31000000 945433631393 77000003 0000000000000000000 00000000000000000 0000000000000000000 0000000000000 6362d322e310 d4000000 22 000000000022

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104a d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4649460001020 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 100c800c80000f 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 fe20c584943435 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 f50524f46494c4 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 500010100000c 2 20. 27.Jawaban: Jawaban: 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 ⇔ 484c696e6f0210 V= πr t2 t 2 3 b c 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 00006d6e74725 = 3,14 × 3 × 4 = 113,04 cm 1 ffd8ffe000104a4649460001020100c800c8 Luas permukaan benda 11 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 247422058595a 0000ffe20c584943435f50524f46494c4500 Volume kerucut= 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 2007ce0002000 = × luas permukaan bola + 2 × luas 010100000c484c696e6f021000006d6e747 3 2 4 πr t 0000000000000000000000000000000000000000000000 19000600310000 25247422058595a2007ce00020009000600 lingkaran ⇔ V616373704d534 = × 3,14 ×132 × 2 2 11 k 310000616373704d534654000000004945 3 = × 4πr2 + 2 × πr2 42 6540000000049 = 3,14 × 3 × 2 = 18,84 cm3 Misalkan 4320735247420000000000000000000000 fxi i Nilai 4543207352474 volume pasir dalam tabung setelah bangun 1 rata-rata = 000000f6d6000100000000d32d485020200 f 8203042 16 1 = × 3,14 ×i+++++ 102 + 2 × × 2000000000000 ruang dibalik = V 9 0000000000000000000000000000000000 s 42= 314 +2 314 = 628 = 3,14 × 10 Vs =0000000000000 V–V = 113,04 – 18,84 = 94,2 0000000000000000000000000000000000 20 cm2 t k 000f6d60001000 2 = 6,25 Nilai kurang dari 6,25 adalah 4, 5, 0000000000000000000000000116370727 3 V= cm ss πr t 00000d32d4850 dan 6. Banyak siswa yang memperoleh 4000001500000003364657363000001840 ⇔94,2 = 3,14 × 32 × t 2020000000000 21. 000006c77747074000001f000000014626b a Jari-jari tabung: r= s 94,2 94,2 nilaiJawaban: kurang dari rata-rata =2+ 4+5= 0000000000000 3,5 707400000204000000147258595a000002 cm 11 Panjang orang. kotak = 5 × diameter ⇔ t = =2 = 3,33 cm s 3,14 ×3 0000000000000 tabung = 5 × 3,5 × 2 = 35 cm Lebar 18000000146758595a0000022c000000146 28. 28,260000000000000 Kedalaman pasir = 258595a0000024000000014646d6e64000 kotak = 4 × diameter tabung = 4 × 3,5 x = 0000000000000 t1 + t n128 =7 s × 20025400000070646d6464000002c4000000 = cm Tinggi kotak = tinggi =2+ 0000000000000 Jawaban: x = 171 cm orang –b cm 88767565640000034c00000086766965770 1 = tabung 10 3,33 = 0000000000000 n = 3 00003d4000000246c756d69000003f80000 2 cm 0000000116370 24. Jawaban: c 5,33 1 00146d6561730000040c000000247465636 x (tinggi badan yang orang – kotak Volume =V V = rata-rata p × A × torang = 2= p cm k k 7274000001500 –x 8000004300000000c725452430000043c00 keluar) Volume kerucut = = 174 cm akhir 3 0000033646573 35 × 28 × 10 = 9.800 cm3 Volume 00080c675452430000043c0000080c62545 πrk2t6300000184000 1 2 −nx = V k – 11 2 2nx seluruh tabung ⇔ Vk = × 3,14 ×36 × 8 2430000043c0000080c7465787400000000 xakhir = b n1 Vb = 0006c77747074 436f7079726967687420286329203139393 −n 2 = 3,14 × 2 × 6 × 8 2 000001f0000000 7 171 −23 p 20πr t × 22 × 8204865776c6574742d5061636b61726420 = 20 (3,5) −× ⇔ 174 = 14626b7074000 73 72 436f6d70616e79000064657363000000000 = 301,44 cm3 10 0020400000014 2 1 0000012735247422049454336313936362 = 200 × ×1.197 3,5 ×−3p 3,5 7 4 3 7258595a00000 174 = 1 d322e31000000000000000000000012735 3 Volume bola = =×200 22 × –0,5 3 2180000001467 174 1 4 =×1.197 3p× 3,5 = 247422049454336313936362d322e31000 πrb 58595a0000022 7.700 cm Volume ruang kosong = Vk – Vb = 3p 3 = 1.197 – 696 0000000000000000000000000000000000 4 3 3 ⇔ Vc000000146258 = × 3,14 × 3 p = 167 cm Jadi, tinggicm badan rata-rata pemain 00000000000000000 9.800 – 7.700 = 2.100 b 3= 4 × 3,14 595a000002400 yang keluar 167 cm. × 32 = 0000014646d6e Jadi, volume ruang kosong tersebut 2.100 cm3. 113,04 cm3 6400000254000 Volume ruang2 di luar bola = Vk – Vb = 22. 2 s= 00070646d6464 3 L1 = luas permukaan 301,44 – 113,04 = 188,4 cm 000002c400000 22 = 0887675656400 5 + 12 Jawaban: balok 1c10 × 5 – × 3,14 × 42 = 10 × 2 00034c0000008 25. = 169 = 13 cm Luas = 500 – 25,12 = 474,88 cm2 L2 6766965770000 permukaan = Rata-rata = kerucut x 03d4000000246 = luas permukaan setengah luasc756d69000003f alas Jawaban: b –+ luas selimut ++++++++ + + k bola 1 –=5879 7679(k 2) + πrs = 78,5 + 8000000146d65 x=78,5 10 = × 4πr2 3,1461730000040c0 × 5 × 13 = 78,5 + 2 x 60 +2kcm2 6 204,1 = 282,6 3 7,6=0000024746563 = 2 × 3,14 × 42 = 100,48 cm2 Luas 4 1 10 6800000430000 19. permukaan bangun = L1 + L2 = 5 76 =0000c72545243 7 60 + 2k Volume jus jeruk = 4 liter 6 2k =0000043c00000 2 16 474,88 + 100,48 = 575,36 cm2 Jadi, 3 7 = 4 dm Jawaban: c k= 880c6754524300 3 luas bangun tersebut 575,36 cm2. = 4.000 cm8 9 00043c0000080 26. Volume gelas = 10 c625452430000 f Vb = πr2t = 3,14 × 32 × 043c0000080c7 1 23. 10 =4657874000000 31,4 Jawaban: b × 9 = 282,6 6 3 10 Diketahui: jari-jari tabung = jari-jari cm 00436f70797269 15 4.000 kerucutc = r tinggi kerucut = t1 tinggi 6768742028632 Jawaban: Banyak gelas =282,6 = 14,15 ≈ 15 10 1 9203139393820 tabung = t2 4 Median = (nilai data ke-24 + nilai 2 Volume pasir = volume tabung 1 diisi 15 buah Jadi,4865776c65747 banyak gelas yang dapat 1 penuh). 42d5061636b61 catatan gelas ke-15 tidak terisi (dengan data ke-25) 1 = (6 + 6) = fk 2 726420436f6d70 1 6,0 616e790000646 7 5736300000000 17 32 0000001273524 42 7422049454336 46 313936362d322 47 e310000000000 48 0000000000001 2735247422049 4543363139363 62d322e310000 0000000000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Banyak kartu bernomor lebih dari 5: 0,2a = 0,6b a 0,6 0 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 n(K) = 20 Peluang terambil kartu = 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 bernomor lebih dari 5 n(K) 20 5 b 0,2 = n(S)5213 == 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 ⇔ a : b = 0,6 : 0,2 = 3 : 1 Jadi, 0 0000000000000000000000000000000000000000000000 perbandingan banyak siswa di kelompok A 36. Jawaban: dan kelompok B adalah 3 : 1 a – mn++25 x= 1 3 mn 25 31. Jawaban: d Besar sudut + ++ pusat Matematika = 360° – 45° – 27 = 2 3 40° – 50° – 95° – 60° = 70° m + n + 25 = 81 m + n =56 40° 16 7 = ⇔n 70°n4

=

× 16 = 28

x= 2

Jadi, banyak siswa yang gemar Matematika ada 28 orang. 32. Jawaban: d Misal frekuensi nilai 7 = 5x maka frekuensi nilai 9 = 4x. ⋅+⋅+ (4 6) (2(12 ⋅+⋅⋅+ (27) 10) ⋅+ (4x 9) 5) 8)(5x ++ 124x ++ + 4 245x

Nilai rata-rata= ⇔

7,8 =

170 +71x 22 + 71x +9x +

41 = 4

++–++ mn25 pq 5 56 25p q + ++ + 5

p + q = (41 × 5) – (56 + 25) = 205 – 81 = 124

+ – pq 124 x= 3 = = 62 22Jadi, rata-rata dari p dan q adalah 62. 37. Jawaban: c Banyak kartu seluruhnya = n(S) = 52 Banyak kartu King = n(K) = 4

7 7,8(22 + 9x) = 170 171,6 + 70,2x = 170 + 71x 1 n(K) 4 1 P(K)= =n(S) = 71x – 70,2x = 171,6 – 170 7 54 13 0,8x = 1,6 0 4 38. Jawaban: c n(S) = 30 Misalkan B = x= 2 Banyak siswa yang mendapat nilai 7 adalah 5 × 2 = 10 orang. Banyak siswa yang kelas yang memiliki sedikitnya memperoleh nilai kurang dari 8 = 2 + 4 + 10 = 16 orang. 3 siswa kidal n(B) = 12 + 8 + 2 = 22 33. Jawaban: a +Mean =

283025232+++++++

n(B) 22 11

P(B) = n(S)3015 == 29. Jawaban: d Ratarata untuk kelas IXA 9 27 39. Jawaban: d Banyak balita yang +  +  +terkena +(4 4) (25)diare (6 6) (5 (38)× 121 diperkirakan = 7) 0,02 2.500 = = == 6,05= ++++ 42653 20 9= 50 orang. Banyak balita yang diperkirakan 3 tidakRata-rata terkena diare 2.500 – 50 = 2.450 untuk=kelas IXB orang.(3 4) +  +  + +(5 (75) 6) (4 7) (28) 34. Jawaban: c Ruang sampel dari = ++++ 37542 pelambungan sebuah dadu dan sekeping uang 40. Jawaban: a Diagram pohon ruang logam sebagai berikut. 121 sampel 3 mata uang logam: = =pelambungan 5,76 ffd8ffe000104a4649460001020100c800c80000ffe 21 20c584943435f50524f46494c4500010100000c484Selisih rata-rata kelas IXA dan IXB = 6,05 – c696e6f021000006d6e74725247422058595a2007c 5,76 = 0,29 e00020009000600310000616373704d5346540000 30. 00004945432073524742000000000000000000000 – x= Untuk kelompok A →nA = a, 0000000f6d6000100000000d32d48502020000000 Misalkan B = kejadian muncul angka A – Untuk kelompok B →nB = b,B x = 00000000000000000000000000000000000000000 Jawaban: b 6,7 genap n(B)= 6 n(S) = 12 –x = 6,9 000000000000000000000000000000000000000007,5 gabungan 00000011637072740000015000000033646573630 + A nx n x BB n(B) 6 1 A = 6,9 00001840000006c77747074000001f00000001462 ⇔ P(B)= n(S) = = nA+n B 12 2 6b707400000204000000147258595a00000218000 6,7a +7,5b 35. 000146758595a0000022c000000146258595a0000 Jawaban: a Banyak ruang ⇔ = 6,9 ab + sampel untuk kartu remi: n(S) = 52 024000000014646d6e640000025400000070646d6 464000002c400000088767565640000034c000000⇔ 6,7a + 7,5b = 6,9a + 6,9b 8676696577000003d4000000246c756d69000003f 8000000146d6561730000040c0000002474656368 000004300000000c725452430000043c0000080c6 75452430000043c0000080c625452430000043c00 00080c7465787400000000436f707972696768742 02863292031393938204865776c6574742d506163 6b61726420436f6d70616e7900006465736300000 00000000012735247422049454336313936362d32 2e310000000000000000000000127352474220494 54336313936362d322e3100000000000000000000 0000000000000000000000000000000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 Daerah yang diarsir tebal adalah kejadian 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 ABCD = 3015 ⇔= muncul tepat dua gambar. n(K) = 3 n(S) = 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 EFCJ EF5 30 8 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 ⇔ EF = 3 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 n(K) 3 = 10 cm Peluang =n(S = 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 8 ) 3 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 GH CI Frekuensi harapan = × 8400 = 150 kali =GH10 ⇔= ABCD 3015 30 × 10 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 B. ⇔ GH = 00000000000000000000000000000000000000000000000 15

1. UraianMisal, kertas I = bangun ABCD dan kertas II = bangun EFGH.

= 20 cm 1

L ABHG (AB 2+ GH) ID = 1 L FEGH (GH + EF)  JI

DC

2

=

80 cm

A

=

100 cm B

AB + GH GH + EF 30 + 20 20 + 10 50 = 5

= Jadi, perbandingan luas ABHG dan FEGH adalah 5 : 3. 100 cm ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 3. ABCD merupakan d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 jajargenjang maka: AD // BC dan AD 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 = BC AB // CD dan AB = CD 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 E x cm F 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 Perhatikan ∆ABE dan ∆CDF. 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 (i) ∠ EBA = ∠FDC (sudut 000000000000000000000000000000000000000000000000000000000000000000000000000000000 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 ABCD sebangun dengan EFGH sehingga siku-siku) (ii) AB = CD (pada jajargenjang sisi 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 sisi-sisi yang bersesuaian sebanding. BC AB berhadapan sama panjang) 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 (iii) ∠ EAB = ∠ FCD (pada jajargenjang sudut 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 = 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 GFEF yang berhadapan sama besar) Oleh karena 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 80 100 memenuhi syarat sudut, sisi, sudut maka 0000000000000000000000000000000000000000000000 ⇔= 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 100 x ∆ABE dan ∆CDF kongruen. H

G

303

0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 100 × 100 4. Volume air yang dipindahkan = ⇔ x= 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 80 volume kerucut 21 2 πr·tab ⇒ · t33 = πrk2 · tk 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 3tab21 = 125 cm Kertas gabungan berbentuk 00000000000000000000000000000000000000000000000 persegi panjang dengan: p = 80 + x 122 · · 36 = · 182 · tk k tk = 32 Jadi, tinggi kerucut 32 cm. = 80 + 125 = 205 cm A = 100 cm L= p ×= A 205 × 100 = 20.500 cm2 Jadi, luas kertas persegi

panjang gabungan tersebut 20.500 cm2. 2. CD dibagi menjadi tiga bagian sama panjang sehingga CJ = JI = ID = 5 cm. ∆ABC, ∆GHJ, dan ∆EFC sebangun sehingga diperoleh:

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c450 0010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006 ffd8ffe000104a4 00310000616373704d53465400000000494543207352474200000000000000000000 64946000102010 0c800c80000ffe20 00000000f6d6000100000000d32d4850202000000000000000000000000000000000 c584943435f5052 00000000000000000000000000000000000000000000000000000000000000116370 2 2 6. s = 4f46494c4500010 9. n(S) = 14 + 6 + 8 = 28 K = kejadian terambil kelereng 7274000001500000003364657363000001840000006c77747074000001f000000014 100000c484c696e 22 kuning = 6f021000006d6e7 15 + 20 = 626b707400000204000000147258595a00000218000000146758595a0000022c0000 P(K) = 82 = M = kejadian 28 7 47252474220585 225 + 400 = 00146258595a0000024000000014646d6e640000025400000070646d6464000002c4 95a2007ce000200 terambil kelereng merah 14 1 09000600310000 625 = 25 cm 00000088767565640000034c0000008676696577000003d4000000246c756d690000 P(M) = 282= L1 =616373704d5346 luas alas belahan 03f8000000146d6561730000040c0000002474656368000004300000000c72545243 2 54000000004945 Fh(K) = P(K) × n ⇔ 140= × n ⇔ n = kerucut 1 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465 7Jadi, frekuensi harapan terambil kelereng =43207352474200 × luas 6 490 00000000000000 merah: 787400000000436f70797269676874202863292031393938204865776c6574742d50 lingkaran 1 2 1 =000000000000f6d × πr 6 Fh(M)= P(M) × n = × 61636b61726420436f6d70616e790000646573630000000000000012735247422049 2 490 = 6000100000000d 1 2 245 =32d48502020000 × π × 15 = 454336313936362d322e310000000000000000000000127352474220494543363139 10. n(S) = 45 O = siswa gemar 6 00000000000000 2 lagu pop, n(O) = 25 B = siswa gemar 36362d322e31000000000000000000000000000000000000000000000000000000 37,5π L = 2 × cm luas penampang 00000000000000 2

00000000000000 segitiga =00000000000000 2 ×1× r × t 2= 15 × 20 = 300 00000000000000 cm2 L3 = luas selimut 00000000000000 kerucut 1 00000001163707 =27400000150000 × πrs 6 00033646573630 1 =00001840000006 × π × 15 × 25 6 c7774707400000 1 375π = 62,5π Luas permukaan = L = 1f000000014626b × 1 6 + L270740000020400 + L3 = 37,5π + 300 + 62,5π = (100π + 0000147258595a 300)00000218000000 cm2 Jadi, luas permukaan belahan 146758595a0000 kerucut tersebut (100π + 300) cm2. 022c0000001462 58595a00000240 00000014646d6e 7. Dua dadu dilambungkan bersama-sama 64000002540000 sebanyak satu kali maka n(S) = 36. A = 0070646d646400 kejadian0002c400000088 muncul jumlah kedua mata dadu 5 atau 10 76756564000003 = {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), 4c000000867669 (6, 4)} n(A) = 7 Peluang muncul jumlah 6577000003d400 0000246c756d69 kedua mata dadu 5 atau 10: 000003f80000001 n(A) 7 P = 46d65617300000 = n(S)36 Jadi, peluang muncul jumlah kedua 40c00000024746 mata dadu 5 7 56368000004300 atau000000c7254524 10 adalah . 36 30000043c00000 80c67545243000 8. Besar sudut daerah SMA 0043c0000080c62 dan SMK5452430000043c = 360° – (150° + 105°) = 105° 0000080c746578 Besar7400000000436f7 sudut daerah SD banyak??????? pelajar SD sudut daerah 23 Besar 2 = * 07972696768742 SMA dan SMK banyak pelajar SMA dan SMK ?????? 33 4 02863292031393 150° 100 ⇒ = 938204865776c6 *105° n ??????? 41 2 574742d5061636 ⇔ n= 70 Jadi, banyak pelajar SMA dan SMK = b61726420436f6d adalah 70 orang. ?????? 22 3 70616e79000064 * 65736300000000 ?????? 33 4 00000012735247* 42204945433631 ?????? 42 1 3936362d322e31 00000000000000 00000000127352= 47422049454336 322 3 313936362d322e * 31000000000000 333 4 00000000000000* 24 1 00000000000000 00000000000000r +t

lagu rock, n(B) = 20 n(tidak gemar keduanya) = 5 n(O ∪ B) = 45 – 5 = 40 n(gemar keduanya) = n(O ∩ B)

= n(O) + n(B) – n(O ∪ B) = 25 + 20 – 40 = 5 Peluang terpanggil anak gemar n(O keduanya: P(O ∩ B)= ∩B) 5=1 = n(S) 459

Fh(O ∩ B) = P(O ∩ B) × 1 1 n 12= × n 1 9

n = 108 Jadi, banyaknya pemanggilan 108 kali.

Bab IV Pangkat dan Bentuk Akar A. Pilihan Ganda 33 1. (–2a)3 = (–2)3a3 = (–2) × (– Jawaban: a × a3 = –8a3 2) × (–2) 5. =V× πr2 × t31k keruc 2. Jawaban: d Oleh ut karena m = 3 diperoleh: m5 15 × 724 × 22 4 3086 = × + m⇒ + m = 337 + 3 + t3k6 4 4 2 ⇔ tk = 6 cm = 3 × 3 + 3 + 3 4 2 × 3 = (3 + 1 + 3 ) Misal tinggi tabung 34 = (4 + 9) 81 = = tb.t2 ⇔ 55 tk = t = tk × = 81 · 6= = 1.053 15 cm b b 13 5 22 = πr2 × 3. Vtabung Jawaban: 4 8 c 16t8b = 22(2 ) = = × 72 × 15 = 2.310 cm3 232 7Jadi, volume tabung 2.310 cm3.

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524f4649 4c450001 4 −2 43 −24 −6 2 × 8 20100000c × (2) 2 × 2 = = 24 8 484c696e −4 −− Jawaban: a 4 (2)6f021000 2 = 24 + (–6) – (–8) = 006d6e74 6 8= 2 2–2 +72524742 3 4 32 * 2 * 3 2058595a 2007ce00 02000900 5. 06003100 00616373 = Jawaban:704d5346 b 54000000 00494543 20735247 42000000 00000000 00000000 000000f6 d6000100 6 22 * 3 000000d3 2d485020 20000000 00000000 00000000 00000000 00000000 5 32 * 3 00000000 00000000 3 00000000 00000000 –5 – (–2) 2 – (–3) 3 = 200000000 3 5 00000000 –2 –5 + 2 2 + 3 1 =00000000 2 3 5 11637072 –3 5 = = 274000001 3 5 50000000 7. 33646573 (2x–3y263000001 z–1)3 = 23(x–3)3= (y2)3(z– 1 3 –9 6 –3 84000000 ) = 8x Jawaban: b y z 6c777470 74000001 8. 3 f0000000 23 − −23 ⎛3a b ⎛ 33(a )b 14626b70 4 43 ⎛c⎛⎛=(c74000002 )

4.

Jawaban:04000000 c 27a−63 3

b 27b 14725859 12 612 = =cac 5a000002 6 22 * 3 18000000 9. 14675859 3 (–8m2n ) × (2k3n4) = 2–8 · 5a000002 3 22c000000 3 4 2k m n n = –16k3m2n3 + 4 = – Jawaban: b 14625859 2 7 16k3m5a000002 n 6. Jawaban: c 40000000 10. 14646d6e −2 −− 32 ⎛ 64000002 ⎛xyz –3 ⎛−5−12 ⎛ 54000000 = (x · 5y–2 ·–2z · x15 · y1 ·–2z–2–2)–2 –3 ⎛xyz ⎛70646d64 = Jawaban: d (x · x · y · y · z · z ) = –1 (x2 · y64000002 · z–1)–2 = x–4y2z2 p c4000000 88767565 = 64000003 ?? ?5 2 32 3 5 4c000000 86766965 11. 77000003 −35 −35 2 () −2 2 −2 () ⎛⎛ ⎛⎛ a b d4000000  ab × × ⎛⎛b ⎛a ⎛ −2a−2 246c756d Jawaban: a ⎛⎛ () = a3 a 3 69000003 ⎛⎛ −2 ⎛⎛ bb f8000000 q 146d6561 73000004 ?6 223 0c000000 24746563 68000004 30000000 0c725452 43000004 3c000008 0c675452 43000004 3c000008 0c625452 43000004 3c000008 0c746578

−6 −10

ab

=

b

a

×

−3 15

−6

a

−2 b

−6 −10 −2

ab b b aa

= ×−3× 15 −6 =

−6

a×b

−12

a×b 9 −3

= a–6 – 9 × b–12 + 3

= a–15b–9

12. Jawaban: a – 22x + 5 = 2–3

2 + 5 = –3 2x 2x = –8 x – x= x= –4

−8 − 2

13. 53.420.00 = 5,342 × 0 Jawaban: b 10.000.0007= 5,342 × 10 14. s= vt 1 Jawaban: c = 0,025 × 0,025 jam =

4

4 0,025 25 0,625 6,25

=×= =

4251001.000

= 6,25 × 10–3 km 15.

x2+ x3 + 3x1++3 +3 = 27 Jawaban:39 b x x2x3 3×3+3×3+3×3 ⇔ = 27 39 x 23 3 (3 + 3 + 3 ) ⇔ = 27 39 3x× 39 3 3 3

=3 3x = 333 x= 3

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2000900060031000 000000000000 2007ce00020009000 32d48502020000 8595a20 4220585 2205859 4220585 00200090 4220585 4220585 4220585 006163737 = 4c696⎛⎛ 00200090 d6e747 74742d5061636b61726420436f6d70616e7900006465736300000000000000127352474220 −2 −1 −1 00000000000 00000000 0000000000 2d4850202 2d4850202 0000000000 2d4850202 0616373704d53465 1352474200 ⎛1 ⎛1⎛742000000 1 524742000 ⎛ 00000000000 000000000f6d 54000000 0000000000 0000000000 000000000000 54000000 54000000 524742000 524742000 54000000 524742000 00000000d32 00000d32d48 54000000 54000000 0000000000 54000000 54000000 000000000 00100000000 524742000 0000000000 54000000 54000000 0000000000 54000000 000f6d60001 65400000 54000000 54000000 00000000000000 = 000000f6 ⎛1 ⎛ +0000000000 ⎛ +⎛ 000000000 4000000004945432 ? ? 95a2007 8 95a2007 ?95a2007 6 15 777 07ce000 95a2007 5a2007c 95a2007 00600310 04d534654 1 +60031000061637370 e6f021 00600310 252474 000000000 0000000000 000000000 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Jawaban: (23 ⎛11⎛ 75 += 348 00000000000 00000f6d60001000 7 –8 + 6 42000000 – (–15) = 7–800000000000 +0000000f6 6 0d32d48502 + 15 = 713 = 2–200000000f + (2242000000 )d–1 + ⎛+0000f6d600 ⎛ 0600310 0090006 0900060 0090006 3704d534 0090006 0090006 0090006 945432073 2d485020 6d6e7 0000000000 000000000 000000000 0000000000 3704d534 95a200 000000000 42000000 0d32d48502 d32d485020 000000000000 42000000 42000000 0000000f6 0000000f6 42000000 0000000f6 00000000000 00000000000 42000000 0d32d48502 42000000 42000000 600010000 42000000 0d32d48502 0000000f6 42000000 0000000000 74200000 42000000 42000000 000000000000 ⎛ 84 ⎛ 00000d32d4850202 00000000000 00000000000000000 00000000000000 2. 316 20000000 0000000000 000000000 000000000 0000000000 000000000 000000000000 00000000 0200000000 2000000000 000000000000 00000000 00000000 d60001000 d60001000 00000000 d60001000 00000000000 00000000000 00000000 0200000000 00000000 00000000 0000d32d4 00000000000 d60001000 6d6000100 00000000 0200000000 010000000 d60001000 00000000 0000000000 00000000 00000000 00000000 0000000000000000 25 3 ⋅ –0031000 363 = 00000000 25· 30031000 + · 0200000000 3 –00000000 00000000000 0006163 0031000 0310000 65400000 0031000 0031000 524742000 47252 65400000 7ce000 00000000000000 −1 000000000 11 000000000f6d60001 00000000 0000000000 000000000 0000000000 000000000 a. 00000000000 000000000000 0000000000000000 00000000011 ⎛12 ⎛ 0d32d4850 00000000 0000000000 0000000000 000000000000 00000000 00000000 00000d32d 00000d32d 00000000 00000d32d 00000000000 00000000000 00000000 00000000 0000000000 00000000 00000000 850202000 00000d32d 0000000000 00000000 000000d32 00000000 0000000000 00000d32d 00000000 0000000000 00000000 00000000 00000000 = + + ⎛ + ⎛ 00000000000000 00000000 000000000 000000000 0000000000 000000000 73704d5 0616373 6163737 0616373 0616373 47422 00049454 200090 0000000000000000 000000000000 44 00000000d32d48502 88 2 −10000000000 5 2 3 31 00049454 2 21 9 ( 4)0616373 *0616373 * (000000000 3) * 2 ?????? ??????? ? ? 36· 63707274000 3000000f6 000000f6 0000000000 0000000000 000000000000 000000f6 000000f6 485020200 485020200 000000f6 485020200 00000000000 00000000000 000000f6 000000f6 0000000000 000000f6 000000f6 000000000 00000000000 485020200 0000000000 000000f6 d48502020 0000000000 202000000 485020200 000000f6 0000000000 0000000f 000000f6 000000f6 0000000000000000 00000000 0000001163 000000000 000000000 0001163707 000000000 000000000000 02000000000000000 00000001163707 = 00150000000 3465400 704d534 04d5346 704d534 32073524 704d534 704d534 704d534 000000000 05859 32073524 006003 ⎛⎛ d6000100 0000000000000000 00000000 7072740000 000000000 000000000 2740000015 000000000 d6000100 0000000000 0000000000 000000000001 d6000100 d6000100 000000000 000000000 d6000100 000000000 00000000000 00000000000 d6000100 d6000100 0000000000 d6000100 d6000100 000000000 00000000000 000000000 0000000000 d6000100 000000000 0000000000 000000000 000000000 d6000100 0000000000 6d600010 d6000100 d6000100 000000000000 = + ⎛⎛ 33646573630 3 5 3 2 2 ( 4) * 2 * ( 3) 9 * 2 ? ? ? ? ? 00000000000000000 27400000150000 0000000000000001 4⎛⎛ 00000000 0150000000 000000000 000000000 0000000336 000000000 0000004 6540000 5400000 6540000 74200000 6540000 6540000 6540000 0000000f6 5a200 74200000 100006 8 18 000000000001 000000d3 0000000000 0000000000 163707274000 000000d3 000000d3 000000000 000000000 000000d3 000000000 00000000000 00000000000 000000d3 000000d3 0000000000 000000d3 000000d3 000000000 00000000000 000000000 0000000000 000000d3 000000000 000000d3 0000000000 000000000 000000000 000000d3 0000000000 0000000d 000000d3 000000d3 00001840000 1637072740000015 00000000000000000 00033646573630 00000000 3364657363 000001163 000001163 4657363000 000001163 163707274000 006c77747074 0000000336465736 9454320 2d485020 0000000000 0000000000 001500000003 0000494 2d485020 0004945 0000494 2d485020 000000000 000000000 2d485020 000000000 00000000000 00000000 00000000000 2d485020 2d485020 0000000000 0000494 2d485020 0000494 2d485020 0000494 000000000 d60001000 00000000000 000000000 0000000000 2d485020 000000000 7ce00 2d485020 0000000000 000000000 00000000 163737 000000000 2d485020 0000000000 32d48502 2d485020 2d485020 = 00000000 + 00001840000006c 0000018400 707274000 707274000 0018400000 707274000 001500000003 3000001840000006 23 00000000000000000 000001f00000 20000000 0000000000 0000000000 364657363000 20000000 20000000 000000000 000000000 20000000 000000000 00001163707 01163707274 20000000 20000000 0000000000 20000000 20000000 000000000 00000001163 000000000 0000000000 20000000 000000000 20000000 0000000000 000000000 000000000 20000000 0000000000 02000000 20000000 20000000 00000000 001500000 001500000 06c7774707 001500000 7352474 5432073 4320735 5432073 00000000 5432073 5432073 5432073 00000d32d 0200000006c7774 00000000 04d534 364657363000 c77747074000001f 00000000000000000 77747074000001f 0014626b707 31619 1 00000000 4. 000000014626b707 Jawaban: c00000000 ( 00000000 00000000 7074000001 003364657 003364657 4000001f000 003364657 00000000 0000000000 0000000000 001840000006 00000000 00000000 000000000 000000000 00000000 000000000 27400000150 00000150000 00000000 0000000000 00000000 000000000 70727400000 000000000 0000000000 00000000 000000000 0000000000 000000000 000000000 00000000 0000000000 00000000 00000000 00000000 001840000006 = 6666 += = 3 000000014626b7 40000020400 00000000000011637 2000000 5247420 2474200 5247420 0000000f 5247420 5247420 5247420 485020200 = 90006 0000000f 654000 4000002040000001 00000000 f000000014 363000001 363000001 000014626b 363000001 c77747074000 00000000 0000000000 0000000000 c77747074000 00000000 00000000 000000000 000000000 00000000 000000000 00000033646 00033646573 00000000 00000000 0000000000 00000000 00000000 000000000 15000000033 000000000 0000000000 00000000 000000000 00000000 0000000000 000000000 000000000 00000000 0000116370 00000000 00000000 00000000 00001472585 07274000001500000 07400000204000 47258595a0000021 00000000 840000006c 840000006c 7074000002 840000006 0000000 0000000 0000000 0000000 0000000 000000000 00310 6d600010 000004 001f00000001 30000000 3 5 3 3 46d600010 2 (1) 4 20000000 ( 00000000 1) 3 364657363000 2000000000 ? ? ? ? ? ? ?? ? ? 1626b707400 95a00000218 00000000 0000000116 0000001163 001f00000001 00000000 00000000 000000000 000000000 00000000 000000000 57363000001 63000001840 00000000 00000000 0000000116 00000000 000000000 0000000116 00000000 000000000 00000000 0000000116 000000000 000000000 00000000 7274000001 00000000 00000000 00000000 8000000146758595 nilai R=3 00336465736300000 000147258595a0 11637072 777470740 777470740 0400000014 c77747074 4626b7074000 11. Jadi,Jawaban: a0002040000 00000014675 6 3707274000 0000000 a0000022c0000001 0000000 0000000 0000000 0000000d 0000000 0000000 0000000 000000000 00061 0000000d 945432 00000000 3707274000 7072740000 4626b7074000 00000000 00000000 000000000 000000000 00000000 000000000 840000006c7 000006c7774 00000000 00000000 3707274000 00000000 00000000 000000000 00184000000 000000000 3707274000 00000000 000000000 00000000 000000000 000000000 00000000 5000000033 00000000 00000000 00000000 74000001 0014725859 00001f0000 00001f0000 7258595a00 000001f000 002040000001 1840000006c777470 00002180000001 8595a000002 46258595a0000024 Ω. 50000000 5a00000218 00014626b 00014626b 0002180000 000014626 00000000 0015000000 0150000000 002040000001 00000000 00000000 000000000 000000000 00000000 000000000 77470740000 7074000001f0 00000000 00000000 0015000000 00000000 00000000 000000000 6c777470740 000000000 0015000000 00000000 000000000 00000000 0015000000 000000000 000000000 00000000 6465736300 00000000 00000000 00000000 0000000 0000000 0000000 0000000 32d48502 0000000 0000000 0000000 000000000 63737 32d48502 073524 47258595a000 000000014646d6e6 2c0000001462 46758595a00000 2852 × 74000001f00000001 5× 0000001467 707400000 707400000 0014675859 b70740000 5 3 00000000 2 2 000000001 2 00000000 002180000001 4000002540000007 00000000 0336465736 3364657363 47258595a000 00000000 00000000 000000000 000000000 00000000 000000000 01f00000001 00000014626 00000000 00000000 0336465736 00000000 00000000 001163707 00001f00000 000000000 0336465736 00000000 000000000 0336465736 000000000 0001840000 00000000 00000000 00000000 58595a00000 5. a. = 2833646573 × 5 × 2 × 5 0f6d6000 000000f 00000f6 000000f 02000000 000000f 000000f 000000f 000000000 04d53 02000000 742000 −24626b707400000204 −2 × 3 22c000000146258 1058595a0000 2 3 63000001 204000000 204000000 5a0000022c 020400000 0646d6464000002c 46758595a000 2 000011637 × 5= 200000000 ×3000001840 57 × 3000116370 = 2006c7774707 24000000014 00000000 3000001840 0000018400 002180000001 00000000 00000000 000116370 000116370 00000000 000116370 4626b707400 b7074000002 00000000 00000000 3000001840 00000000 00000000 274000001 0014626b707 000116370 3000001840 00000000 163707274 00000000 00000000 00000000 00000000 4000000887675656 84000000 000000147258595a0 595a0000024000 147258595 0000001462 014725859 1000000 6d60001 d600010 6d60001 00000000 6d60001 6d60001 6d60001 000000000 = 46540 00000000 000000 2 7022c000000 7 147258595 3 646d6e64000 20022c0000001 ×072740000 3 × 2 × 5 = 2 × 3 00000000 000006c777 00006c7774 46758595a000 00000000 727400000 00000000 727400000 00020400000 00000000 00000000 000006c777 00000000 00000000 500000003 727400000 00000000 000006c777 000001500 727400000 00000000 4000001f000 00000000 00000000 00000000 40000034c0000008 3 3 2 300000000 5727400000 3 404000000147 2 (1) ( 1) (2 ) 2 340000020400 3 2000006c777 ? ? 00000000 ?? ? ? ? ? 6c777470 146258595a a00000218 a00000218 58595a0000 5a0000021 46258595a000 00002180000001467 000014646d6e64 00254000000 00d32d4 0000000 0000000 0000000 00000000 0000000 0000000 0000000 000000000 00000 00000000 000000 676696577000003d 7 4707400000 700000000 74000001 0000024000 000000146 000000146 0240000000 800000014 00000000 4707400000 7074000001 0022c0000001 00000000 00000000 150000000 150000000 00000000 150000000 0147258595a 258595a0000 00000000 00000000 4707400000 00000000 00000000 364657363 00001472585 150000000 4707400000 00000000 015000000 00000000 000003364 150000000 000014626b 00000000 00000000 00000000 × (2 × 5) = 72 × 10 = 002400000001 70646d64640 58595a0000022c000 00000254000000 4000000246c756d6 f0000000 000014646d 758595a00 758595a00 14646d6e64 6758595a0 8502020 0d32d48 d32d485 0d32d48 00000000 0d32d48 0d32d48 0d32d48 000000000 00494 00000000 000000 8 4646d6e64000 b 0,4 × 00000000 1f00000001 f000000014 46258595a000 00000000 00000000 336465736 336465736 00000000 336465736 00000218000 02180000001 00000000 00000000 1f00000001 00000000 00000000 000001840 95a00000218 336465736 1f00000001 00000000 033646573 00000000 1f00000001 657363000 336465736 00000000 7074000002 00000000 00000000 00000000 00002c400000 9000003f80000001 7,2 × 10 6e64000002 000146258595a0000 70646d64640000 14626b70 00022c0000 00022c0000 0000025400 000022c00 002540000007 46d6561730000040 08876756564 00000000 4626b70740 626b707400 002400000001 00000000 00000000 300000184 300000184 00000000 300000184 00014675859 46758595a00 00000000 00000000 4626b70740 00000000 00000000 000006c777 00000014675 300000184 4626b70740 00000000 630000018 00000000 4626b70740 001840000 300000184 00000000 0400000014 00000000 00000000 00000000 0000000 5020200 0202000 5020200 00000000 5020200 5020200 5020200 000000000 54320 00000000 000000 Sehingga diperoleh a = 10 dan b 74000002 5400000070 001462585 001462585 000070646d 000014625 024000000014646d6 02c400000088767 0646d6464000 c000000247465636 0000034c0000 ⇔ b 11637072 0000204000 0002040000 4646d6e64000 11637072 11637072 0000006c7 0000006c7 11637072 0000006c7 5a0000022c0 00022c00000 11637072 11637072 0000204000 11637072 11637072 470740000 8595a000002 0000006c7 0000204000 11637072 40000006c 11637072 0000204000 006c77747 0000006c7 11637072 7258595a000 01163707 11637072 11637072 04000000 95a000002 95a000002 6464000002 8595a0000 8000004300000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 000000000 73524646d646400 00000000 0f6d600 002c40000008 = –6. Jadi, 00000000 a+b= 10 + (–6) = 4 e6400000254000000 565640000034c00 00867669657 = c725452430000043 14725859 0002c40000 400000001 400000001 c400000088 024000000 2 74000001 0001472585 0014725859 002540000007 74000001 74000001 774707400 774707400 74000001 774707400 00000146258 0146258595a 74000001 74000001 0001472585 74000001 74000001 01f0000000 2c000000146 774707400 0001472585 74000001 777470740 74000001 0001472585 074000001f 774707400 74000001 0021800000 27400000 74000001 74000001 876756564000 33 ⎛⎛ 7000003d400 11 ⎛⎛ 2 70646d6464000002c 00008676696577 ⎛⎛ 0000000 0000000 0000000 00000000 0000000 0000000 74200 00000000 010000 a 5. 0000000 Jawaban: b Perhatikan 60000000 5 3000000000 4 2 22 3 sifat 3 2 ???? ? ? c0000080c6754524 2014646d6e 5a000002 0088767565 4646d6e64 4646d6e64 7675656400 ⎛⎛⎛⎛ × 500001f000 × ⎛⎛ ⎛⎛ 0034c0000008 50000000 95a0000021 5a00000218 0646d6464000 50000000 50000000 0001f0000 0001f0000 50000000 0001f0000 595a0000024 00000240000 50000000 50000000 95a0000021 50000000 50000000 14626b707 258595a0000 0001f0000 95a0000021 50000000 95a0000021 000000014 0001f0000 50000000 0146758595a 15000000 50000000 50000000 22 = ( 0,4 × 0,04 ) −5 4 ⇔50000000 0000246c756d 30000043c0000080 40000008876756564 000003d4000000 b ⎛⎛ 18000000 640000034c 000002540 00034c0000 640000025 operasi bentuk aljabar (a + b)(a – b) 0000000 0000000 0000000 0000000 00000000 0000000 0000000 0000000 000116370 00000 39 000002540 00000000 0000d3 676696577000 b. ⎛⎛ 69000003f800 22 c625452430000043 33646573 8000000146 0000001467 002c40000008 33646573 33646573 00014626b 00014626b 33646573 00014626b 00000001464 00014646d6e 33646573 33646573 8000000146 33646573 33646573 400000204 02400000001 00014626b 8000000146 33646573 000014626 33646573 8000000146 626b70740 00014626b 33646573 0000022c000 03364657 33646573 33646573 ⎛⎛ 2 14675859 0000008676 000007064 000007064 0086766965 400000070 0000034c000000867 246c756d6900000 003d40000002 ⎛⎛ 3 = a20000000 –758595a000 b0000000 . 0000000 0000146d656 c0000080c7465787 0000000 00000000 0000000 0000000 0000000 727400000 00000 00000000 2d4850 a a 63000001 58595a0000 876756564000 63000001 63000001 707400000 707400000 63000001 707400000 6d6e6400000 64000002540 63000001 63000001 758595a000 63000001 63000001 000000147 4646d6e6400 707400000 758595a000 63000001 b70740000 63000001 758595a000 000020400 707400000 63000001 0001462585 36300000 63000001 63000001 5a000002 6965770000 6d6464000 6d6464000 77000003d4 646d64640 46c756d69000 6696577000003d400 3f8000000146d65 400000000436f707 = 0,4 × a 1730000040c0 2= 0,016 a 3 0,04 = 2 6150000000 – 500025400000 + 20022c00000 E 84000000 33 – (–4) 80022c00000 2c000000 03d4000000 002c400000 002c400000 000000246c 00002c400 84000000 0022c00000 022c000000 0034c0000008 84000000 84000000 204000000 204000000 84000000 204000000 25400000070 0000070646d 84000000 84000000 0022c00000 84000000 84000000 258595a000 204000000 020400000 84000000 000014725 84000000 95a00000240 18400000 84000000 84000000 0000000 0000000 0000000 0000000 00000000 0000000 0000000 0000000 00000 00000000 202000 b 23204000000 003f80000001 9726967687420286 ⎛⎛ 10000246c756d69000 1 ⎛⎛ 261730000040c000 00000247465 =14625859 ⎛⎛ × × ⎛⎛ : 088767565 246c756d69 088767565 756d690000 000088767 3292031393938204 46d656173000 6c777470 0146258595 146258595a 676696577000 6c777470 6c777470 147258595 147258595 6c777470 147258595 646d6464000 6464000002c 6c777470 6c777470 0146258595 6c777470 6c777470 002180000 070646d6464 147258595 0146258595 6c777470 014725859 0146258595 ⎛⎛ 8595a0000 3−54 147258595 6c777470 0000001464 06c77747 6c777470 6c777470 ⎛⎛ 225 6c777470 63680000043 = 0,016 0000000 0000000 0000000 0000000 00000000 0000000 0000000 0000000 336465736 00000 00000000 000000 003f8000000146d65 00024746563680 865776c6574742d5 B. b 5a000002 000003f800 640000034c 640000034c 03f80000001 565640000 0040c0000002 00000000c725 74000001 a000002400 0000024000 003d40000002 74000001 74000001 a00000218 a00000218 74000001 a00000218 002c4000000 40000008876 74000001 74000001 a000002400 74000001 74000001 001467585 000002c4000 a00000218 a000002400 74000001 5a0000021 74000001 a000002400 021800000 a00000218 74000001 6d6e6400000 07400000 74000001 74000001 4 6 061636b617264204 40000000 000000867 000000867 46d6561730 034c00000 161730000040c00000 2 3 5 0000146d65 00004300000000 0000000 0000000 0000000 0000000 00000000 0000000 0000000 0000000 300000184 00000 00000000 000000 474656368000 45243000004 36f6d70616e79000 =14646d6e × 2 × × f0000000 0000014646 000014646d 46c756d69000 f0000000 f0000000 000000146 000000146 f0000000 000000146 88767565640 75656400000 f0000000 f0000000 0000014646 f0000000 f0000000 95a0000022 00088767565 000000146 0000014646 f0000000 800000014 f0000000 0000014646 000000146 f0000000 2540000007 1f000000 f0000000 f0000000 2 8014675859 3–5 6173000004 669657700 669657700 000040c000 086766965 004300000000 6 –2 –5 + 6 – (–2) = 2 3 E 3 7 1. a. –5 + 6 – (–2) 02474656368000004 2 2 3 3c0000080c67 0646573630000000 0000000 0000000 0000000 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6. Jawaban: c 2 ⎛111 ⎛⎛   3 ⎛⎛1 ⎛ ⎛⎛× 2 : ⎛⎛ = ⎛⎛⎛⎛ × 42 ⎛⎛⎛⎛ 2 ⎛⎛⎛⎛ ⎛⎛ 2 ⎛⎛

2 −52 −5

23 :

⎛⎛

4 −3 −5

⎛⎛11 1⎛⎛ ⎛⎛ ⎛⎛ × ⎛⎛ ⎛⎛22 2⎛⎛⎛⎛

= ⎛⎛×

4(3) +− + 56

⎛⎛1 1 ⎛⎛ 22 ⎛⎛⎛⎛

= ⎛⎛ =

⎛⎛

7. Jawaban: c 34 −5

34 −5

⎛⎛⎛⎛ ⎛⎛ 3 2332 3 ⎛⎛× ×⎛⎛⎛⎛34 = 5 × × − 52555 ⎛⎛⎛⎛⎛⎛ 2

=

4 -5 23× 3×3 3 4 -5 5×2×5

=

2×3 4 −2 2×5

=

52 2 ×3

3 −1

− 43

52 25

= 236 =⋅ 8. Jawaban: b −3 −3

32 ⎛ ⎛(2) − 5×

32 ⎛ ⎛ 2 −×5 ⎛⎛ =−1 −1 ⎛⎛ 65× ⎛23××5 ⎛

31 − 2( 1) −−

−3

⎛−2 ×5 ⎛ = ⎛⎛ ⎛3⎛ ⎛−2−×523 = ⎛⎛ −3 –2 ⎛ 3 ⎛⎛ = (–2 –6

= –2 6 3 –9 3 5

× 3–1 × 53)–3 × 3 × 5–9 = –2– 3

9. 3x + 5 23x + 5=⇒ = 2–4 412 2 Jawaban: c4 3x + 5 = –4 3 3x = –9 3 x= –3

10. Jawaban: a −2 3 m × 5= 5 15 3 5

3

⇒m = × 2

1

35− 5 5 3 35

⇔m = × 2

35− 3 55 ⋅

⇔m = m=

5−

5

2 2 2

42

5 −

4 – (–2) m = 54 m = 56

⋅

ffd8ffe00010 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe000 ffd8ffe00 ffd8ffe000 ffd8ffe00 ffd8ffe0001 ffd8ffe000 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe000 ffd8ffe00 ffd8ffe0001 ffd8ffe000 ffd8ffe00 ffd8ffe000 ffd8ffe00 ffd8ffe00 ffd8ffe000104 ffd8ffe000 ffd8ffe00 ffd8ffe00 ffd8ffe000 ffd8ffe00 ffd8ffe0001 ffd8ffe000 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe000 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe ffd8ffe0 ffd8ffe0 4a464946000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0104a464 0104a464 0104a464 104a46494 0104a464 104a46494 0104a464 104a46494 04a4649460 0104a464 0104a464 0104a464 0104a464 0104a464 104a464946 0104a464 04a46494600 0104a464 104a46494 104a46494 0104a464 0104a464 a464946000102 104a464946 0104a464 0104a464 104a46494 0104a464 04a4649460 104a464946 0104a4649 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 104a46494 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 000104a 00104a4 00104a4 1020100c800 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 94600010 94600010 94600010 600010201 94600010 600010201 94600010 600010201 001020100c8 001020100c 94600010 94600010 94600010 94600010 94600010 0001020100 94600010 01020100c80 94600010 600010201 600010201 94600010 94600010 0100c800c8000 0001020100 94600010 94600010 600010201 94600010 001020100c 0001020100 460001020 94600010 94600010 94600010 94600010 94600010 94600010 94600010 600010201 94600010 94600010 94600010 94600010 94600010 94600010 94600010 4649460 6494600 6494600 c80000ffe20c 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 20100c80 20100c80 20100c80 00c800c80 20100c80 00c800c80 20100c80 00c800c80 00c80000ffe 800c80000ff 20100c80 20100c80 20100c80 20100c80 20100c80 c800c80000 20100c80 0c80000ffe20 20100c80 00c800c80 00c800c80 20100c80 20100c80 0ffe20c5849434 c800c80000 20100c80 20100c80 00c800c80 20100c800 800c80000ff c800c80000 100c800c8 20100c80 20100c80 20100c80 20100c80 20100c80 20100c80 20100c80 00c800c80 20100c80 20100c80 20100c80 20100c80 20100c80 20100c80 20100c80 584943435f5 0010201 0102010 0102010 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0c80000ff 0c80000ff 0c80000ff 000ffe20c5 0c80000ff 000ffe20c5 0c80000ff 000ffe20c5 20c58494343 e20c584943 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff ffe20c58494 0c80000ff c584943435f 0c80000ff 000ffe20c5 000ffe20c5 0c80000ff 0c80000ff 35f50524f4649 ffe20c58494 0c80000ff 0c80000ff 000ffe20c5 c80000ffe2 e20c584943 ffe20c58494 0000ffe20c 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff 000ffe20c5 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0524f46494c 00c800c 0c800c8 0c800c8 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 e20c5849 e20c5849 e20c5849 84943435f e20c5849 84943435f e20c5849 84943435f 5f50524f464 435f50524f4 e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 3435f50524 e20c5849 50524f46494 e20c5849 84943435f 84943435f e20c5849 e20c5849 4c45000101000 3435f50524f e20c5849 e20c5849 84943435f 0c5849434 435f50524f4 3435f50524 584943435 e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 84943435f e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 45000101000 2 43435f50 17. Jawaban: 23. Jawaban: c 3 tahun cahaya 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 43435f50 43435f50 43435f50 50524f464 43435f50 50524f464 43435f50 50524f464 94c45000101 6494c45000 43435f50 43435f50 43435f50 43435f50 43435f50 f46494c450 43435f50 c4500010100 43435f50 50524f464 50524f464 43435f50 43435f50 00c484c696e6f 46494c4500 43435f50 43435f50 50524f464 35f50524f 6494c45000 f46494c450 f50524f464 43435f50 43435f505 43435f50 43435f50 43435f50 43435f50 43435f50 50524f464 43435f50 43435f50 43435f50 43435f505 43435f50 43435f50 80000ff 0000ffe2 11. 0000ffe2 Jawaban: b ⎛236 ⎛ 00c484c696e 12 524f4649 b 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 524f4649 524f4649 524f4649 94c450001 524f4649 94c450001 524f4649 94c450001 00000c484c6 10100000c4 524f4649 524f4649 524f4649 524f4649 524f4649 0010100000 524f4649 000c484c696 524f4649 94c450001 94c450001 524f4649 524f4649 021000006d6e7 010100000c 524f4649 524f4649 94c450001 46494c450 10100000c4 0010100000 94c450001 524f4649 24f46494c 524f4649 524f4649 524f4649 524f4649 94c450001 524f4649 524f4649 524f4649 524f4649 24f46494c 524f4649 524f4649 =3× 9,46 × 10 2 km 2 ffd8ffe000104a464946000 e20c584 0c58494 0c58494 6f021000006 ⎛26323,6 ⎛ 2 + 2,63 12 =  ⎛ +   1020100c800c80000ffe20c5 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 4c450001 4c450001 4c450001 0100000c4 4c450001 0100000c4 4c450001 0100000c4 96e6f021000 84c696e6f02 4c450001 4c450001 4c450001 4c450001 4c450001 c484c696e6 4c450001 e6f02100000 4c450001 0100000c4 0100000c4 4c450001 4c450001 472524742205 484c696e6f 4c450001 4c450001 0100000c4 00101000 84c696e6f02 c484c696e6 0100000c4 4c450001 45000101 4c450001 4c450001 4c450001 4c450001 4c450001 0100000c4 4c450001 4c450001 4c450001 4c450001 45000101 4c450001 4c450001 = 28,38 × 10 km =⎛ 25×7 ⎛1 0 ⎛1 00 + ⎛ d6e74725247 943435f 3435f505 3435f505 84943435f50524f46494c450 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0100000c 0100000c 0100000c 84c696e6f 0100000c 84c696e6f 0100000c 84c696e6f 006d6e7472 1000006d6e 0100000c 0100000c 0100000c 0100000c 0100000c f021000006 0100000c 6d6e7472524 0100000c 84c696e6f 84c696e6f0 0100000c 0100000c 8595a2007ce00 021000006d 0100000c 0100000c 84c696e6f 00c484c69 1000006d6e f021000006 84c696e6f 0100000c 00000c48 0100000c 0100000c 0100000c 0100000c 0100000c 84c696e6f 0100000c 0100000c 0100000c 0100000c 00000c48 0100000c 0100000c – × 47 13 0010100000c484c696e6f02 69.169 55.696 422058595a2 = 2,838 ×7472524742 10484c696e km =484c696e 25 7484c696e +484c696e 50524f4 24f46494 24f46494 = +×021000006 0000000000000000000000000000000000000000000000 484c696e 484c696e 484c696e 021000006 484c696e 021000006 484c696e 021000006 5247422058 7472524742 484c696e 484c696e 484c696e 484c696e 484c696e d6e7472524 484c696e 7422058595a 484c696e 021000006 21000006d 484c696e 484c696e 020009000600 6e74725247 484c696e 484c696e 021000006 6e6f02100 d6e7472524 021000006 4c696e6f0 484c696e 484c696e 484c696e 484c696e 484c696e 484c696e 4c696e6f0 484c696e 484c696e 1000006d6e7472524742205 10010.000 007ce000200 24. c450001 8595a2007ce000200090006 6494c45 c450001 6f021000 6f021000 6f021000 d6e747252 6f021000 d6e747252 6f021000 d6e747252 595a2007ce0 2058595a20 6f021000 6f021000 6f021000 6f021000 6f021000 7422058595 6f021000 2007ce00020 6f021000 d6e747252 6e7472524 6f021000 6f021000 310000616373 422058595a 6f021000 6f021000 d6e747252 0006d6e7 2058595a20 7422058595 d6e747252 6f021000 21000006 6f021000 6f021000 6f021000 6f021000 6f021000 d6e747252 6f021000 6f021000 6f021000 6f021000 21000006 6f021000 6f021000 47 – ffd8ffe00010 09000600310 00310000616373704d5346 6.916.900 55.696 4a464946000 006d6e74 006d6e74 006d6e74 474220585 006d6e74 474220585 006d6e74 474220585 0020009000 07ce000200 006d6e74 006d6e74 006d6e74 006d6e74 006d6e74 a2007ce000 006d6e74 0090006003 006d6e74 474220585 742205859 006d6e74 006d6e74 704d53465400 2007ce0002 006d6e74 006d6e74 474220585 47252474 07ce000200 a2007ce000 474220585 006d6e74 d6e74725 006d6e74 006d6e74 006d6e74 006d6e74 006d6e74 474220585 006d6e74 006d6e74 006d6e74 006d6e74 d6e74725 006d6e74 006d6e74 = + 5400000000494543207352 0001010 0100000 0100000 = 57 + 47 – 37 22 00061637370 10.00010.000 1020100c800c Jawaban: d L–672524742 4742000000000000000000 72524742 72524742 72524742 95a2007ce 72524742 95a2007ce 72524742 95a2007ce 6003100006 0900060031 72524742 72524742 72524742 72524742 72524742 2000900060 72524742 1000061637 72524742 95a2007ce 5a2007ce0 72524742 72524742 000000494543 0009000600 72524742 72524742 95a2007ce 22058595 0900060031 2000900060 95a2007ce 24742205 72524742 72524742 72524742 72524742 72524742 95a2007ce 72524742 72524742 72524742 72524742 24742205 72524742 72524742 2 4d534654000 80000ffe20c58 0000c48 c484c69 c484c69 154 × 10 = × r 0000000000f6d6000100000 2 6.972.536 = 67 72058595a = πr⇔ 4943435f5052 2058595a 2058595a 2058595a 000200090 2058595a 000200090 2058595a 000200090 16373704d5 0000616373 2058595a 2058595a 2058595a 2058595a 2058595a 0310000616 2058595a 3704d53465 2058595a 000200090 002000900 2058595a 2058595a 207352474200 3100006163 2058595a 2058595a 000200090 a2007ce00 0000616373 0310000616 000200090 2058595a 8595a200 2058595a 2058595a 2058595a 2058595a 000200090 2058595a 2058595a 2058595a 2058595a 8595a200 2058595a 2058595a 00000494543 = 000d32d485020200000000 4c696e6 6e6f0210 6e6f0210 ffd8ffe00010 ffd8ffe00010 4f46494c4500 –6 27ce00020 0000000000000000000000 2007ce00 2007ce00 2007ce00 006003100 2007ce00 006003100 2007ce00 006003100 3465400000 704d534654 2007ce00 2007ce00 2007ce00 2007ce00 2007ce00 373704d534 2007ce00 4000000004 2007ce00 006003100 060031000 2007ce00 2007ce00 000000000000 73704d5346 2007ce00 2007ce00 006003100 02000900 704d534654 373704d534 006003100 2007ce00 7ce00020 2007ce00 2007ce00 2007ce00 2007ce00 2007ce00 006003100 2007ce00 2007ce00 2007ce00 2007ce00 2007ce00 2007ce00 10.000 = 20735247420 ⇔ 154 × 7 × 10 = 22 × r 18. Jawaban: 4a4649460001 4a4649460001 010100000c48 0000000000000000000000 f021000 00006d6 00006d6 02000900 02000900 02000900 006163737 02000900 006163737 02000900 006163737 0004945432 0000000049 02000900 02000900 02000900 02000900 02000900 6540000000 02000900 9454320735 02000900 006163737 061637370 02000900 02000900 000000000000 5400000000 02000900 02000900 006163737 06003100 0000000049 6540000000 006163737 02000900 00900060 02000900 02000900 02000900 02000900 02000900 006163737 02000900 02000900 02000900 02000900 00900060 02000900 02000900 020100c800c8 020100c800c8 4c696e6f0210 00000000000 –6 2 697,2596 c 0000000000000000000000 ⇔ 7 × 7 × 10 = r 32 ffd8ffe000104a46 0000ffe20c584 0000ffe20c584 00006d6e747 3x 006d6e7 e747252 e747252 06003100 06003100 06003100 04d534654 06003100 04d534654 06003100 04d534654 0735247420 4543207352 06003100 06003100 06003100 06003100 06003100 0494543207 06003100 2474200000 06003100 04d534654 4d5346540 06003100 06003100 00f6d60001000 4945432073 06003100 06003100 04d534654 00616373 4543207352 0494543207 04d534654 06003100 03100006 06003100 06003100 06003100 06003100 06003100 04d534654 06003100 06003100 06003100 06003100 03100006 06003100 06003100 00000000000 0000000000000000000001 12. Jawaban: –3 943435f5052449460001020100c8 943435f50524 25247422058 a 54 = 1637072740000015000000 ⇔ r= 7 × 10 = 3x 00c80000ffe20c584 00000f6d600 00616373 00616373 00616373 000000004 00616373 000000004 00616373 000000004 0000000000 4742000000 00616373 00616373 00616373 00616373 00616373 3524742000 00616373 0000000000 00616373 000000004 000000049 00616373 00616373 00000d32d485 5247420000 00616373 00616373 000000004 704d5346 4742000000 3524742000 000000004 00616373 16373704 00616373 00616373 00616373 00616373 00616373 000000004 00616373 00616373 00616373 00616373 16373704 00616373 00616373 f46494c45000 f46494c45000 595a2007ce00 4725247 4742205 4742205 0336465736300000184000 943435f50524f464 10100000c484 10100000c484 02000900060 0100000000d 704d5346 704d5346 704d5346 945432073 704d5346 945432073 704d5346 945432073 0000000000 704d5346 704d5346 704d5346 704d5346 704d5346 0000000000 704d5346 0000000000 704d5346 945432073 454320735 704d5346 704d5346 020200000000 0000000000 704d5346 704d5346 945432073 54000000 0000000000 0000000000 945432073 704d5346 d5346540 704d5346 704d5346 704d5346 704d5346 704d5346 945432073 704d5346 704d5346 704d5346 704d5346 d5346540 704d5346 704d5346 3 3 0006c77747074000001f000 94c4500010100000 4220585 8595a20 25. Jawaban: a = 6x8595a20 y c696e6f02100 c696e6f02100 03100006163 c484c696e6f02100 000014626b707400000204 32d48502020 54000000 54000000 54000000 524742000 54000000 524742000 54000000 524742000 0000000f6d6 0000000000 54000000 54000000 54000000 54000000 54000000 0000000000 54000000 000f6d60001 54000000 524742000 247420000 54000000 54000000 000000000000 0000000000 54000000 54000000 524742000 00494543 0000000000 0000000000 524742000 54000000 00000004 54000000 54000000 54000000 54000000 54000000 524742000 54000000 54000000 54000000 54000000 00000004 54000000 54000000 13. Jawaban: 0006d6e747250006d6e747252474 0006d6e74725 73704d53465 000000147258595a0000021 95a2007 07ce000 07ce000 Bentuk sekawan dari 00000000000 247422058595 247422058595 40000000049 00494543 00494543 00494543 000000000 00494543 000000000 00494543 000000000 0001000000 00f6d600010 00494543 00494543 00494543 00494543 00494543 00000f6d60 00494543 00000000d3 00494543 000000000 000000000 00494543 00494543 000000000000 0000f6d600 00494543 00494543 000000000 20735247 00f6d600010 00000f6d60 000000000 00494543 94543207 00494543 00494543 00494543 00494543 00494543 000000000 00494543 00494543 00494543 00494543 94543207 00494543 00494543 22058595a2007ce0 d 8000000146758595a000002 a2007ce00020 a2007ce00020 45432073524 0020009000600310 00000000000 ce00020 2000900 2000900 2c000000146258595a00000 20735247 20735247 20735247 000000000 20735247 000000000 20735247 000000000 00d32d4850 0000000d32 20735247 20735247 20735247 20735247 20735247 0010000000 20735247 2d48502020 20735247 000000000 000000000 20735247 20735247 000000000000 0100000000 20735247 20735247 000000000 42000000 0000000d32 0010000000 000000000 20735247 35247420 20735247 20735247 20735247 20735247 20735247 000000000 20735247 20735247 20735247 20735247 35247420 20735247 20735247 2 −4 19. Jawaban: 009000600310000616373704d534 009000600310 ( 74200000000 35 0000000f6 = (35 24000000014646d6e640000 00000000000 0090006 0600310 0600310 42000000 42000000 42000000 0000000f6 42000000 0000000f6 42000000 0000000f6 2020000000 d485020200 42000000 42000000 42000000 42000000 42000000 0d32d48502 42000000 0000000000 42000000 0000000f6 000000f6d 42000000 42000000 000000000000 d32d485020 42000000 42000000 00000000 d485020200 0d32d48502 0000000f6 42000000 00000000 42000000 42000000 42000000 42000000 42000000 0000000f6 42000000 42000000 42000000 42000000 00000000 42000000 42000000 000616373704 000616373704 00000000000 6540000000049454 d 025400000070646d646400 ffd8ffe000104a4 −4 00000000000 3207352474200000 d53465400000 d53465400000 000000000f6d ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f5 5 d60001000 ) d60001000 00000000 00000000 00000000 d60001000 00000000 d60001000 00000000 d60001000 0000000000 00000000 00000000 00000000 00000000 00000000 0200000000 00000000 0000000000 00000000 d60001000 600010000 00000000 00000000 000000000000 2000000000 00000000 00000000 00000000 0000000000 0200000000 d60001000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 0002c400000088767565640 64946000102010 0031000 0006163 0006163 5 0000000000000000 000494543207 000494543207 60001000000 −2 00000000000 000034c0000008676696577 2 0c800c80000ffe2 = 35 = 0524f46494c4500010100000c484c696e6f021000006d6e747252474 00000000 00000000 00000000 00000d32d 00000000 00000d32d 00000000 00000d32d 0000000000 00000000 00000000 00000000 00000000 00000000 0000000000 00000000 0000000000 00000000 00000d32d 0000d32d4 00000000 00000000 000000000000 0000000000 00000000 00000000 00000d32d 000000f6d 0000000000 0000000000 00000d32d 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000d32d 00000000 00000000 00000000 00000000 00000000 00000000 00000000 0000000f6d600010 352474200000 352474200000 00d32d48502 0616373 73704d5 73704d5 000003d4000000246c756d6 0c584943435f50 ⇔ 000000000000 m × (23 00000000000 5 0000000d32d48502 2 000000000000 02000000000 3485020200 000000f6 000000f6 000000f6 485020200 000000f6 485020200 000000f6 485020200 0000000000 000000f6 000000f6 000000f6 000000f6 000000f6 0000000000 000000f6 0000000000 000000f6 485020200 850202000 000000f6 000000f6 000000000000 0000000000 000000f6 000000f6 485020200 60001000 0000000000 0000000000 000000f6 000f6d600 000000f6 000000f6 000000f6 000000f6 000000f6 485020200 000000f6 000000f6 000000f6 000000f6 000f6d600 000000f6 000000f6 524f46494c4500 9000003f8000000146d6561 22058595a2007ce00020009000600310000616373704d5346540000 0200000000000000 00000000000 00000000000f 00000000000f⋅ 00000000000 704d534 3465400 3465400 2 2 d6000100 730000040c0000002474656 010100000c484c 0000000000000000 d6000100 d6000100 d6000100 000000000 d6000100 000000000 d6000100 000000000 0000000000 d6000100 d6000100 d6000100 d6000100 d6000100 0000000000 d6000100 0000000000 d6000100 000000000 000000000 d6000100 d6000100 000116370727 0000000000 d6000100 d6000100 000000000 00000d32 0000000000 0000000000 000000000 01000000 d6000100 d6000100 d6000100 d6000100 d6000100 000000000 d6000100 d6000100 d6000100 d6000100 01000000 d6000100 d6000100 14. Jawaban: ⇔ m × (23 – 7)= ( 12) – ( 6d6000100000 6d6000100000 00000000000 000049454320735247420000000000000000000000000000f6d6000 00000000000 696e6f02100000 368000004300000000c7254 0000000000000000 6540000 0000004 0000004 000d32d48502 000d32d48502 00000000000 000000d3 000000d3 000000d3 000000000 000000d3 000000000 000000d3 000000000 0000000000 000000d3 000000d3 000000d3 000000d3 000000d3 0000000000 000000d3 0000000000 000000d3 000000000 000000000 000000d3 000000d3 400000150000 0000000000 000000d3 000000d3 000000000 d4850202 0000000000 0000000000 000000000 000000d3 00d32d48 000000d3 000000d3 000000d3 000000d3 000000d3 000000000 000000d3 000000d3 000000d3 000000d3 00d32d48 000000d3 000000d3 b 0000000000000000 52430000043c0000080c675 6d6e7472524742 00000011637 100000000d32d485020200000000000000000000000000000000000 020000000000 020000000000 00000000000 1 1 0000000000000000 2058595a2007ce 452430000043c0000080c62 2d485020 2d485020 2d485020 000000000 2d485020 0000494 000000000 9454320 2d485020 000000000 0000000000 2d485020 2d485020 2d485020 2d485020 2d485020 0000000000 2d485020 0000000000 2d485020 000000000 000000000 2d485020 2d485020 000336465736 0000000000 9454320 2d485020 2d485020 000000000 00000000 0000000000 0000000000 000000000 2d485020 50202000 2d485020 2d485020 2d485020 2d485020 2d485020 000000000 2d485020 2d485020 2d485020 2d485020 50202000 2d485020 2d485020 000000000000 000000000000 00000000000 07274000001 0000000000000000 5452430000043c0000080c7 00020009000600 0000000000000000000000000000000000000000000000000000000 () : () 1 20000000 20000000 20000000 20000000 000000000 20000000 000000000 20000000 000000000 0000000000 20000000 20000000 20000000 20000000 20000000 0000000000 20000000 0000000000 20000000 000000000 000000000 20000000 20000000 300000184000 0000000000 20000000 20000000 000000000 00000000 0000000000 0000000000 000000000 20000000 00000000 20000000 20000000 20000000 20000000 20000000 000000000 20000000 20000000 20000000 00000000 20000000 20000000 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()

() ()

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f0000000 f0000000 003f800000014 f0000000 00014626 f0000000 f0000000 000000146 f0000000 000000146 f0000000 f0000000 000040c000 086766965 000040c000 000040c000 086766965 000040c000 086766965 086766965 400000070646d6464000 8676696577000003d4 088767565640000034c 52430000043c000 00043c00000 0000000 00000 3x 3 +2 = 3x = 0002474656 77000003d 0002474656 0002474656 77000003d 0002474656 77000003d 77000003d 14626b70 002c4000000887675656 14626b70 6d6561730000 14626b70 b7074000 14626b70 14626b70 758595a00 14626b70 758595a00 14626b70 14626b70 1 280c67545243 000000246c756d6900 0000008676696577000 0080c6754524300 b Ek =0000000 mv 040c000000247 1 32 2+ 3 2 32 2 + + 3680000043 400000024 3680000043 3680000043 400000024 3680000043 400000024 400000024 40000034c00000086766 74000002 74000002 74000002 00204000 74000002 74000002 00022c000 74000002 00022c000 74000002 74000002 0000043c000 00000 2 0003f8000000146d65 003d4000000246c756d 00043c0000080c6 00000000c7 6c756d690 00000000c7 00000000c7 6c756d690 00000000c7 6c756d690 6c756d690 . 96577000003d40000002 0080c625452 04000000 04000000 465636800000 04000000 00014725 04000000 04000000 000146258 04000000 000146258 04000000 04000000 0000000 2545243000 00003f8000 2545243000 2545243000 00003f8000 2545243000 00003f8000 00003f8000 1 61730000040c000000 69000003f8000000146 46c756d69000003f8000 00000 –325452430000043c 430000043c0 = · 1,2 × 10 · (8 × 14725859 14725859 4300000000c72 14725859 8595a000 14725859 14725859 595a00000 14725859 595a00000 14725859 14725859 0043c00000 000146d65 0043c00000 0043c00000 000146d65 0043c00000 000146d65 000146d65 000146d6561730000040 247465636800000430 d6561730000040c0000 000080c7465 20000000 = 0,6 × 10–3 ·0000080c7465787 6,4 × 00000 80c6754524 617300000 80c6754524 80c6754524 617300000 80c6754524 617300000 617300000 c0000002474656368000 5a000002 5a000002 545243000004 5a000002 00218000 5a000002 5a000002 240000000 5a000002 240000000 5a000002 5a000002 2 2 78740000000 0000000c7254524300 0024746563680000043 4) –3 400000000436f70 +4 10 0000000 30000043c0 40c000000 30000043c0 30000043c0 40c000000 30000043c0 40c000000 10 =3c0000080c675 38,410 =40c000000 004300000000c7254524 18000000 18000000 18000000 00014675 18000000 18000000 14646d6e6 18000000 14646d6e6 18000000 18000000 0436f707972 0000f6 00043c0000080c6754 00000000c7254524300 797269676874202 247465636 000080c625 000080c625 247465636 000080c625 247465636 247465636 30000043c0000080c675 1000080c625 0000000 69676874202 38,410 = 3,84 × 14675859 14675859 452430000043c 14675859 8595a000 14675859 14675859 400000254 14675859 400000254 14675859 14675859 4524300000 800000430 4524300000 4524300000 800000430 4524300000 800000430 800000430 52430000043c000008 00043c0000080c67545 863292031393938 452430000043c0000080 ? 86329203139 d6000 2 0000080c62545 0011637 5a000002 5a000002 5a000002 0022c000 5a000002 5a000002 000000706 5a000002 000000706 5a000002 5a000002 43c0000080 0000000c7 43c0000080 43c0000080 0000000c7 43c0000080 0000000c7 0000000c7 10 c625452430000043c000 0c625452430000043c 2430000043c0000080c 204865776c65747 39382048657 3x = c746578740 254524300 c746578740 c746578740 254524300 c746578740 254524300 254524300 0080c746578740000000 2c000000 2c000000 2430000043c00 2c000000 00014625 2c000000 2c000000 46d646400 2c000000 46d646400 2c000000 2c000000 0727400 10000 76c6574742d 3 2 32 2 + + 0000080c7465787400 625452430000043c000 42d5061636b6172 0000000436f 00043c000 0000000436f 0000000436f 00043c000 0000000436f 00043c000 00043c000 0436f707972696768742 2 14625859 14625859 00080c7465787 14625859 8595a000 14625859 14625859 0002c4000 14625859 0002c4000 14625859 14625859 5061636b617 Jadi,0080c74657874000000 energi 0001500 kinetik benda 3,84 × 10 joule. E 000000436f70797269 6420436f6d70616e 0000d 7079726967 0080c6754 7079726967 7079726967 0080c6754 7079726967 0080c6754 0080c6754 02863292031393938204 26420436f6d 5a000002 5a000002 400000000436f 5a000002 00240000 5a000002 5a000002 000887675 5a000002 000887675 5a000002 5a000002 32 2 32 2?? 6874202863 524300000 6874202863 6874202863 524300000 6874202863 524300000 524300000 676874202863292031 00436f7079726967687 790000646573630 865776c6574742d50616 0000033 70616e79000 32d48 40000000 40000000 707972696768 40000000 00014646 40000000 40000000 656400000 40000000 656400000 40000000 40000000 2920313939 43c000008 2920313939 2920313939 43c000008 2920313939 43c000008 43c000008 36b61726420436f6d706 393938204865776c65 4202863292031393938 000000000000012 06465736300 6465736 3820486577 0c6254524 3820486577 3820486577 0c6254524 3820486577 0c6254524 0c6254524 16e79000064657363000 14646d6e 14646d6e 742028632920 14646d6e d6e64000 14646d6e 14646d6e 34c000000 14646d6e 34c000000 14646d6e 14646d6e 00000000000 50202 74742d5061636b6172 204865776c6574742d5 735247422049454 6c6574742d 30000043c 6c6574742d 6c6574742d 30000043c 6c6574742d 30000043c 30000043c 00000000000127352474 3000001 64000002 64000002 313939382048 64000002 00254000 64000002 64000002 867669657 64000002 867669657 64000002 64000002 01273524742 6420436f6d70616e79 061636b61726420436f 336313936362d32 ? 3x = 96 2 3 2 4 98 ? + ? ? ? 3x = 53 2 1 ? ? x = 5 – 32 ? x= 1 3 (5 – 5061636b61 0000080c7 5061636b61 5061636b61 0000080c7 5061636b61 0000080c7 0000080c7 22049454336313936362 00000 20494543363 54000000 54000000 65776c6574742 54000000 00070646 54000000 54000000 7000003d4 54000000 7000003d4 54000000 54000000 8400000 726420436f6 465787400 726420436f6 726420436f6 465787400 726420436f6 465787400 465787400 000064657363000000 6d70616e79000064657 2e3100000000000 d322e310000000000000 13936362d32 70646d64 70646d64 d5061636b617 70646d64 d6464000 70646d64 70646d64 000000246 70646d64 000000246 70646d64 70646d64 00000 d70616e790 000000436f d70616e790 d70616e790 000000436f d70616e790 000000436f 000000436f 00000000012735247422 06c7774 000000001273524742 3630000000000000012 000000000001273 2e310000000 0006465736 707972696 0006465736 0006465736 707972696 0006465736 707972696 707972696 049454336313936362d3 64000002 64000002 26420436f6d70 64000002 002c4000 64000002 64000002 c756d6900 64000002 c756d6900 64000002 64000002 00000000000 204945433631393636 7352474220494543363 524742204945433 7074000 00000 3000000000 768742028 3000000000 3000000000 768742028 3000000000 768742028 768742028 22e31000000000000000 c4000000 c4000000 616e79000064 c4000000 00088767 c4000000 c4000000 0003f8000 c4000000 0003f8000 c4000000 c4000000 00001273524 2d322e310000000000 13936362d322e310000 6313936362d322e 0000012735 632920313 0000012735 0000012735 632920313 0000012735 632920313 632920313 00000000000000000000 001f000 74220494543 00000 88767565 88767565 657363000000 88767565 56564000 88767565 88767565 000146d65 88767565 000146d65 88767565 88767565 2474220494 939382048 2474220494 2474220494 939382048 2474220494 939382048 939382048 000000000000127352 0000000000000000001 00000000000000000002 310000000000000 36313936362 0000146 64000003 64000003 000000001273 64000003 0034c000 64000003 64000003 617300000 64000003 617300000 64000003 64000003 5433631393 65776c657 5433631393 5433631393 65776c657 5433631393 65776c657 65776c657 0 20 00000 474220494543363139 2735247422049454336 000000000000000 d322e310000 6362d322e3 4742d5061 6362d322e3 6362d322e3 4742d5061 6362d322e3 4742d5061 4742d5061 4c000000 4c000000 524742204945 4c000000 00086766 4c000000 4c000000 40c000000 4c000000 40c000000 4c000000 4c000000 26b7074 00000000000 36362d322e31000000 313936362d322e31000 000000000000000 00000 1000000000 636b61726 1000000000 1000000000 636b61726 1000000000 636b61726 636b61726 86766965 86766965 433631393636 86766965 96577000 86766965 86766965 247465636 86766965 247465636 86766965 86766965 00000000000 0000020 000000000000000000 0000000000000000000 00000000000)(4 0000000000 420436f6d7 0000000000 0000000000 420436f6d7 0000000000 420436f6d7 420436f6d7 – 00000000000 77000003 77000003 2d322e310000 77000003 003d4000 77000003 77000003 800000430 77000003 800000430 77000003 00000 0001273524 0616e7900 0001273524 0001273524 0616e7900 0001273524 0616e7900 0616e7900 000000000000000000 0000000000000000000 53) =4 · 4 – 4 ·77000003 53 – 4000000 00000000000 d4000000 d4000000 000000000000 d4000000 000246c7 d4000000 d4000000 0000000c7 d4000000 0000000c7 d4000000 d4000000 7422049454 006465736 7422049454 7422049454 006465736 7422049454 006465736 006465736 0000000000000= 000000000000= 53 × × 000000 00000

ffd8ffe000 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe00ffd8ffe00 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 104a46494 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 0104a464 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 600010201 94600010 94600010 94600010 94600010 94600010 94600010 94600010 94600010 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 00c800c80 20100c80 20100c80 20100c80 20100c80 20100c80 20100c800 20100c80 20100c80 000000000000000000000000000000000000000000000000000000000000000000000000000000000 000ffe20c5 0c80000ff 0c80000ff 0c80000ff 0c80000ff 0c80000ff c80000ffe 0c80000ff 0c80000ff 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 84943435f e20c5849 e20c5849 e20c5849 e20c5849 e20c5849 20c584943 e20c5849 e20c5849 50524f464 43435f50 43435f50 43435f50 43435f50 43435f50 435f50524 43435f50 43435f50 6. 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 10. a. Luas = panjang × 94c450001 524f4649 524f4649 524f4649 524f4649 524f4649 f46494c45 524f4649 524f4649 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 lebar = (5 + 0100000c4 4c450001 4c450001 4c450001 4c450001 4c450001 00010100 4c450001 4c450001 Jawaban: c 5)44 (22)n = (2 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 84c696e6f 0100000c 0100000c 0100000c 0100000c 0100000c 000c484c6 0100000c 0100000c n 4 4 2 4 = 32 22n = 220 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 021000006 484c696e 484c696e 484c696e 484c696e 484c696e 96e6f0210 484c696e 484c696e 2 = 20 = 18 d6e747252 Jadi, luas persegi 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 6f021000 6f021000 6f021000 6f021000 6f021000 00006d6e 6f021000 6f021000 2n 2 panjang 18 cm . 474220585 006d6e74 006d6e74 006d6e74 006d6e74 006d6e74 74725247 006d6e74 006d6e74 n= 10 Jadi, n = 10. 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 2 2 95a2007ce 72524742 72524742 72524742 72524742 72524742 42205859 b. Diagonal = panjang + 72524742 lebar2 72524742 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 000200090 2058595a 2058595a 2058595a 2058595a 2058595a 5a2007ce0 2058595a 2058595a 2 2 00000000000000000000000000000000000000000000000 d2 =p + Affd8ffe00010 =(5 + 006003100 2007ce00 2007ce00 2007ce00 2007ce00 2007ce00 00200090 2007ce00 2007ce00 4a4649460001

006163737 02000900 02000900 02000900 02000900 02000900 00600310 02000900 02000900 020100c800c8 0000ffe20c584 04d534654 06003100 06003100 06003100 06003100 06003100 00061637 06003100 06003100 943435f50524f 000000004 00616373 00616373 00616373 00616373 00616373 3704d534 00616373 00616373 46494c450001 +( 0100000c484c 945432073 704d5346 704d5346 704d5346 704d5346 704d5346 65400000 704d5346 704d5346 696e6f021000 524742000 54000000 54000000 54000000 54000000 54000000 00049454 54000000 54000000 006d6e747252 = 64 000000000 d=00494543 47422058595a 00494543 00494543 00494543 00494543 32073524 00494543 00494543 2007ce000200 000000000 20735247 20735247 20735247 20735247 20735247 74200000 20735247 20735247 090006003100 0000000f6 42000000 42000000 42000000 42000000 42000000 00000000 42000000 42000000 00616373704d Jadi, panjang diagonalnya 8 cm. 534654000000 d60001000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 004945432073 00000d32d 00000000 00000000 00000000 00000000 00000000 0000000f6 00000000 00000000 524742000000 000000000000 485020200 000000f6 000000f6 000000f6 000000f6 000000f6 d6000100 000000f6 000000f6 0000000000f6 000000000 d6000100 d6000100 d6000100 d6000100 d6000100 000000d3 d6000100 d6000100 d60001000000 Latihan Ulangan Tengah 00d32d485020 000000000 000000d3 000000d3 000000d3 000000d3 000000d3 2d485020 000000d3 000000d3 200000000000 Semester 22d485020 000000000 2d485020 2d485020 2d485020 2d485020 20000000 2d485020 2d485020 000000000000 A. Pilihan 000000000 20000000 20000000 20000000 20000000 20000000 00000000 20000000 20000000 000000000000 Ganda 000000000000 000000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 1. 000000000000 6 000000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 000000000000 3 =3×3×3×3×3 000000000000 000000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 × 3 = 279 000000000000 Jawaban: d000000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 116370727400 2. 000150000000 000000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 2 8 336465736300 38 × (6 +000000000 3)00000000 = 00000000 3 × 00000000 00000000 00000000 00000000 00000000 00000000 000184000000 2 2 8 Jawaban: a 000116370 92 = 38 × (300000000 )00000000 = 36c7774707400 00000000 00000000 00000000 00000000 00000000 00000000 0001f0000000 × 34 = 312727400000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 14626b707400 150000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 000204000000 147258595a00 336465736 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 3. Jawaban: c 3000218000000 5 × 33 300000184 00000000 00000000 00000000 00000000 00000000 11637072 00000000 00000000 146758595a00 (i 8 15 = 3)5 + 3 = 3 ≠ 3 00022c000000 0000006c7 11637072 11637072 11637072 11637072 11637072 74000001 11637072 11637072 2 –2 146258595a00 44 : 4774707400 =74000001 4474000001 = 42 74000001 50000000 74000001 74000001 2 3 274000001 × 374000001 6 (iii) (6 ) = 6 =000240000000 6 Jadi, 0001f0000 50000000 50000000 50000000 50000000 50000000 33646573 50000000 50000000 14646d6e6400 pernyataan yang benar (ii) dan 000254000000 00014626b 33646573 33646573 33646573 33646573 33646573 63000001 33646573 33646573 70646d646400 (iii). 3 2 7 1Jawaban: 707400000 a (2 63000001 63000001 ) 63000001 63000001 – (–2) 63000001 = 84000000 63000001 63000001 0002c4000000 684000000 6 204000000 84000000 84000000 84000000 84000000 6c7774707 84000000 84000000 23 × 2 – (–2) × (–2) = 887675656400 2 + 2 00034c000000 147258595 6c777470 6c777470 6c777470 4000001f0 6c777470 6c777470 867669657700 × 26 = 26(1 + 2) =6c777470 3× 26 6c777470 0003d4000000 a00000218 74000001 74000001 74000001 74000001 74000001 00000014 74000001 74000001 2Jawaban: d a = 12246c756d6900 dan b f0000000 f0000000 f0000000 626b7074 f0000000 f0000000 2 2000000146 3 f0000000 20003f8000000 2f0000000 3 = 18 (a – b ) = (12 – 18 ) 758595a00 14626b70 14626b70 14626b70 14626b70 14626b70 00000204 14626b70 14626b70 146d65617300 00040c000000 00022c000 74000002 74000002 74000002 74000002 74000002 00000014 74000002 74000002 247465636800 000146258 04000000 04000000 04000000 04000000 04000000 7258595a 04000000 04000000 000430000000 0c7254524300 595a00000 14725859 14725859 14725859 14725859 14725859 00000218 14725859 14725859 00043c000008 240000000 5a000002 5a000002 5a000002 5a000002 5a000002 00000014 5a000002 5a000002 0c6754524300 2 = ((6 × 2)00043c000008 – (618000000 × 14646d6e6 18000000 18000000 18000000 18000000 6758595a 18000000 18000000 2 3 3) ) 14675859 =14675859 (620c6254524300 × 2214675859 – 400000254 14675859 14675859 0000022c0 14675859 14675859 00043c000008 2 2 6000000706 ×5a000002 325a000002 )3 = (625a000002 (2 – 5a000002 5a000002 00000146 5a000002 5a000002 0c7465787400 2 2c000000 3 2c000000 2c000000 258595a0 2c000000 2c000000 346d646400 ))32c000000 =2c000000 66000000436f70 (–5) = – 797269676874 6 202863292031 0002c4000 14625859 14625859 14625859 00002400 14625859 14625859 125 × 14625859 614625859 393938204865 000887675 5a000002 5a000002 5a000002 5a000002 5a000002 00000146 5a000002 5a000002 776c6574742d 656400000 40000000 40000000 40000000 40000000 40000000 46d6e640 40000000 40000000 5061636b6172 6420436f6d70 34c000000 14646d6e 14646d6e 14646d6e 14646d6e 14646d6e 00002540 14646d6e 14646d6e 616e79000064 867669657 64000002 64000002 64000002 64000002 64000002 00000706 64000002 64000002 657363000000 7000003d4 54000000 54000000 54000000 54000000 54000000 46d64640 54000000 54000000 000000001273 524742204945 000000246 70646d64 70646d64 70646d64 70646d64 70646d64 00002c400 70646d64 70646d64 433631393636 64000002 64000002 64000002 64000002 00008876 64000002 64000002 ?2c756d6900 ?125 ? ?27 ?? ?: 64000002 2d322e310000 000000000000 ??0003f8000 49 c4000000 5???? c4000000 c4000000 c4000000 c4000000 75656400 c4000000 c4000000 000000127352 =000146d65 88767565 88767565 88767565 88767565 88767565 00034c000 88767565 88767565 474220494543 ?2?35 ?3 ??? ???? ???? ? 36313936362d 617300000 64000003 64000003 64000003 64000003 64000003 00086766 64000003 64000003 : 322e31000000 40c000000 4c000000 4c000000 4c000000 4c000000 4c000000 96577000 4c000000 4c000000 32 4 5 000000000000 ?? 247465636 86766965 86766965 86766965 86766965 86766965 003d4000 86766965 86766965 000000000000 800000430 77000003 77000003 77000003 77000003 77000003 000246c75 77000003 77000003 ? ?000000000000 000000000000 0000000c7 d4000000 d4000000 d4000000 d4000000 d4000000 6d690000 d4000000 d4000000 + (5 –

?65

3 ?? ????

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 * ? ?4 2 5 3? ?? ? 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 = 000000000000000000000000000000000000000000000000000000000000000000000000000000000 ?6 ?6 5 3 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 * 13. Jawaban: 485 3 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 d 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 645?+ 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 = 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 683?+ 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 7. 00000000000000000000000000000000000000000000000 3 2 3 2 2 1

3 + 9 = 3 + (3 ) = Jawaban: c 33(1 + 3) = 33 + 34 = 33 × 4 = 22 × 33

1

8.

?25

18.348.000 = 1,8348 × Jawaban: b = 1,835 × 107 = 10.000.000 1,84 × 107 = (23)–3 : (2–= 9. Jawaban: b Pembulatan 2 6 ) = 2–9 : 2– sampai dua tempat desimal: 12 = 2–9 – (–12)= 4.450.000.000 = 4,45 × 109 Pembulatan sampai satu tempat = 23 = 8 4.450.000.0 = 4,5 × 9 desimal: 15. Jawaban: a –2 3 a a 00 10 (64 ) = 8 10. Jawaban: b (5 × 10– 6 –2 3 3 a a ((2 ) ) = (2 ) 5 ) × (4 × 10–4) × 1012 = 5 × 23 3 26 × (–2) × 3 = 23a 10–5 × 4 × 10–4 × 1012 = 5 × –36 3a 2 23 2 = 2 ⇔ –36 =*53a 4 × 10–5 × 10–4 × 1012 a = –12 Jadi, nilai a = – = 20 × 10–5 – 4 + 12 225 12. = 20 × 14. Jawaban: a 103 = 2 × 101 + 3 = 2 11. × 104 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000 

a : b =8–3 : ⎛⎛ 4 

1=

6 1 006d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000 –6 (42)–3 × −2 = 4 × 000000000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000 Jawaban: a32 0000000000000000000000000000000000000000000000000000116370727400000150000000336465736300000184 = (22)–6 × 322 0000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a000002 (25)2 = 2–12 2c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000 10 1 × 034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000 =2 2–2 = 4 004300000000c725452430000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400 12. 000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e7900006 46573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494 2–3 = =3c2118 54336313936362d322e31000000000000000000000000000000000000000000000000000000 Jawaban:

(

1 –2 ) 3 23 3

9 (

= 32 =

)–4 = (( )–1)–4 = ( )4 Jadi, pernyataan yang benar adalah pernyataan (ii) dan (iii). 32 2

ffd8ffe0001 ffd8ffe000 ffd8ffe000 ffd8ffe0001 ffd8ffe0001 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00010 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ffe000104 ffd8ffe00 ffd8ffe0001 ffd8ffe000 ffd8ffe000104 ffd8ffe0 ffd8ffe00010 ffd8ffe000 ffd8ffe0001 ffd8ffe000 ffd8ffe00 ffd8ffe0 ffd8ffe0001 ffd8ffe000 ffd8ffe00 ffd8ffe0001 ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe00 ffd8ffe000104 ffd8ffe00 ffd8ffe00 ffd8ffe00 ffd8ff ffd8ffe000 04a4649460 104a46494 104a46494 04a4649460 04a46494600 04a4649460 ffd8ffe0 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 4a464946000 0104a464 0104a464 0104a464 a46494600010 0104a464 04a4649460 104a46494 a46494600010 00104a46 4a464946000 104a46494 04a4649460 104a46494 0104a464 00104a46 04a46494600 104a46494 0104a464 04a46494600 104a46494 104a464946 104a46494 0104a464 a46494600010 0104a464 104a464946 0104a464 0104a464 ffd8ffe00010 001020100c 600010201 600010201 001020100c01020100c80 001020100ce0001 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 1020100c800 94600010 94600010 94600010 20100c800c80 94600010 001020100c 600010201 20100c800c80 49460001 1020100c800 600010201 001020100c 600010201 94600010 49460001 01020100c80 600010201 94600010 01020100c80 600010201 000102010 600010201 94600010 20100c800c80 946000100001020100 94600010 94600010 4a464946000 800c80000ff 00c800c800 00c800c800 800c80000ff 0c80000ffe20 800c80000ff 00104a4 04a46 c800c80000 1020100c800 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 c80000ffe20c 20100c80 20100c80 20100c80 000ffe20c5849 20100c80 800c80000ff 00c800c80 000ffe20c5849 020100c8 c80000ffe20c 00c800c800 800c80000ff 00c800c800 20100c80 020100c8 0c80000ffe20 00c800c80 20100c80 0c80000ffe20 00c800c80 0c800c8000 00c800c80 20100c80 000ffe20c5849 20100c80 20100c80 20100c80 e20c584943 00ffe20c58 00ffe20c58 e20c5849434 c584943435f e20c584943 ffd8ffe000104a46 c80000ffe20c 49460 ffe20c58494 435f50524f4 4943435f50 4943435f50 35f50524f46 50524f46494 435f50524f4 6494600 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 584943435f5 0c80000ff 0c80000ff 0c80000ff 43435f50524f4 0c80000ff e20c584943 000ffe20c5 43435f50524f4 00c80000 584943435f5 00ffe20c58 e20c584943 00ffe20c58 0c80000ff 00c80000 c584943435f 000ffe20c5 0c80000ff c584943435f 000ffe20c5 0ffe20c5849 000ffe20c5 0c80000ff 43435f50524f4 0c80000ff 0c80000ff 0c80000ff 49460001020100c8 584943435f50 6494c45000 524f46494c 524f46494c 494c450001c4500010100 6494c4500000102 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 0524f46494c e20c5849 e20c5849 e20c5849 6494c4500010 e20c5849 435f50524f4 84943435f 6494c4500010 ffe20c584 0524f46494c 4943435f50 435f50524f4 4943435f50 e20c5849 ffe20c584 50524f46494 84943435f e20c5849 50524f46494 84943435f 43435f5052 84943435f e20c5849 6494c4500010 e20c5849 3435f50524f e20c5849 e20c5849 524f46494c45 00c80000ffe20c584 10100000c4 450001010 450001010 0100000c48 10100000c4 0102010000c484c696 25. Jawaban: 19. Jawaban: 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 45000101000 43435f505 43435f505 43435f50 100000c484c6 43435f50 6494c45000 50524f464 100000c484c6 943435f5 45000101000 524f46494c 6494c45000 524f46494c 43435f50 943435f5 c4500010100 50524f464 43435f50 c4500010100 50524f464 4f46494c45 50524f464 43435f50 100000c484c6 43435f50 46494c4500 43435f50 43435f50 00010100000 0100c 2 84c696e6f02 0000c484c6 0000c484c6 4c696e6f021 e6f02100000 84c696e6f02 ffd8ffe000104a46 943435f50524f464 1 1 c484c696e6f0 = 2 94c450001 b c 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 00c484c696e 24f46494c 24f46494c 524f4649 96e6f0210000 524f4649 10100000c4 96e6f02100000 0524f464 00c484c696e 450001010 10100000c4 450001010 524f4649 0524f464 000c484c696 94c450001 524f4649 000c484c696 94c450001 000101000 94c450001 524f4649 96e6f02100000 524f4649 010100000c 524f4649 524f4649 1000006d6e 96e6f02100 96e6f02100 000006d6e7 6d6e7472524 1000006d6e 0c800c8 33 2 ffd8ffe000104a46 49460001020100c 94c4500010100000 800c8 3 5pq= 521000006d6e × 284c696e6f02 × 94c45000 3 4c450001 7472524742 0006d6e74 0006d6e74 4725247422 7422058595a 7472524742 2 200c484c696 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 6f021000006 45000101 45000101 4c450001 06d6e7472524 4c450001 0100000c4 6d6e74725247 6f021000006 0000c484c6 84c696e6f0 0000c484c6 4c450001 94c45000 e6f02100000 0100000c4 4c450001 e6f02100000 0100000c4 0100000c4 6d6e74725247 4c450001 484c696e6f 4c450001 4c450001 x=36 = (6) = 6 = 216 49460001020100c8 800c80000ffe20c5 11 c484c696e6f02100 74725247422 0000ff 021000006d 725247422 058595a200 2058595a20 3 0000ffe22007ce00020 52058595a20 2 725247422 8 2 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 d6e74725247 00000c48 00000c48 0100000c 7422058595a2 0100000c 1000006d6e 84c696e6f 422058595a20 10100000 d6e74725247 96e6f02100 21000006d6 96e6f02100 0100000c 10100000 6d6e7472524 84c696e6f 0100000c 6d6e7472524 84c696e6f e6f0210000 84c696e6f 0100000c 422058595a20 0100000c 0100000c 0100000c 3 2 058595a2007 00c80000ffe20c584 =5 × 2 0006d6e747252474 × 2 84943435f50524f4 ×5 =7ce00020009 07ce000200 058595a20 058595a20 0090006003 07ce000200 22 e20c5 0000000000000000000000000000000000000000000000 422058595a2 4c696e6f0 ce000200090 4c696e6f0 484c696e 007ce0002000 484c696e 7472524742 021000006 07ce00020009 c484c696 422058595a2 0006d6e74 e747252474 0006d6e74 484c696e c484c696 7422058595a 021000006 484c696e 7422058595a 021000006 06d6e7472 021000006 6 484c696e 07ce00020009 484c696e 6e74725247 484c696e 484c696e 3 4 2 3 3 = (21000061637 0900060031 07ce000200 07ce000200 0006003100 0900060031 943435f50524f464 6494c4500010100 ) 22058595a2007ce0 y= 64 = 2 = 16 0c58494 2 0061637370 = 6 00600310000 6 007ce000200 84943 007ce000200 21000006 = 20 21000006 6f021000 900060031000 × 6f021000 2058595a20 2 d6e747252 000600310000 × e6f02100 5 725247422 22058595a2 725247422 6f021000 e6f02100 2007ce00020 d6e747252 6f021000 2007ce00020 d6e747252 524742205 d6e747252 6f021000 000600310000 6f021000 422058595a 6f021000 6f021000 0000616373 090006003 090006003 3704d53465 0000616373 3 94c4500010100000 000c484c696e6f02 0020009000600310 616373704d5 43435f505 1 704d534654 100006163 100006163 4d53465400 4000000004 704d534654 09000600310 d6e74725 d6e74725 006d6e74 0616373704d5 006d6e74 07ce000200 474220585 616373704d53 0006d6e7 09000600310 058595a20 007ce00020 058595a20 006d6e74 0006d6e7 00900060031 474220585 006d6e74 00900060031 474220585 8595a2007c 474220585 006d6e74 616373704d53 006d6e74 2007ce0002 006d6e74 006d6e74 Jadi, nilai 435f50 c484c696e6f02100 1000006d6e74725 23 000616373704d534 34654000000 0000000049 73704d534 73704d534 00000049459454320735 0000000049 00061637370 24742205 24742205 72524742 346540000000 72524742 0900060031 95a2007ce 465400000000 00061637370 07ce00020 0090006003 07ce000200 72524742 47252474 00006163737 95a2007ce 72524742 00006163737 95a2007ce e00020009 95a2007ce 72524742 465400000000 72524742 0009000600 72524742 72524742 647252474 00494543207 = 20 (2 ×5 ) = 20 0006d6e74725247 247422058595a20 6540000000049454 524f46 4543207352 654000000 654000000 4320735247 2474200000 4543207352 20. 24f46494 30. Jawaban: 35247420000 4d534654000 8595a200 8595a200 2058595a 049454320735 2058595a 0000616373 000200090 494543207352 22058595 4d534654000 009000600 1000061637 090006003 2058595a 22058595 04d53465400 000200090 2058595a 04d53465400 000200090 000600310 000200090 2058595a 494543207352 2058595a 3100006163 2058595a 2058595a 4742000000 004945432 004945432 4200000000 0000000000 4742000000 422058595a2007ce 07ce00020009000 3207352474200000 a 22122 1 33933 9 494c4 73704d5346 b 00000000000 00000494543 7ce00020 7ce00020 2007ce00 247420000000 2007ce00 704d534654 006003100 474200000000 a2007ce0 00000494543 310000616 3704d53465 100006163 2007ce00 a2007ce0 00000049454 006003100 2007ce00 00000049454 006003100 000616373 006003100 2007ce00 474200000000 2007ce00 2007ce00 2007ce00 0000000000 073524742 073524742 0000000000 0000000000 23 3(9 +27 ) = 3((3) +(3 ) ) c450001 000200090006003 600310000616373 0000000000000000 ffd8ffe000104a4649 ffd8ffe000104a4649460 2 00000000000 50001 0000000000 000000000 000000000 0000000000f 000f6d60001 0000000000 20735247420 00900060 00900060 02000900 000000000000 02000900 0000000049 006163737 000000000000 00200090 20735247420 373704d53 4000000004 73704d534 02000900 32073524742 02000900 32073524742 006163737 704d53465 006163737 02000900 000000000000 02000900 5400000000 02000900 02000900 4006163737 00200090 460001020100c800c8  001020100c800c80000ff 4 5 10000616373704d 704d53465400000 45 0000000f6d600010 00f6d6000100 26. 00f6d600010 Jawaban: c ⎛ ⎛00000000d3 × ⎛⎛= ⎛⎛ 24 1 × ⎛⎛ 000000000 000000000 0000ffe20c584943435 6d60001000 e20c584943435f50524f4 00f6d600010 0100000 9 27 9 27 01000 00000000000 03100006 03100006 06003100 000000000f6d 06003100 4543207352 04d534654 00000000f6d6 00600310 00000000000 465400000 9454320735 654000000 06003100 00600310 00000000000 04d534654 06003100 0000000000 04d534654 400000000 04d534654 06003100 00000000f6d60 06003100 4945432073 06003100 06003100      f50524f46494c450001 6494c4500010100000c4 = 3(3 +3)33 3 000000d32d4 534654000000004 000494543207352 0000000d32d48502 0000000d32 000000000 000000000 00000d32d4 2d48502020 0000000d32 ffd8ffe000104a46494600 0100000c484c696e6f 84c696e6f021000006d6e 85020200000 00000000000 16373704 16373704 00616373 600010000000 00616373 4742000000 000000004 000100000000 00061637 00000000000 000494543 2474200000 004945432 00616373 00061637 00000000000 000000004 00616373 0000000000 000000004 494543207 000000004 00616373 00100000000d 00616373 5247420000 00616373 00616373 24+21 33 33 d485020200 0f6d600010 0f6d600010 8502020000 0000000000 d485020200 +00c48 021000006d6e747252 74725247422058595a20 945432073524742 474200000000000 0200000000000000 01020100c800c80000ffe2 c484c69 00000000000 47422058595a2007ce 07ce0002000900060031 1704d5346 = 3 +3 2 0000000000 00000f6d600 d5346540 0000000d3 0000000d3 d5346540 704d5346 0d32d4850202 704d5346 0000000000 945432073 d32d48502020 3704d534 00000f6d600 207352474 0000000000 073524742 0000000000 704d5346 3704d534 000000f6d60 945432073 704d5346 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ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c450 0010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006 00310000616373704d53465400000000494543207352474200000000000000000000 00000000f6d6000100000000d32d4850202000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000116370 Bab V Barisan dan 7274000001500000003364657363000001840000006c77747074000001f000000014 Deret 626b707400000204000000147258595a00000218000000146758595a0000022c0000 00146258595a0000024000000014646d6e640000025400000070646d6464000002c4 00000088767565640000034c0000008676696577000003d4000000246c756d690000 03f8000000146d6561730000040c0000002474656368000004300000000c72545243 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465 787400000000436f70797269676874202863292031393938204865776c6574742d50 61636b61726420436f6d70616e790000646573630000000000000012735247422049 454336313936362d322e310000000000000000000000127352474220494543363139 36362d322e31000000000000000000000000000000000000000000000000000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe0001 ffd8ffe000 ffd8ffe0001 ffd8ffe00 ffd8ffe0001 ffd8ffe0001 ffd8ffe0 ffd8ffe0001 ffd8ffe00 ffd8ffe0001 ffd8ffe0 ffd8ffe0001 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 04a4649460 104a46494 04a4649460 0104a464 04a4649460 04a4649460 00104a46 04a4649460 0104a464 04a4649460 00104a46 04a4649460 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 001020100c 600010201 001020100c 94600010 001020100c 001020100c 49460001 001020100c 94600010 001020100c 49460001 001020100c 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 800c80000ff 00c800c800 800c80000ff 20100c80 800c80000ff 800c80000ff 020100c8 800c80000ff 20100c80 800c80000ff 020100c8 800c80000ff 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 e20c584943 00ffe20c58 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 435f50524f4 4943435f50 e20c584943 0c80000ff e20c584943 e20c584943 00c80000 e20c584943 0c80000ff e20c584943 00c80000 e20c584943 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 6494c45000 524f46494c 435f50524f4 e20c5849 435f50524f4 435f50524f4 ffe20c584 435f50524f4 e20c5849 435f50524f4 ffe20c584 435f50524f4 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 10100000c4 450001010 6494c45000 43435f50 6494c45000 6494c45000 943435f5 6494c45000 43435f50 6494c45000 943435f5 6494c45000 1Jawaban: b Banyak kelereng pada setiap pola 62d322e31000000000000000000000000000000000000000000000000000000 84c696e6f02 0000c484c6 10100000c4 524f4649 10100000c4 10100000c4 0524f464 10100000c4 524f4649 10100000c4 0524f464 10100000c4 1000006d6e 96e6f02100 sebagai berikut. Pola ke-1 = 1 Pola ke-2 = 1 + 2 7472524742 0006d6e74 84c696e6f0 4c450001 84c696e6f0 84c696e6f0 94c45000 84c696e6f0 4c450001 84c696e6f0 94c45000 84c696e6f0 Pola ke-3 = 1 + 2 + 3 Pola ke-4 = 1 + 2 + 3 + 4 2058595a20 725247422 21000006d6 0100000c 21000006d6 21000006d6 10100000 21000006d6 0100000c 21000006d6 10100000 21000006d6 Pola ke-5 = 1 + 2 + 3 + 4 + 5 Pola ke-6 = 1 + 2 + 07ce000200 058595a20 e747252474 484c696e e747252474 e747252474 c484c696 e747252474 484c696e e747252474 c484c696 e747252474 3 + 4 + 5 + 6 Pola ke-7 = 1 + 2 + 3 + 4 + 5 + 6 + 0900060031 07ce000200 22058595a2 6f021000 22058595a2 22058595a2 e6f02100 22058595a2 6f021000 22058595a2 e6f02100 22058595a2 0000616373 090006003 7 = 28 Jadi, banyak kelereng pada pola ke-7 ada A. Pilihan 704d534654 100006163 007ce00020 006d6e74 007ce00020 007ce00020 0006d6e7 007ce00020 006d6e74 007ce00020 0006d6e7 007ce00020 28. 0000000049 73704d534 Ganda 0090006003 72524742 0090006003 0090006003 47252474 0090006003 72524742 0090006003 47252474 0090006003 2Jawaban: c Jika pola di atas diteruskan akan 4543207352 654000000 1. Jawaban: a 3, 1000061637 2058595a 1000061637 1000061637 22058595 1000061637 2058595a 1000061637 22058595 1000061637 4742000000 004945432 diperoleh pola ke-8 sebagai berikut. 3704d53465 2007ce00 3704d53465 3704d53465 a2007ce0 3704d53465 2007ce00 3704d53465 a2007ce0 3704d53465 0000000000 073524742 0000000000 000000000 4000000004 02000900 4000000004 4000000004 00200090 4000000004 02000900 4000000004 00200090 4000000004 +3 +3 +3 +3+3 +3 Pola bilangan di atas 00f6d60001 000000000 9454320735 06003100 9454320735 9454320735 00600310 9454320735 06003100 9454320735 00600310 9454320735 adalah menambahkan 3 pada 00000000d3 000000000 bilangan 2474200000 00616373 2474200000 2474200000 00061637 2474200000 00616373 2474200000 2474200000 2d48502020 0f6d600010 sebelumnya. a= 15 + 300061637 =18 b= a + 3= 18 0000000000 704d5346 0000000000 0000000000 3704d534 0000000000 704d5346 0000000000 0000000000 0000000d3 3704d534 0000000000 + 3 = 21 Jadi, dua bilangan selanjutnya 0000000000 2d4850202 0000000000 54000000 0000000000 65400000 0000000000 54000000 0000000000 65400000 0000000000 adalah 18 0000000000 dan 21. 0000000000 000000000 000f6d6000 00494543 000f6d6000 000f6d6000 00049454 000f6d6000 00494543 000f6d6000 00049454 000f6d6000 0000000000 000000000 100000000d 20735247 100000000d 100000000d 32073524 100000000d 20735247 100000000d 32073524 100000000d 0000000000 000000000 ffd8ffe000104a4649460001 2. 32d4850202 Jawaban: a 42000000 75, 70, 65, a, 55, 50, 45, 0000000000 000000000 42000000 32d4850202 32d4850202 74200000 32d4850202 32d4850202 74200000 32d4850202 0000000000 000000000 b, · 020100c800c80000ffe20c584 ·0000000000 · 00000000 0000000000 0000000000 00000000 0000000000 00000000 0000000000 00000000 0000000000 0000000000 000000000 ffd8ffe000104a4649460001020100c800c 0000000000 00000000 0000000000 0000000000 00000000 0000000000 00000000 0000000000 00000000 0000000000 943435f50524f46494c45000 0000000000 000000000 –5 –5 –5–5–5 –5–5 Pola bilangan di atas adalah 80000ffe20c584943435f50524f46494c450 0000000000 000000f6 0000000000 0000000000 0000000f 0000000000 000000f6 0000000000 0000000f 0000000000 0000116370 000000000 10100000c484c696e6f02100 mengurangkan 5 pada bilangan 0010100000c484c696e6f021000006d6e7 7274000001 000000000 0000000000 d6000100 0000000000 0000000000 6d600010 0000000000 d6000100 0000000000 6d600010 0000000000 0006d6e7472524742205859 sebelumnya. a= 65 –0000000000 5000000000 =0000000000 60 b= 45 – 5 = 5000000033 4725247422058595a2007ce00020009000 0000000000 000000d3 0000000000 0000000000 0000000d 0000000000 000000d3 0000000d 6465736300 000001163 40 Jadi, nilai a dan b berturut-turut adalah 5a2007ce0002000900060031 600310000616373704d53465400000000 0000000000 2d485020 0000000000 0000000000 32d48502 0000000000 2d485020 0000000000 32d48502 0000000000 0001840000 707274000 60 dan 40. 0000616373704d534654000 494543207352474200000000000000000 0000000000 20000000 0000000000 0000000000 02000000 0000000000 20000000 0000000000 02000000 0000000000 006c777470 001500000 00000000000f6d6000100000000d32d485 74000001f0 003364657 0000000000 00000000 0000000000 0000000000 00000000 0000000000 00000000 0000000000 00000000 0000000000 00000494543207352474200 0000001462 363000001 020200000000000000000000000000000 0000000000 00000000 0000000000 0000000000 00000000 0000000000 00000000 0000000000 00000000 0000000000 3. 00000000000000000000000 Jawaban: c 1, 3, 6, 11, 18, 6b70740000 840000006 000000000000000000000000000000000 0000011637 00000000 0000011637 0000011637 00000000 0000011637 00000000 0000011637 00000000 0000011637 29, 000f6d6000100000000d32d4 ··· 0204000000 c77747074 000000000000000000000000000000000 0727400000 00000000 0727400000 0727400000 00000000 0727400000 00000000 0727400000 00000000 0727400000 147258595a 000001f000 85020200000000000000000 116370727400000150000000336465736 0000021800 000014626 1500000003 1500000003 00000000 1500000003 00000000 1500000003 00000000 1500000003 +2 00000000 +3 +5 1500000003 +7+11 0000146758 b70740000 00000000000000000000000 3000001840000006c77747074000001f00 3646573630 00000000 3646573630 3646573630 00000000 3646573630 00000000 3646573630 00000000 3646573630 595a000002 020400000 2, 3,00000000 5, 7, 11, · ·0000184000 ·00000000 merupakan bilangan 0000014626b7074000002040000001472 0000184000 0000184000 0000184000 00000000 0000184000 00000000 0000184000 00000000000000000000000 2c00000014 014725859 58595a00000218000000146758595a0000 0006c77747 00000000 0006c77747 00000000 00000000 0006c77747 00000000 0006c77747 prima Jadi,0006c77747 pola0006c77747 bilangan 1, 3, 6, 11, 18, 6258595a00 5a0000021 00000000000000000000000 022c000000146258595a00000240000000 0002400000 800000014 074000001f 00000000 074000001f 074000001f 00000000 074000001f 00000000 074000001f 00000000 074000001f 29, · · · adalah menambahkan bilangan 00000000011637072740000 0014646d6e 6758595a0 14646d6e640000025400000070646d6464 0000000146 00000000 0000000146 0000000146 00000000 0000000146 00000000 0000000146 00000000 0000000146 prima berurutan. 6400000254 000022c00 01500000003364657363000 1Jawaban: a26b7074000 Banyak noktah pada pola ke-1 = 4 000002c400000088767565640000034c00 26b7074000 00000000 26b7074000 00000000 26b7074000 00000000 26b7074000 00000000 26b7074000 0000007064 000014625 Banyak noktah pada pola ke-2 = 8 Banyak noktah 001840000006c77747074000 00008676696577000003d4000000246c75 0020400000 11637072 0020400000 0020400000 01163707 0020400000 11637072 0020400000 01163707 0020400000 6d64640000 8595a0000 pada pola ke-3 = 12 Banyak noktah pada pola ke-4 02c4000000 024000000 6d69000003f8000000146d656173000004 0147258595 74000001 0147258595 0147258595 27400000 0147258595 74000001 0147258595 27400000 0147258595 001f000000014626b7074000 8876756564 014646d6e = 16 Banyak noktah pada pola ke-5 = 20 Jadi, pola 0c0000002474656368000004300000000c a000002180 50000000 a000002180 a000002180 15000000 a000002180 50000000 a000002180 15000000 a000002180 00204000000147258595a00 0000034c00 640000025 bilangan yang sesuai adalah 4, 8, 12, 16, 20. 725452430000043c0000080c6754524300 0000014675 33646573 0000014675 0000014675 03364657 0000014675 33646573 0000014675 03364657 0000014675 0000867669 400000070 000218000000146758595a0 2Jawaban: d8595a00000 Pola8595a00000 ke-1 =646d64640 1 8595a00000 × 3 Pola ke-2 = 2 × 4 00043c0000080c625452430000043c0000 8595a00000 63000001 8595a00000 36300000 63000001 8595a00000 36300000 6577000003 d400000024 00002c400 080c7465787400000000436f7079726967 Pola000022c000000146258595a0 ke-3 = 322c0000001 × 5 Pola ke-4 = 4 × 6 Pola ke-5 = 5 22c0000001 84000000 22c0000001 18400000 22c0000001 84000000 22c0000001 18400000 22c0000001 6c756d6900 000088767 6874202863292031393938204865776c65 × 7000024000000014646d6e64 Pola6c777470 ke-646258595a0 =6× 8 Pola ke-7 = 7 × 9 Pola ke-7 46258595a0 46258595a0 06c77747 46258595a0 6c777470 46258595a0 06c77747 46258595a0 0003f80000 565640000 74742d5061636b61726420436f6d70616e = 80000025400000070646d646 × 10 Pola0000240000 ke-90000240000 = 9 × 11 Pola ke-10= 10 × 12 0000240000 74000001 0000240000 07400000 74000001 0000240000 07400000 0000240000 00146d6561 034c00000 790000646573630000000000000012735 = 120 Jadi, banyak lingkaran pada 00014646d6 f0000000 00014646d6 00014646d6 1f000000 00014646d6 f0000000 00014646d6 1f000000 00014646d6 730000040c 086766965 pola ke-10 4000002c4000000887675656 0000002474 77000003d 247422049454336313936362d322e31000 adalah e640000025 14626b70 e640000025 e640000025 014626b7 e640000025 14626b70 e640000025 014626b7 e640000025 40000034c000000867669657 6563680000 400000024 000000000000000000012735247422049 4000000706 74000002 4000000706 4000000706 07400000 4000000706 74000002 4000000706 07400000 4000000706 0430000000 6c756d690 7000003d4000000246c756d6 454336313936362d322e31000000000000 46d6464000 04000000 46d6464000 46d6464000 20400000 46d6464000 04000000 46d6464000 20400000 46d6464000 0c72545243 00003f8000 11 11 −002c400000 − 9000003f8000000146d65617 000000000000000000000000000000000 002c400000 14725859 002c400000 01472585 002c400000 14725859 002c400000 01472585 002c400000 0000043c00 000146d65 2 2 2 2 7. x + x = 2(x –x ) 00080c6754 617300000 000000000 0887675656 5a000002 0887675656 0887675656 95a00000 0887675656 5a000002 0887675656 95a00000 0887675656 30000040c000000247465636 ⇔ ab = c(b + 5243000004 40c000000 40000034c0 18000000 40000034c0 40000034c0 21800000 40000034c0 18000000 40000034c0 21800000 1 40000034c0 8000004300000000c7254524 2a) 3c0000080c 247465636

⇔ 0000086766 14675859 0000086766 0000086766 01467585 0000086766 14675859 0000086766 01467585 0000086766 6254524300 800000430 30000043c0000080c6754524 xx 1 9657700000 9657700000 5a000002 9657700000 9657700000 95a00000 5a000002 9657700000 95a00000 9657700000 00043c0000 0000000c7 30000043c0000080c6254524 222c00000 ⇔ x 2c000000 +3d40000002 =3d40000002 22c000000 080c746578 254524300 3d40000002 3d40000002 22c00000 3d40000002 3d40000002 7400000000 00043c000 x 30000043c0000080c7465787 46c756d690 14625859 46c756d690 46c756d690 01462585 46c756d690 14625859 46c756d690 01462585 46c756d690 436f707972 0080c6754 00003f8000 5a000002 00003f8000 00003f8000 95a00000 00003f8000 5a000002 00003f8000 95a00000 00003f8000 400000000436f70797269676 6967687420 524300000 120. 000146d656 40000000 000146d656 000146d656 24000000 000146d656 40000000 000146d656 24000000 000146d656 2863292031 43c000008 87420286329203139393820 3939382048 0c6254524 1730000040 14646d6e 1730000040 1730000040 014646d6 1730000040 14646d6e 1730000040 014646d6 1730000040 4865776c6574742d5061636b 65776c6574 30000043c c000000247 64000002 c000000247 c000000247 e6400000 c000000247 64000002 c000000247 e6400000 c000000247 742d506163 0000080c7 61726420436f6d70616e7900 4656368000 54000000 4656368000 4656368000 25400000 4656368000 54000000 4656368000 25400000 4656368000 6b61726420 465787400 00646573630000000000000 0043000000 70646d64 0043000000 0043000000 070646d6 0043000000 70646d64 0043000000 070646d6 0043000000 436f6d7061 000000436f 6e79000064 707972696 00c7254524 64000002 00c7254524 00c7254524 46400000 00c7254524 64000002 00c7254524 46400000 00c7254524 01273524742204945433631 6573630000 768742028 30000043c0 c4000000 30000043c0 30000043c0 2c400000 30000043c0 c4000000 30000043c0 2c400000 30000043c0 3936362d322e31000000000 0000000000 632920313 000080c675 88767565 000080c675 000080c675 08876756 000080c675 88767565 000080c675 08876756 000080c675 1273524742 939382048 00000000000001273524742 4524300000 64000003 4524300000 4524300000 56400000 4524300000 64000003 4524300000 56400000 4524300000 2049454336 65776c657 2049454336313936362d322 313936362d 4742d5061 43c0000080 4c000000 43c0000080 43c0000080 34c00000 43c0000080 4c000000 43c0000080 34c00000 43c0000080 322e310000 636b61726 e3100000000000000000000 c625452430 86766965 c625452430 c625452430 08676696 c625452430 86766965 c625452430 08676696 c625452430 0000000000 420436f6d7 000043c000 77000003 000043c000 000043c000 57700000 000043c000 77000003 000043c000 57700000 000043c000 00000000000000000000000 0000000012 0616e7900 0080c74657 d4000000 0080c74657 0080c74657 3d400000 0080c74657 d4000000 0080c74657 3d400000 0080c74657 006465736 00000000000 7352474220

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe00 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe0001 ffd8ffe0001 ffd8ffe0001 ffd8ffe0001 ffd8ffe0 ffd8ffe000104 ffd8ffe000104 ffd8ffe00 ffd8ffe000104a46 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 0104a464 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 04a4649460 04a4649460 04a4649460 04a4649460 00104a4 a46494600010 a46494600010 0104a464 49460001020100c 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 94600010 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 001020100c 001020100c 001020100c 001020100c 6494600 20100c800c80 20100c800c80 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 94600010 800c80000ffe20c5 20100c80 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 800c80000ff 800c80000ff 800c80000ff 800c80000ff 0102010 000ffe20c5849 000ffe20c5849 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 0c80000ff 20100c80 84943435f50524f4 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 e20c584943 e20c584943 e20c584943 e20c584943 0c800c8 43435f50524f4 43435f50524f4 e20c5849 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 0c80000ff 6494c4500010100 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 435f50524f4 435f50524f4 435f50524f4 435f50524f4 0000ffe2 6494c4500010 6494c4500010 43435f50 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 6494c45000 6494c45000 6494c45000 6494c45000 0c58494 100000c484c6 100000c484c6 28 36 P = 23 + 10= 33 Q =e20c5849 P + 12 = 33 + 12= 000c484c696e6f02 62d322e31000000000000000000000000000000000000000000000000000000 524f4649

Jadi, 3435f505 bilangan selanjutnya96e6f0210000 adalah .dan 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 10100000c4 10100000c4 96e6f0210000 4c450001 343512 45 R =Q +10100000c4 14 =10100000c4 45 + 14 = 59 Jadi, tiga 43435f50 1000006d6e74725 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 84c696e6f0 84c696e6f0 84c696e6f0 84c696e6f0 24f46494 06d6e7472524 06d6e7472524 0100000c bilangan berikutnya adalah 33, 45, dan 14. Jawaban: d Bilangan- 247422058595a20 524f4649 484c696e 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 21000006d6 21000006d6 21000006d6 21000006d6 c450001 7422058595a2 7422058595a2 bilangan pada pembilang: 1, 2,07ce00020009000 4, 4c450001 6f021000 0000000000000000000000000000000000000000000000 e747252474 e747252474 e747252474 e747252474 0100000 007ce0002000 007ce0002000 59. 8, 16, 32, 006d6e74 0100000c 600310000616373 22058595a2 22058595a2 22058595a2 22058595a2 c484c69 900060031000 900060031000 72524742 10. 007ce00020 Jawaban: d007ce00020 2, 50, 74, ffd8ffe000104a4649460001 007ce00020 007ce00020 6e6f0210 0616373704d5 0616373704d5 ×2 ×2 ×2×2×2 ×2 484c696e 704d53465400000 2058595a 86, x,0090006003 95 0090006003 020100c800c80000ffe20c58 0090006003 0090006003 00006d6 346540000000 346540000000 Bilangan di urutan ke-7 pada pembilang 6f021000 000494543207352 2007ce00 ffd8ffe000104a464946000102 1000061637 1000061637 1000061637 e747252 049454320735 049454320735 +48 +24 +12 +61000061637 +3 4943435f50524f46494c4500 adalah 64. Bilangan-bilangan pada 02000900 006d6e74 474200000000000 0100c800c80000ffe20c5849434 3704d53465 3704d53465 3704d53465 3704d53465 4742205 247420000000 247420000000 06003100 010100000c484c696e6f0210 penyebut: 1.024, 512, 256,000000000000000 128, 64, 32, x4000000004 = 864000000004 + 64000000004 = 924000000004 Jadi,72524742 isian 35f50524f46494c45000101000 8595a20 000000000000 000000000000 00616373 00006d6e747252474220585 yang tepat adalah 92. ffd8ffe000104a4649460001 00c484c696e6f021000006d6e7 2058595a 9454320735 9454320735 9454320735 9454320735 07ce0001 1 1 1 00f6d6000100000 704d5346 1000000000f6d 1000000000f6d × × × × ×× 2 22222 95a2007ce00020009000600 020100c800c80000ffe20c58 54000000 2474200000 2474200000 2474200000 2000900 600010000000 600010000000 11. 4725247422058595a2007ce000 Jawaban: c 2474200000 x, 2007ce00 000d32d48502020 2000900060031000061637370 Bilangan di urutan ke-7 pada0d32d4850202 penyebut adalah 16. 00494543 310000616373704d5346540 0000000000 0000000000 0000000000 0000000000 0600310 0d32d4850202 4943435f50524f46494c4500 02000900 000000000000000 4d53465400000000494543207 20735247 Jadi, bilangan pecahan di urutan ke-7 0000000000 0000000000 0000000000 0000000000 0006163 000000000000 000000000000 00000004945432073524742 010100000c484c696e6f0210 . ffd8ffe000104a46494600010201 ×3 ×3×3 ×3 ×3 06003100 000000000000000 42000000 3524742000000000000000000 64 16 000f6d6000 000f6d6000 000f6d6000 000f6d6000 73704d5 000000000000 000000000000 00000000000000000000000 ffd8ffe000104a464946 00c800c80000ffe20c584943435f5 adalah 00006d6e747252474220585 00000000 0000000000f6d6000100000000 000000000000000 100000000d 100000000d 100000000d 100000000d 3465400 000000000000 000000000000 Pada pola bilangan di00616373 atas, diperoleh: 15. 0524f46494c4500010100000c484 0001020100c800c8000 00000000 00000f6d6000100000000d32 95a2007ce00020009000600 d32d485020200000000000000 704d5346 000000000000000 32d4850202 32d4850202 32d4850202 32d4850202 0000004 000000000000 000000000000 36 000000f6 0ffe20c584943435f5052 c696e6f021000006d6e747252474 36 = 3y ⇔0000000000 y = 0000000000 = 0000000000000000000000000 d4850202000000000000000 310000616373704d5346540 0000000000 0000000000 9454320 000000000000 000000000000 3 54000000 000000000000000 d6000100 Jawaban: c 4f46494c450001010000 22058595a2007ce000200090006 0000000000000000000000000 135 00000000000000000000000 0000000000 0000000000 0000000000 0000000000 7352474 000000000000 000000000000 00000004945432073524742 12 y = 3x ⇒ 12 = 3x 000000d3 +3 6 0c484c696e6f02100000 Pola bilangan: 3 00494543 000000000000000 00310000616373704d534654000 0000000000000000000000000 +8 = 231 2000000 +17 0000000000 0000000000 0000000000 0000000000 000000000000 000000000000 2d485020 00000000000000000000000 00000000000000000000000 6d6e747252474220585 12 +2 = 2 +5 11 000004945432073524742000000 000011637072740 0000001163707274000001500 ⇔ 0000000000 x= = 20735247 4 20000000 0000000000 0000000000 0000000000 0000000 000000000001 000000000001 00000000000000000000000 0000000000000000000000f6d600 00000f6d6000100000000d32 395a2007ce0002000900 Jadi, nilai x = 4 dan42000000 y 0000033646573630000018400 000015000000033 00000000 0000000000 0000000000 0000000000 0000000000 0000000 163707274000 163707274000 0600310000616373704 0100000000d32d4850202000000 = 12. 00000000000116370727400 +4 = 22 0000000 +9 20 d4850202000000000000000 00006c77747074000001f00000 00000000 0000000000 0000000000 0000000000 0000000000 00000000 646573630000018 001500000003 001500000003 d534654000000004945 000000000000000000000000000 00000000 0014626b70740000020400000 00015000000033646573630 12. Jawaban: 00000000000000000000000 0000000000 0000000000 0000000000 0000000000 0f6d6000 364657363000 364657363000 00000000 40000006c777470 4320735247420000000 000000000000000000000000000 00000000 0147258595a00000218000000 Banyak segitiga putih pada pola d 00001840000006c77747074 0000000000 0000000000 0000000000 0000000000 1000000 001840000006 001840000006 2 5 8 1114 00000000000000000000000 0000000000000000000 ffd8ffe000104a4649460001 000000f6 74000001f000000 000000000000000000000000000 00000000 ,,,,,··· 146758595a0000022c00000014 ke-8 = 8 + 7 + 6 + 5 + 4 + 3 + 0000000000 0000000000 0000000000 0000000000 00d32d4 c77747074000 c77747074000 14916 25 000001f000000014626b7074 00000000000000000000000 00f6d6000100000000d3 020100c800c80000ffe20c584 00000000 000000001163707274000001500 4 d6000100 014626b70740000 6258595a00000240000000146 +n x 2+16 +00000000 1= =2 36 0000011637 0000011637 0000011637 0000011637 8502020 001f00000001 001f00000001 2d48502020000000000 00000204000000147258595 000003364657363000001840000 Bilangan pada 943435f50524f46494c45000 00000000000116370727400 46d6e64000002540000007064 000000d3 020400000014725 0727400000 0727400000 0727400000 0727400000 0000000 4626b7074000 4626b7074000 00000000 8. a0000021800000014675859 Jawaban: c Jika5 pola diteruskan akan 0000000000000000000 10100000c484c696e6f02100 006c77747074000001f000000014 Bilangan-bilangan pembilang: pembilang: 2, 5,1500000003 8,pada 11, 00015000000033646573630 6d6464000002c4000000887675 +32 = 2 1500000003 1500000003 1500000003 0000000 002040000001 002040000001 2d485020 8595a0000021800 00000000 diperoleh pola ke-10 sebagai berikut. 0000000000000000000 0006d6e7472524742205859 626b70740000020400000014725 1, 65640000034c00000086766965 3, 6, 10,3646573630 5a0000022c0000001462585 00001840000006c77747074 14, 00000000 +p0000000 3646573630 3646573630 3646573630 47258595a000 47258595a000 +3 +3 +3 +3 +3+3 20000000 0000146758595a0 0000000000000000000 5a2007ce000200090006003 8595a0000021800000014675859 77000003d4000000246c756d69 00000000 95a0000024000000014646d 0000184000 0000184000 0000184000 0000184000 0000000 002180000001 002180000001 000001f000000014626b7074 0000000000000000000 10000616373704d53465400 00000000 000022c00000014 5a0000022c000000146258595a0 000003f8000000146d65617300 Bilangan pada pembilang berpola 11637072 0006c77747 0006c77747 0006c77747 0006c77747 0000000 46758595a000 46758595a000 Nilai n = 17 + 16 = 33 6e640000025400000070646 00000204000000147258595 0000000001163707274 00000049454320735247420 000024000000014646d6e640000 00000000 6258595a0000024 74000001 00040c00000024746563680000 ditambah 3. Dua bilangan pembilang 074000001f 074000001f 074000001f 074000001f 0000000 0022c0000001 0022c0000001 Nilai x = 37 + n = 37 + 33 = 70 0000015000000033646 ffd8ffe000104a464946000102010 00000000000000000000000 d6464000002c40000008876 025400000070646d6464000002c a0000021800000014675859 50000000 04300000000c72545243000004 00000000 000000014646d6e 0000000146 0000000146 0000000146 0000000146 0000000 46258595a000 46258595a000 selanjutnya 17 dan 20. Bilangan pada Jadi, isian yang tepat adalah 70. 5736300000184000000 0c800c80000ffe20c584943435f505 0000f6d6000100000000d32d ffd8ffe000104a4649460001020100 400000088767565640000034c00 33646573 7565640000034c000000867 5a0000022c0000001462585 3c0000080c675452430000043c 26b7074000 26b7074000 26b7074000 26b7074000 0000000 002400000001 002400000001 00000000 640000025400000 penyebut: 6c77747074000001f000 24f46494c4500010100000c484c69 48502020000000000000000 c800c80000ffe20c584943435f50524 63000001 00008676696577000003d400000 1, 4, 9, 16, 25, Bilangan pada penyebut 0000080c625452430000043c00 6696577000003d400000024 95a0000024000000014646d 0020400000 0020400000 0020400000 0020400000 0000000 4646d6e64000 4646d6e64000 B. 00000000 070646d64640000 84000000 000014626b707400000 6e6f021000006d6e7472524742205 00000000000000000000000 f46494c4500010100000c484c696e6f 0246c756d69000003f8000000146 berpola bilangan persegi. 00080c7465787400000000436f 6c756d69000003f800000014 0147258595 0147258595 0147258595 0147258595 0000000 002540000007 002540000007 6e640000025400000070646 6c777470 204000000147258595a 8595a2007ce000200090006003100 00000000000000000000000 021000006d6e74725247422058595 00000000 02c400000088767 d6561730000040c000000247465 7079726967687420286329203 1. a. Jika pola tersebut dilanjutkan, akan a000002180 a000002180 a000002180 a000002180 0000000 0646d6464000 0646d6464000 74000001 6d6561730000040c0000002 Uraian d6464000002c40000008876 0000021800000014675 00616373704d5346540000000049 00000000000000000000000 a2007ce0002000900060031000061 ffd8ffe000104a4649460001020100c800c 6368000004300000000c7254524 00000000 565640000034c00 1393938204865776c6574742d5 diperoleh pola ke-4, ke-5, ke-6 sebagai f0000000 0000014675 0000014675 0000014675 0000014675 0000000 002c40000008 002c40000008 Dua bilangan penyebut selanjutnya 36 dan 49. 8595a0000022c000000 4543207352474200000000000000 00000000001163707274000 6373704d534654000000004945432 80000ffe20c584943435f50524f46494c450 47465636800000430000000 30000043c0000080c6754524300 7565640000034c000000867 061636b61726420436f6d70616 00000000 000086766965770 14626b70 berikut. 8595a00000 8595a00000 8595a00000 8595a00000 0000000 876756564000 876756564000 17 146258595a000002400 00000000000000f6d600010000000 00150000000336465736300 07352474200000000000000000000 0010100000c484c696e6f021000006d6e7 00043c0000080c6254524300000 Jadi, dua bilangan selanjutnya dan 0c725452430000043c00000 e790000646573630000000000 6696577000003d400000024 74000002 22c0000001 22c0000001 22c0000001 22c0000001 0001163 0034c0000008 0034c0000008 00000000 00003d400000024 36 49 20 0000014646d6e640000 0d32d48502020000000000000000 0001840000006c777470740 00000000f6d6000100000000d32d48 4725247422058595a2007ce00020009000 43c0000080c7465787400000000 04000000 . 46258595a0 0000127352474220494543363 80c675452430000043c0000 6c756d69000003f800000014 46258595a0 46258595a0 46258595a0 7072740 676696577000 676696577000 00000000 6c756d69000003f8 025400000070646d646 0000000000000000000000000000 00001f000000014626b70740 50202000000000000000000000000 600310000616373704d53465400000000 13. Jawaban: 436f707972696768742028632920 14725859 13936362d322e310000000000 080c625452430000043c000 0000240000 0000240000 0000240000 0000240000 0000150 003d40000002 003d40000002 6d6561730000040c0000002 0000000000000000000000000000 0000204000000147258595a 00000000000000000000000000000 494543207352474200000000000000000 00000000 31393938204865776c6574742d5 000000146d65617 5a000002 d 4000002c40000008876 0000000000001273524742204 00014646d6 00014646d6 00014646d6 00014646d6 0000003 46c756d69000 46c756d69000 0080c746578740000000043 47465636800000430000000 7565640000034c00000 0000000000000000000000011637 00000218000000146758595 00000000000000000000000000000 00000000000f6d6000100000000d32d485 18000000 061636b61726420436f6d70616e7 11637072 30000040c000000 9454336313936362d322e3100 Banyak batang korek api pada pola ke-10 e640000025 e640000025 e640000025 e640000025 3646573 003f80000001 003f80000001 14675859 08676696577000003d4 0727400000150000000336465736 a0000022c00000014625859 00000000000001163707274000001 020200000000000000000000000000000 6f7079726967687420286329 900006465736300000000000000 0c725452430000043c00000 0000000000000000000000000 74000001 247465636800000 +2 4000000706 +3+4 +5+6 +7 +8 = banyak batang korek api yang mendatar 4000000706 4000000706 4000000706 6300000 46d656173000 46d656173000 5a000002 000000246c756d69000 3000001840000006c77747074000 5a0000024000000014646d6 50000000336465736300000184000 000000000000000000000000000000000 127352474220494543363139363 2031393938204865776c657 0000000000000000000000000 80c675452430000043c0000 Pola ke-4 Pola ke-5 Pola ke-6 + banyak batang tegak = 11 ×0040c0000002 10 + 11 × 46d6464000 46d6464000 46d6464000 46d6464000 1840000 0040c0000002 50000000 4300000000c7254 2c000000 003f8000000146d65617 001f000000014626b707400000204 e640000025400000070646d 0006c77747074000001f0000000146 000000000000000000000000000000000 Bilangan-bilangan pada pembilang 62d322e31000000000000000000 00 4742d5061636b6172642043 080c625452430000043c000 10 = 110006c777 + 110 = 220 Jadi, banyak batang 002c400000 002c400000 002c400000 002c400000 14625859 474656368000 474656368000 33646573 30000040c0000002474 000000147258595a000002180000 6464000002c400000088767 26b70740000020400000014725859 116370727400000150000000336465736 berpola bilangan segitiga. Dua bilangan b.000012735247422049454336313 Banyak kubus pada 52430000043c000 setiap pola 5a000002 korek api pada pola ke-10 adalah 220 0887675656 0887675656 0887675656 0887675656 6f6d70616e79000064657363 4707400 004300000000 004300000000 0080c746578740000000043 6563680000043000000 00146758595a0000022c000000146 565640000034c0000008676 5a00000218000000146758595a000 3000001840000006c77747074000001f00 936362d322e3100000000000000 63000001 0080c6754524300 pembilang selanjutnya adalah 28 dan 36. sebagai berikut. Pola ke-1 = 12c72545243000 = 1 Pola 40000000 40000034c0 40000034c0 40000034c0 40000034c0 0001f000 c72545243000 batang. 00000000000000127352474 6f7079726967687420286329 2 2 2 00c725452430000043c 258595a0000024000000014646d6 696577000003d4000000246 0022c000000146258595a00000240 0000014626b7074000002040000001472 000000000000000000000000000 84000000 00043c0000080c6 ke-2 14646d6e = 1 + 2 = 5 Pola ke-3 = 1 + 22 + Bilangan-bilangan pada penyebut: 1, 8, 0000086766 0000086766 0000086766 0000086766 0000146 0043c0000080 0043c0000080 0000080c67545243000 e640000025400000070646d64640 c756d69000003f8000000146 00000014646d6e640000025400000 58595a00000218000000146758595a0000 22049454336313936362d32 0000000000000 2 2031393938204865776c657 2 2 2 64000002 3 = 14 Pola ke-4 = 1 + 2 25452430000043c + 3c67545243000 + 42 = 30 9657700000 9657700000 9657700000 9657700000 26b7074 c67545243000 27, 64, 125, 216, · · ·6c777470 0043c0000080c625452 00002c4000000887675656400000 d6561730000040c00000024 070646d6464000002c40000008876 9. 022c000000146258595a00000240000000 Jawaban: c 3, 5, 9,15, 23, P, Q, 54000000 2 2 2 2 2e310000000000000000000 4742d5061636b6172642043 Pola ke-50000020 = 1 + 2 + 3 + 0000080c7465787 40043c0000080 +0043c0000080 52 = 55 3d40000002 3d40000002 77 7 7 3d40000002 7 7 73d40000002 7 74000001 430000043c0000080c7 34c0000008676696577000003d40 74656368000004300000000 7565640000034c000000867669657 70646d64 2 R, 14646d6e640000025400000070646d6464 2 00012735247422049454336 6f6d70616e79000064657363 46c756d690 46c756d690 Pola ke-64000000 = 1 + 22 + 32 + 400000000436f70 4c62545243000 +c62545243000 52 + 62 = 1465787400000000436f7 32333 44 546c756d690 3 63 746c756d690 3 83 Bilangan-bilangan f0000000 64000002 00000246c756d69000003f8000000 c725452430000043c0000080 7000003d4000000246c756d690000 000002c400000088767565640000034c00 00003f8000 00003f8000 00003f8000 00003f8000 1472585 0043c0000080 0043c0000080 313936362d322e310000000 91 pada penyebut berpola bilangan kubik. 00000000000000127352474 c4000000 +2 +4 +6 +8 +10 +12 +14 0797269676874202863 146d6561730000040c0000002474 c675452430000043c0000080 03f8000000146d6561730000040c00 00008676696577000003d4000000246c75 14626b70 797269676874202 000146d656 000146d656 000146d656 000146d656 95a0000 c74657874000 c74657874000 88767565 Dua bilangan penyebut selanjutnya adalah 00000000000000000000000 22049454336313936362d32 2920313939382048657 656368000004300000000c725452 c625452430000043c0000080 00002474656368000004300000000 6d69000003f8000000146d656173000004 74000002 863292031393938 1730000040 1730000040 1730000040 1730000040 0218000 00000436f707 00000436f707 64000003 343 dan 512. c7465787400000000436f707 c725452430000043c0000080c67545 76c6574742d5061636b 430000043c0000080c67545243000 0c0000002474656368000004300000000c 00000000000000000000000 2e310000000000000000000 04000000 204865776c65747 4c000000 c000000247 c000000247 c000000247 c000000247 0001467 972696768742 972696768742 61726420436f6d70616e 0043c0000080c625452430000043c 97269676874202863292031 2430000043c0000080c6254524300 725452430000043c0000080c6754524300 0 00012735247422049454336 86766965 4656368000 4656368000 4656368000 4656368000 58595a0 028632920313 028632920313 14725859 42d5061636b6172 7900006465736300000 0000080c7465787400000000436f7 393938204865776c6574742 00043c0000080c746578740000000 00043c0000080c625452430000043c0000 77000003 313936362d322e310000000 0043000000 0043000000 0043000000 0043000000 000022c 939382048657 939382048657 5a000002 6420436f6d70616e 0000000001273524742 0797269676874202863292031393 d5061636b61726420436f6d7 0436f7079726967687420286329203 080c7465787400000000436f7079726967 d4000000 00c7254524 00c7254524 00c7254524 00c7254524 0000001 76c6574742d5 76c6574742d5 00000000000000000000000 2049454336313936362 938204865776c6574742d5061636 0616e790000646573630000 1393938204865776c6574742d5061 6874202863292031393938204865776c65 18000000 790000646573630 246c756d 30000043c0 30000043c0 30000043c0 30000043c0 4625859 061636b61726 061636b61726 00000000000000000000000 d322e31000000000000 b61726420436f6d70616e79000064 00000000001273524742204 636b61726420436f6d70616e790000 74742d5061636b61726420436f6d70616e 69000003 14675859 000000000000012 000080c675 000080c675 000080c675 000080c675 5a00000 420436f6d706 420436f6d706 f8000000 0000000000127352474 6573630000000000000012735247 9454336313936362d322e31 64657363000000000000001273524 790000646573630000000000000012735 0 5a000002 735247422049454 4524300000 4524300000 4524300000 4524300000 2400000 16e790000646 16e790000646 146d6561 2204945433631393636 422049454336313936362d322e31 00000000000000000000001 7422049454336313936362d322e31 247422049454336313936362d322e31000 43c0000080 43c0000080 43c0000080 43c0000080 0014646 573630000000 573630000000 2c000000 336313936362d32 73000004 2d322e3100000000000 0000000000000000000000127352 27352474220494543363139 00000000000000000000001273524 000000000000000000012735247422049 c625452430 c625452430 c625452430 c625452430 d6e6400 000000012735 000000012735 0c000000 14625859 2e3100000000000 0000000000000000000 47422049454336313936362d322e 36362d322e3100000000000 7422049454336313936362d322e31 454336313936362d322e31000000000000 24746563 000043c000 000043c000 000043c000 000043c000 0002540 247422049454 247422049454 0000000000000000000 3100000000000000000000000000 00000000000000000000000 00000000000000000000000000000 000000000000000000000000000000000 5a000002 000000000001273 68000004 0080c74657 0080c74657 0080c74657 0080c74657 0000070 336313936362 336313936362 00000 0000000000000000000000000000 00000000000000000000· · · 0000000000000000000000000 000000000

ffd8ffe00 0104a464 94600010 ffd8ffe000104a4 20100c800 64946000102010 c80000ffe 0c800c80000ffe2 20c584943 0c584943435f50 435f50524 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000 c. Pola bilangan yang 524f46494c4500 248 16 b. , , ,,··· f46494c45 006d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000 010100000c484c terbentuk 1, 5, 14, 30, 55, 1 8 27 64 00010100 000000000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000 696e6f02100000 91, ··· Bilangan-bilangan pada pembilang: 2. a. 71, 000c484c6 0000000000000000000000000000000000000000000000000000116370727400000150000000336465736300000184 6d6e7472524742 ffd8ffe000104a46 ffd8ffe000104a46494600 96e6f0210 0000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a000002 49460001020100c 2058595a2007ce –701020100c800c80000ffe20 –7 –7 –7–7 –7 x = 50 – 7 = 43 y= x – 7 ffd8ffe000104a4649460001020100c8 00006d6e 800c80000ffe20c5 2c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000 00c80000ffe20c584943435f50524f464 =c584943435f50524f46494c 43 – 7 = 36 z= y – 00020009000600 7 = 36 – 7 = 29 84943435f50524f4 74725247 034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000 94c4500010100000c484c696e6f02100 Pembilang pada bilangan urutan6494c4500010100 ke-7 adalah 31000061637370 Jadi, tiga bilangan selanjutnya adalah 4500010100000c484c696e 42205859 0006d6e74725247422058595a2007ce 004300000000c725452430000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400 4d534654000000 000c484c696e6f02 128. 43, 36, dan 29. 6f021000006d6e74725247 5a2007ce0 00020009000600310000616373704d5 000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e7900006 1000006d6e74725 00494543207352 3465400000000494543207352474200 422058595a2007ce000200 00200090 Bilangan-bilangan pada 247422058595a20 46573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494 47420000000000 00000000000000000000000000f6d600 090006003100006163737 07ce00020009000 00600310 54336313936362d322e31000000000000000000000000000000000000000000000000000000 0100000000d32d48502020000000000 penyebut: 1, 8, 27, 64, 125, 600310000616373 04d534654000000004945 b. 16, 8, 4, 2, x, y, z 00000000000000 00061637 0000000000000000000000000000000 7 777777 0000f6d6000100 704d53465400000 216, ffd8ffe000104a4649460001020100 432073524742000000000 3704d534 111 11 1 10000000000000000000000000000000 3 23 3343 5363 73 Penyebut pada 000494543207352 000000d32d4850 ××× ×× × 0000000000000000000000011637072 c800c80000ffe20c584943435f50524f 0000000000000000000f6d 222 22 2 65400000 474200000000000 20200000000000 bilangan 7400000150000000336465736300000 urutan ke-7 adalah 46494c4500010100000c484c696e6f 6000100000000d32d4850 000000000000000 00049454 1 1840000006c77747074000001f000000 00000000000000 343. x= 2 × 2=1111 00f6d6000100000 021000006d6e74725247422058595 202000000000000000000 014626b707400000204000000147258 32073524 00000000000000 Jadi, bilangan pada urutan000d32d48502020 ke-7 a2007ce00020009000600310000616 000000000000000000000 595a00000218000000146758595a000 . 74200000 000000000000000 y= x × = 1 × = 128 00000000000000 0022c000000146258595a0000024000 343 373704d5346540000000049454320 000000000000000000000 2 2 2 adalah 00000000 000000000000000 00000000000000 000014646d6e6400000254000000706 3 79 3579 73524742000000000000000000000 1 111 000000000000000000000 00000000 c. , 1,46d6464000002c40000008876756564 , , · · ·sama dengan 000000000000000 , , , ,· · · 2 00000000000000 z= y × = × = 8 11 2000000000000000 5 8 0000000f6d6000100000000d32d485 2 224 1 000000000000001163707 0000034c0000008676696577000003d 0000000f6 ffd8ffe000104a46494600010 11 00000000000000 000000000000000 02020000000000000000000000000 274000001500000003364 4000000246c756d69000003f80000001 Bilangan-bilangan pada 000000000000000 d6000100 Jadi, tiga bilangan selanjutnya adalah 20100c800c80000ffe20c58494 11637072740000 46d6561730000040c00000024746563 00000000000000000000000000000 657363000001840000006 pembilang: 3, 5, 7, 9, 11, 13, 000000d3 000011637072740 1,3435f50524f46494c45000101 , 01500000003364 68000004300000000c7254524300000 2 1 00000000000000000000000000000 ffd8ffe000104a4649460001020100c8 c77747074000001f000000 000015000000033 . 2d485020 +2 +2 +2 +2+2 +2 dan 4 43c0000080c675452430000043c00000 65736300000184 00c80000ffe20c584943435f50524f464 646573630000018 00000c484c696e6f021000006 00000000000011637072740000015 014626b70740000020400 20000000 80c625452430000043c0000080c74657 94c4500010100000c484c696e6f02100 0000006c777470 40000006c777470 00000003364657363000001840000 c. 1, 16, 81,256, x, y, z 0000147258595a0000021 00000000 87400000000436f70797269676874202 Pembilang pada bilangan urutan ke-7 d6e74725247422058595a200 0006d6e74725247422058595a2007ce 74000001f000000 74000001f00000 863292031393938204865776c657474 006c77747074000001f00000001462 77ce0002000900060031000061 777777 8000000146758595a0000 00020009000600310000616373704d5 00000000 014626b70740000 adalah 15. Bilangan-bilangan pada 0014626b707400 2d5061636b61726420436f6d70616e79 3465400000000494543207352474200 16b707400000204000000147258595 022c000000146258595a00 4 24 3444 5464 74 020400000014725 00000000 0000646573630000000000000012735 6373704d5346540000000049 penyebut: 2, 5, 8, 11, 14, 17, 20 +3 +3 00020400000014 00000000000000000000000000f6d600 4 4 4 8595a0000021800 a00000218000000146758595a0000 00024000000014646d6e6 ffd8ffe000104a4649460001020100c8 x= 5 = 625 y= 6 =7258595a000002 1.296 z= 7 = 247422049454336313936362d322e31 00000000 0100000000d32d48502020000000000 0000146758595a0 454320735247420000000000 +3 +3+3 +3 00c80000ffe20c584943435f50524f464 022c000000146258595a0000024000 40000025400000070646d 0000000000000000000000127352474 2.401 Jadi, tiga bilangan selanjutnya 00000000 0000000000000000000000000000000 000022c00000014 18000000146758 ffd8ffe000104a464946000102 94c4500010100000c484c696e6f02100 22049454336313936362d322e310000 000000000000000000f6d6000 000014646d6e64000002540000007 6464000002c4000000887 0000000000000000000000000000000 Penyebut pada bilangan urutan ke-7 adalah 00000000 6258595a0000024 adalah 625, 1.296, dan 2.401. 0006d6e74725247422058595a2007ce 595a0000022c00 0100c800c80000ffe20c5849434 0000000000000000000000000000000 0646d6464000002c40000008876756 0000000000000000000000011637072 67565640000034c000000 000000014646d6e 100000000d32d48502020000 20. 00000000 00020009000600310000616373704d5 0000000000000000000 0000146258595a 35f50524f46494c45000101000 7400000150000000336465736300000 5640000034c0000008676696577000 8676696577000003d4000 3465400000000494543207352474200 Jadi, bilangan pada urutan640000025400000 ke-7 00000000 000000000000000000000000 1840000006c77747074000001f000000 00000240000000 00c484c696e6f021000006d6e7 . 070646d64640000 00000000000000000000000000f6d600 003d4000000246c756d69000003f80 000246c756d69000003f80 00000000 15 20 014626b707400000204000000147258 02c400000088767 000000000000000000000000 14646d6e640000 4725247422058595a2007ce000 0100000000d32d48502020000000000 d.1, 6, 12, 20, x, y, z adalah 00000146d6561730000040c0000002 00000146d656173000004 00000000 595a00000218000000146758595a000 2 −48 565640000034c00 02540000007064 2000900060031000061637370 000000000000000000000000 7777777 d. –1,0000000000000000000000000000000 , , , · · ·sama dengan 6 10 474656368000004300000000c72545 0022c000000146258595a0000024000 0c0000002474656368000 000086766965770 00000000 0000000000000000000000000000000 3 4d53465400000000494543207 1 000000000000000000011637 × 2 2 × 3 3 × 4 4 ×6d6464000002c4 55×66×77 000014646d6e6400000254000000706 00003d400000024 2430000043c0000080c67545243000 004300000000c72545243 0000000000000000000000011637072 11637072 46d6464000002c40000008876756564 −1 2 −48 3524742000000000000000000 6c756d69000003f ×072740000015000000033646 8 x = 5 × 6 = 30 y 00000088767565 = 6 × 7 74000001 = 42 z = 7400000150000000336465736300000 0043c0000080c625452430000043c0 0000043c0000080c675452 , , , ,· · · 0000034c0000008676696577000003d 8000000146d6561 1 610 640000034c0000 0000000000f6d6000100000000 1840000006c77747074000001f000000 3 000080c7465787400000000436f707 7 57363000001840000006c7774 × 8 = 56 Jadi, tiga bilangan 50000000 430000043c0000080c6254 4000000246c756d69000003f80000001 730000040c00000 014626b707400000204000000147258 00867669657700 d32d485020200000000000000 97269676874202863292031393938 46d6561730000040c00000024746563 52430000043c0000080c74 Bilangan-bilangan pada pembilang: – 024746563680000 selanjutnya adalah 30, 42, dan33646573 56. 595a00000218000000146758595a000 7074000001f000000014626b7 0003d400000024 0000000000000000000000000 68000004300000000c7254524300000 04300000000c725 204865776c6574742d5061636b6172 65787400000000436f7079 0022c000000146258595a0000024000 1, 2, –4, 8, –16, 32, ×(–2) ×(–2) ×(–2) ×(– 63000001 43c0000080c675452430000043c00000 6c756d69000003 0000000000000000000000000 452430000043c00 074000002040000001472585 6420436f6d70616e79000064657363 726967687420286329203 2)000014646d6e6400000254000000706 x(–2) x(–2) Pembilang pada bilangan 84000000 80c625452430000043c0000080c74657 00080c675452430 f8000000146d65 0000000000000000000000000 46d6464000002c40000008876756564 00000000000000127352474220494 95a000002180000001467585 1393938204865776c6574 6c7774707 87400000000436f70797269676874202 urutan ke-7 adalah – 000043c0000080c 0000034c0000008676696577000003d 61730000040c00 0000001163707274000001500 54336313936362d322e3100000000 863292031393938204865776c657474 742d5061636b617264204 625452430000043 3. a. 4, 95a0000022c00000014625859 , 444 ,,··· 4000001f0 4000000246c756d69000003f80000001 3 5 7 00002474656368 0000033646573630000018400 2d5061636b61726420436f6d70616e79 c0000080c746578 00000000000000127352474220494 64. Bilangan-bilangan pada 36f6d70616e79000064657 46d6561730000040c00000024746563 00000014 5a0000024000000014646d6e 0000646573630000000000000012735 00000430000000 00006c77747074000001f00000 7400000000436f7 Bilangan-bilangan pada pembilang 68000004300000000c7254524300000 54336313936362d322e3100000000 363000000000000001273 626b7074 penyebut: 247422049454336313936362d322e31 079726967687420 640000025400000070646d64 0c725452430000 0014626b70740000020400000 43c0000080c675452430000043c00000 00000000000000000000000000000 524742204945433631393 tetap yaitu 4. Bilangan-bilangan pada 0000000000000000000000127352474 286329203139393 00000204 1, 3, 6, 10, 15, 21, 28 · · · merupakan 80c625452430000043c0000080c74657 043c0000080c67 0147258595a00000218000000 64000002c4000000887675656 00000000000000000 22049454336313936362d322e310000 6362d322e310000000000 8204865776c6574 penyebut: 1, 3, 5, 7, 9,11, 87400000000436f70797269676874202 00000014 5452430000043c 146758595a0000022c00000014 pola bilangan segitiga. Penyebut pada 0000000000000000000000000000000 742d5061636b617 40000034c0000008676696577 000000000000127352474 863292031393938204865776c657474 7258595a 0000000000000000000 +2 +2 +2 +2+2 +2 0000080c625452 6258595a00000240000000146 26420436f6d7061 2d5061636b61726420436f6d70616e79 bilangan urutan ke-7 adalah 22049454336313936362d 00000218 000003d4000000246c756d690 6e7900006465736 28. 64 430000043c0000 46d6e64000002540000007064 0000646573630000000000000012735 Penyebut pada bilangan urutan ke-7 adalah 322e31000000000000000 300000000000000 00000014 00003f8000000146d65617300 247422049454336313936362d322e31 080c7465787400 6d6464000002c4000000887675 13. 000000000000000000000 6758595a Jadi, bilangan pada urutan127352474220494 ke-7 adalah 0000000000000000000000127352474 00040c0000002474656368000 000000436f7079 65640000034c00000086766965 54336313936362d Jadi, 00000000000000000064, bilangan pada urutan ke-7 – 22049454336313936362d322e310000 . 0000022c0 . 322e31000000000 72696768742028 77000003d4000000246c756d69 28 004300000000c725452430000 0000000000000000000000000000000 13 57, 50,4 x, y, z 00000146 000000000000012 adalah 63292031393938 000003f8000000146d65617300 0000000000000000000 043c0000080c6754524300000 735247422049454 258595a0 204865776c6574 00040c00000024746563680000 336313936362d32 43c0000080c62545243000004 00002400 742d5061636b61 04300000000c72545243000004 2e3100000000000 00000146 3c0000080c746578740000000 000000000000000 726420436f6d70 3c0000080c675452430000043c 46d6e640 000000000000000 0436f70797269676874202863 616e7900006465 0000080c625452430000043c00 00002540 00000000000002, 73630000000000 00080c7465787400000000436f 292031393938204865776c657 00000706 4, 8, 16, 32, 64, 00001273524742 7079726967687420286329203 4742d5061636b61726420436f ×2 ×2 ×2 ×2×2 46d64640 20494543363139 1393938204865776c6574742d5 106 Kunci Jawaban dan Pembahasan ×2 6d70616e7900006465736300 00002c400 36362d322e3100 061636b61726420436f6d70616 00008876 000000000000127352474220 00000000000000 e790000646573630000000000 75656400 49454336313936362d322e31 00000012735247 0000127352474220494543363 00034c000 42204945433631 13936362d322e310000000000 000000000000000000000012 00086766 3936362d322e31 0000000000001273524742204 735247422049454336313936 96577000 00000000000000 9454336313936362d322e3100 362d322e3100000000000000 003d4000 00000000000000 0000000000000000000000000 000246c75 000000000000000000000000 00000000000000 0000000000000000000000000 6d690000 0000000000000000 000000000000 00

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe00 ffd8ffe00010 ffd8ffe00 ffd8ffe00010 ffd8ffe00010 ffd8ffe00010 ffd8ffe0 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 0104a4649 4a464946000 0104a4649 4a464946000 4a464946000 4a464946000 00104a4 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 460001020 1020100c800c 460001020 1020100c800c 1020100c800 1020100c800c 1020100c800 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 6494600 100c800c8 80000ffe20c5 100c800c8 80000ffe20c5 c80000ffe20c 80000ffe20c5 c80000ffe20c 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 0102010 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 0000ffe20c 84943435f505 0000ffe20c 84943435f505 584943435f5 84943435f505 584943435f5 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 0c800c8 584943435 24f46494c450 584943435 24f46494c450 0524f46494c 24f46494c450 0524f46494c 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 f50524f46 0010100000c f50524f46 0010100000c 45000101000 0010100000c 4. Banyak garis pada pola 0000ffe245000101000 62d322e31000000000000000000000000000000000000000000000000000000

494c45000 494c45000 484c696e6f02 00c484c696e 484c696e6f02 00c484c696e ke-1= 1484c696e6f02 Banyak garis pada pola 0c58494 10100000c 1000006d6e7 6f021000006 6f021000006 ke-210100000c = 1000006d6e7 3 Banyak garis pada1000006d6e7 pola 3435f505 484c696e6 47252474220 d6e74725247 d6e74725247 ke-3484c696e6 = 47252474220 6 Banyak garis pada47252474220 pola 24f46494 f02100000 58595a2007ce 422058595a2 58595a2007ce 422058595a2 ke-4f02100000 = 58595a2007ce 10 Banyak garis pada c450001 6d6e74725 00020009000 00020009000 007ce000200 00020009000 007ce000200 pola ke-5 = 156d6e74725 Pola bilangan banyak garis pada setiap 247422058 60031000061 247422058 60031000061 09000600310 60031000061 09000600310 0100000 pola: 1, 595a2007c 6373704d534 595a2007c 6373704d534 00061637370 6373704d534 00061637370 c484c69 +265400000000 +3e00020009 +4 +5 +6 +7 +8 +9+10 e00020009 65400000000 4d534654000 65400000000 4d534654000 6e6f0210 000600310 49454320735 00000494543 00000494543 p= 15 + 6 000600310 =21 q=49454320735 p + 749454320735 =21 + 7 =28 r 00006d6 000616373 000616373 24742000000 20735247420 20735247420 = q 24742000000 + 8 =28 + 8 =36 s=24742000000 r + 9 =36 + 9 e747252 704d53465 00000000000 704d53465 00000000000 00000000000 00000000000 =45 t = s + 10= 45 + 10= 55 Jadi, 400000000 00000000000f 400000000 00000000000f 00000000000 00000000000f 00000000000 4742205 banyak garis pada pola ke-10 adalah 55. 494543207 6d600010000 494543207 6d600010000 00000f6d600 6d600010000 8595a2000000f6d600 352474200 0000d32d485 352474200 0000d32d485 0100000000d 0000d32d485 0100000000d 07ce000 000000000 02020000000 000000000 02020000000 32d48502020 02020000000 32d48502020 5. 000000000 Misal L000000000 = luas 2000900 00000000000 00000000000 00000000000 00000000000 segitiga hitam00000000f L1 = 000000000000 0600310 00000000f 00000000000 00000000000 00000000000 6d6000100 00000000000 6d6000100 00000000000 00000000000 00000000000 1 0006163 L2 = 4 000000d32 00000000000 000000d32 00000000000 00000000000 73704d500000000000 d48502020 00000000000 d48502020 00000000000 00000000000 00000000000 ffd8ffe000104a46494600 1 111 3 3465400 L = + 3 × × = + 000000000 00000000000 000000000 00000000000 00000000000 00000000000 3 01020100c800c80000ffe2 4444 42 0000004 000000000 00000000000 000000000 00000000000 00000000000 00000000000 0c584943435f50524f4649 1 3 111 00000000000 9454320 00000000000 00000000000 L000000000 = +000000000 2 + 3× 3 × ×00000000000 × 4c4500010100000c484c6 4 44444 000000000 11637072740 000000000 11637072740 00000011637 11637072740 00000011637 7352474 96e6f021000006d6e7472 13 32 00001500000 000000000 07274000001 00001500000 07274000001 5247422058595a2007ce0 =00001500000 + 2+ 3 000000000 2000000 44 4 000000000 000000000 00336465736 00336465736 50000000336 00336465736 50000000336 002000900060031000061 0000000 Luas segitiga hitam pada gambar ke-5: 000000000 30000018400 000000000 30000018400 46573630000 30000018400 46573630000 6373704d5346540000000 0000000 2 3000000000 000000000 00006c77747 00006c77747 01840000006 00006c77747 01840000006 13 3 3 049454320735247420000 L000000000 = 000000000 + 2 0000000 + 3074000001f00 + c7774707400 074000001f00 c7774707400 5 074000001f00 444 4 000000000000000000000 000000000 0000014626b 000000000 0000014626b 0001f000000 0000014626b 0001f000000 0f6d6000 000f6d6000100000000d3 4 +×70740000020 + ×+175 43 22 3 000011637 34 343 000011637 70740000020 014626b7074 70740000020 = = 44 1000000014626b7074 2d4850202000000000000 25 072740000 40000001472 072740000 40000001472 00000204000 40000001472 00000204000 000000000000000000000 6 00d32d4 015000000 58595a00000 015000000 58595a00000 00014725859 58595a00000 00014725859 000000000000000000000 8502020 033646573 21800000014 033646573 21800000014 5a000002180 21800000014 5a000002180 000000000000000000000 630000018 6758595a000 630000018 6758595a000 00000146758 6758595a000 00000146758 0000000 000000000000000000001 40000006c 0022c000000 40000006c 0022c000000 595a0000022 0022c000000 0000000595a0000022 163707274000001500000 777470740 146258595a0 777470740 146258595a0 c0000001462 146258595a0 c0000001462 003364657363000001840 0000000 A. 000006c77747074000001 Pilihan 00001f000 00002400000 00001f000 00002400000 58595a00000 00002400000 58595a00000 0000000 Ganda 000014626 0014646d6e6 000014626 0014646d6e6 24000000014 0014646d6e6 24000000014 1. f000000014626b7074000 Jawaban:0000000 d Barisan bilangan tersebut b70740000 40000025400 b70740000 40000025400 646d6e64000 40000025400 646d6e64000 00204000000147258595a merupakan barisan Fibonacci. Aturan barisan 020400000 000070646d6 020400000 000070646d6 00254000000 000070646d6 00254000000 0000000 000002180000001467585 Fibonacci adalah mulai suku ketiga, diperoleh 014725859 464000002c4 014725859 464000002c4 70646d64640 464000002c4 70646d64640 0000000 95a0000022c0000001462 dengan cara menjumlahkan dua suku 5a0000021 00000088767 5a0000021 00000088767 00002c40000 00000088767 58595a000002400000001 000000000002c40000 sebelumnya. 800000014 56564000003 800000014 56564000003 00887675656 56564000003 00887675656 4646d6e64000002540000 0000000 6758595a0 4c000000867 6758595a0 4c000000867 40000034c00 4c000000867 1, 1, 2, 3, 5, 8, 13, x, y, z40000034c00 x = 8 + 13 0070646d6464000002c40 000022c00 66965770000 00008676696 00008676696 = 2166965770000 y= 13000022c00 + 0000000 x=13 + 2166965770000 = 34 z= x 000008876756564000003 000014625 03d40000002 000014625 03d40000002 577000003d4 03d40000002 577000003d4 0000000 + y= 21 + 34= 55 Jadi, tiga 4c0000008676696577000 suku 8595a0000 46c756d6900 8595a0000 000000246c7 46c756d6900 000000246c7 berikutnya 21,0000000 34,46c756d6900 dan 55. 003d4000000246c756d69 024000000 0003f8000000 024000000 0003f8000000 56d69000003 0003f8000000 000003f8000000146d656 000000056d69000003 014646d6e 146d6561730 014646d6e 146d6561730 f8000000146 146d6561730 f8000000146 1730000040c0000002474 0001163 640000025 000040c0000 640000025 000040c0000 d6561730000 000040c0000 d6561730000 2. 656368000004300000000 Jawaban:7072740 c 400000070 00247465636 400000070 00247465636 040c0000002 00247465636 040c0000002 c725452430000043c0000 646d64640 80000043000 646d64640 80000043000 47465636800 80000043000 47465636800 0000150 080c675452430000043c0 00002c400 00000c72545 00002c400 00000c72545 00043000000 00000c72545 00043000000 0000003 000080c62545243000004 000088767 2430000043c 000088767 2430000043c 00c72545243 2430000043c 3c0000080c74657874000 364657300c72545243 565640000 0000080c675 565640000 0000080c675 0000043c000 0000080c675 0000043c000 00000436f707972696768 6300000 034c00000 45243000004 034c00000 45243000004 0080c675452 45243000004 0080c675452 742028632920313939382 1840000 086766965 3c0000080c62 086766965 3c0000080c62 430000043c0 3c0000080c62 430000043c0 04865776c6574742d5061 77000003d 54524300000 77000003d 54524300000 000080c6254 54524300000 006c777 636b61726420436f6d706000080c6254 400000024 43c0000080c7 400000024 43c0000080c7 52430000043 43c0000080c7 52430000043 4707400 16e790000646573630000 6c756d690 46578740000 6c756d690 46578740000 c0000080c74 46578740000 000000000012735247422 0001f000c0000080c74 00003f800 0000436f7079 00003f800 0000436f7079 65787400000 0000436f7079 65787400000 049454336313936362d32 0000146 0000146d6 72696768742 0000146d6 72696768742 000436f7079 72696768742 000436f7079 2e3100000000000000000 26b7074 561730000 02863292031 561730000 02863292031 72696768742 02863292031 72696768742 000001273524742204945 040c00000 39393820486 040c00000 39393820486 02863292031 39393820486 02863292031 0000020 4336313936362d322e310 024746563 5776c657474 024746563 5776c657474 39393820486 5776c657474 39393820486 4000000 000000000000000000000 680000043 2d5061636b6 680000043 2d5061636b6 5776c657474 2d5061636b6 000000000000000000000 14725855776c657474 00000000c 1726420436f6 00000000c 2d5061636b6 1726420436f6 2d5061636b6 000000000002, 5,1726420436f6 10, 17,

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 104a464946 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 0001020100 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c c800c80000 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 ffe20c58494 0000000000000000000000000000000000000000000

3435f50524f 46494c4500 9. Jawaban: d Rumus 2 010100000c suku ke-n: U = n (n + 1)n 2 U1 = 1 (1 + 1) = 1484c696e6f × 2 =2 U2 = 22(2 + 1) 021000006d = 4 × 3 =12 U3 =6e74725247 32(3 + 1) = 9 × 4 =36 U4 = 42(4 + 1) = 16 422058595a × 5=80 U5 = 52(5 + 1) = 25 × 6=150 2007ce0002 Jadi, barisan bilangan 0009000600 dengan rumus U = n2(n + n 3100006163 1) adalah 2, 12, 36, 80, 150, · · · . 73704d5346 10. Jawaban: b 5400000000 4945432073 Untuk U = nn2 + 2n – 4: 2 U1 = 1 + 2 · 1 – 45247420000 = –1, U2 = 4, U3 = 11, 0000000000 U4 = 20 diperoleh0000000000 barisan bilangan –1, 4,

11, 20, n· · · Untuk 0000f6d600 U = n2 – 2n + 4: 2 U1 = 1 – 2 · 1 + 40100000000 = 3, U2 = 4, U3 = 7, U4 d32d485020 = 12 diperoleh barisan bilangan 3, 4, 7, 2000000000 12, · · · nUntuk U =0000000000 2n + 1: U1 = 2 · 1 + 1 = 3, U2 = 5, U3 = 7, 0000000000 U4 = 9 diperoleh barisan bilangan 3, 0000000000 2 0000000000 5, 7, 9, n· · · Untuk U = 2n + n: U1 = 2 · 12 + 1 = 0000000000 3, U2 = 10, U3 = 21, U4 = 0000000000 36 diperoleh barisan bilangan 3, 10, 21, 0000000000 36, · · · Jadi, rumus suku ke-n barisan 0000000000 12, · · · adalah nU = n2 – 2n + 4. bilangan 3, 4, 7, 0000001163 11. Jawaban: c U17072740000 =1 1 = 2 – 1 =2 – 1 U2 = 30150000000 =4 3364657363 2 – 1 =2 – 1 U3 = 7 = 80000018400 –1 =23 – 1 U4 = 15= 16 –00006c7774 1= 7074000001 4 2 – 1 U5 = 31= 32 – 1 = f000000014 25 – 1 # U = 2n – 1 626b707400 n 0002040000 Suku ke-10 barisan bilangan 10 0014725859 U tersebut: = 2 – 1 = 1.023 bilangan 10Jadi, suku ke-10 barisan 5a00000218 tersebut 1.023. 0000001467 12. Jawaban: a Barisan 58595a0000 bilangan 2, 8, 18, 32, 022c000000 · · · . U1 = 2 = 2 × 1 = 2 × 12 U2 =146258595a 8=2×4= 0000024000 2 2 × 2 Uffd8ffe000104a4649460001020100c80 = 18= 2 × 9 = 2 × 32 U4 3 000014646d 0c80000ffe20c584943435f50524f46494 2 = 32= 2 × 16= 2 × 4 # U = 2n2 6e64000002 c4500010100000c484c696e6f02100000 5400000070 6d6e74725247422058595a2007ce0002 0009000600310000616373704d534654 646d646400 n 00000000494543207352474200000000 0002c40000 Un 00000000000000000000f6d600010000 = 2n2 ⇒ 578 = 2n2 0088767565 0000d32d485020200000000000000000 640000034c 5 578 00000000000000000000000000000000 n2 = = 00000000000000000000000000000000 0000008676 2 00000000000000116370727400000150 n2 = 289 6965770000 0000003364657363000001840000006c 03d4000000 77747074000001f000000014626b7074 ⇔ 246c756d69 n= 00000204000000147258595a00000218 000000146758595a0000022c00000014 000003f800 6258595a0000024000000014646d6e64 0000146d65 0000025400000070646d6464000002c4 6173000004 00000088767565640000034c00000086 0c00000024 76696577000003d4000000246c756d69 000003f8000000146d6561730000040c 7465636800 0000002474656368000004300000000c 0004300000 725452430000043c0000080c67545243 000c725452 0000043c0000080c625452430000043c 430000043c 0000080c7465787400000000436f7079 72696768742028632920313939382048 0000080c67 65776c6574742d5061636b6172642043 5452430000 6f6d70616e7900006465736300000000 043c000008 00000012735247422049454336313936 0c62545243 362d322e310000000000000000000000 12735247422049454336313936362d32 0000043c00 2e310000000000000000000000000000 00080c7465 00000000000000000000000000

13. U5 = Jawaban: 17 + 9 =b Barisan bilangan 3, 7, 11, 15, · · · , 79, 83 U1= 3 =4 – 1 = 4 U = U5 + 11 = 26 + 11 = 37 26 × 1 – 61 U2= 7 =8 – 1 = 4 × 2 – 1 U3 = Jadi, dua 26 16 – 1 11= 12 – 1 = suku 4 × 3berikutnya – 1 U4 = 15= dan 37. =3.4 × 4 – 1 ffd8ffe000104a4649460001020100 # c800c80000ffe20c584943435f50524f Jawaban: c – 1 = 4n – 1 7777 7777 U n= 4 × n46494c4500010100000c484c696e6f U 1 U 2 U 3 U U⇒ 5U6 U7U 83 =84n – 1 5, 14, n= 4n –41021000006d6e74725247422058595 n 4n = 83 +a2007ce00020009000600310000616 8= 27 dan U7 = Diperoleh U13 84 n n= 373704d5346540000000049454320 119. Nilai U3 + U7 = 27 + 119 = n= 21 Jadi, banyak suku pada barisan bilangan 73524742000000000000000000000 146. tersebut adalah 21. 0000000f6d6000100000000d32d485 4. Jawaban: 02020000000000000000000000000 n c U= n 00000000000000000000000000000 + n2 11 00000000000000000000000000000 U = =+ 00000000000011637072740000015 1123 2 2 00000003364657363000001840000 U2= 224 = + 006c77747074000001f00000001462 3 3 6b707400000204000000147258595 U3= 325 = + a00000218000000146758595a0000 12 3 022c000000146258595a0000024000 Jadi, barisan bilangannya , , , · · · . 3 45 000014646d6e64000002540000007 0646d6464000002c40000008876756 5. 5640000034c0000008676696577000 n 003d4000000246c756d69000003f80 Jawaban: c U U4 = 2(4)2 – 5 = 2 × 16 – 5 Diperoleh: 00000146d6561730000040c0000002 2 2 = 2n – 5 U5 = 2(5) = 27– 5 = 2 × 25 – 5 474656368000004300000000c72545 Nilai U4 + 2430000043c0000080c67545243000 U5 = 27 + 45 = 45 Jadi, U4 + U5 adalah 0043c0000080c625452430000043c0 = 72nilai dari 000080c7465787400000000436f707 6. 72. 97269676874202863292031393938 n 2 204865776c6574742d5061636b6172 U10= Jawaban: U = 10 11a2 –6420436f6d70616e79000064657363 – 22 × × 10 11 = = 100 121 –– 20 22 = 80 2 U = n=1–99 2n 00000000000000127352474220494 U10Jadi, + U11jumlah = 80 + 99 = 1 suku ke-10 dan ke-11 54336313936362d322e3100000000 179179. adalah 00000000000000127352474220494 7. Jawaban: a 54336313936362d322e3100000000 U1 = 2 = 3 ×00000000000000000000000000000 1–1 U2 = 5 = 3 ×00000000000000000 2–1 U3 = 8 = 3 × 3 – 1 U4 = # 11= 3 × 4 – 1 Un =3×n–1 = 3n – 1 Jadi, rumus suku ke-n barisan bilangan 2, 5, 8, 11, · · · adalah U = 3n – 1. n

8. Jawaban: d U1 = 2 × 3=(1 + 1) × (1 + 2) U2 = 3 × 4=(2 + 1) × (2 + 2) U3 = 4 × 5=(3 + 1) × (3 + 2) U4 = 5 × 6=(4 # + 1) × (4 + 2) Un= (n + 1) × (n + 2) = (n + 1)(n + 2) Jadi, rumus suku ke-n adalah U = (n + 1)(n + 2). n

b. Barisan bilangan 7, 3, –1, –5, –9, · · · U1= 7 = –4 × 1 + 11 U2= 3 = –4 × 2 + 11 U3 = –1= –4 × 3 + 11 U4

c. Barisan 1 × 3, 4 × 4, 9 × 5, 16 × 6, · · · U1= 1 × 3 =12 × (1 + 2) U2=

= –5= –4 × 4 + 11

(3 + 2) U4 = 16 × 6= 42 × (4 + 2) # Un= n2 × (n + 2) 2 U12Jadi, = 12nilai × (12 2) = 144 × 14 U +adalah

# Un= –4n + 11 Jadi, rumus suku ke-n adalah U = –4n + 11. n2345

c.

4 × 4 =22 × (2 + 2) U3= 9 × 5 =32 ×

12

1, , , , , · · · 93115 7

U1=

d.

1

=2 =2 U =2322×−1 U3= 21×−1

144 × 14. 357 9 , , ,,··· 5 9 13 bilangan dari 17 Barisan 4 pembilang: 3, 5, 7, 9,. . .

3

ffd8ffe000104a46494600 U1= 3 01020100c800c80000ffe2 3 14. Jawaban: = 2 × 1 + 1 U2= = 0c584943435f50524f4649 5 e (1) 54c4500010100000c484c6 1,2 2, 8,116, 32, 64, 128, dan 256 = × 4, 2+ 4 4 merupakan n – dengan 1 barisan U4724×−1 96e6f021000006d6e7472 U nrumus U. = 23×−1= U34= =7 9= ==222× × 34++ # 5247422058595a2007ce0 1 1002000900060031000061 U = n2n + 1 (2) 1, nn ffd8ffe000104a46494600 ffd8ffe000104a4649460001020100c800 Diperoleh pada U= = n +1 + 2pembilang +3+4+5+ 6 + 7suku + 8 ke-n adalah 2n1 2n −1 6373704d5346540000000 ×− 01020100c800c80000ffe2 c80000ffe20c584943435f50524f46494c4 2n + 1. Barisan bilangan 049454320735247420000 ffd8ffe000104a4649460001020100c800c800 Jadi, rumus suku ke-n barisan bilangan +1+1+1+1+1+1+1 0c584943435f50524f4649 500010100000c484c696e6f021000006d6 dari penyebut 00ffe20c584943435f50524f46494c450001010 000000000000000000000 n 4c4500010100000c484c6 e74725247422058595a2007ce00020009 1, 2, 11,. 16, 0000c484c696e6f021000006d6e74725247422 tersebut adalahnU = 5, 9, 4, 13,7,17, . . 22, 29, dan 37 000f6d6000100000000d3 96e6f021000006d6e7472 000600310000616373704d53465400000 merupakan 058595a2007ce000200090006003100006163 barisan bertingkat dua. 2d4850202000000000000 5247422058595a2007ce0 000494543207352474200000000000000 d. Barisan bilangan 64,125, 216, 343, 73704d5346540000000049454320735247420 +4 +4 +4 U = 5 = . 000000000000000000000 (3) 1,2, 4, 2, 9,1 2, 16, 2, 25 3 −1 000000000000000000000000000f6d60001000 00000000000000f6d6000100000000d32 512, · · · U1= 64 = 42n = (1 + 3)3 U2= 125= 4002000900060031000061 × 1 + 1 U2=77 9777 00000d32d4850202000000000000000000000 000000000000000000000 6373704d5346540000000 d48502020000000000000000000000000 3 3 3 3 0000000000000000000000000000000000000 5 = (2 + 3) U3= 216= 6 = (3 + 3) U4= 000000000000000000000 1 Ubilangan = 1222= 324 4× 2 522 + Pola pada suku 049454320735247420000 000000000000000000000000000000000 3 0000000000000000000000000000000000000 3 3 000000000000000000001 343= 7 = (4 + 3) ganjil adalah 13= 4 × 3bilangan + 1 U4 persegi dan suku 000000000000000000000 000000000000000000000000000000000 1163707274000001500000003364657363000 163707274000001500000 001840000006c77747074000001f0000000146 000f6d6000100000000d3 000011637072740000015000000033646 genap selalu bernilai U = 4n = 17= n 4 × 4++11 2. Jadi, ketiga # 26b707400000204000000147258595a000002 003364657363000001840 2d4850202000000000000 57363000001840000006c777470740000 Diperoleh penyebut pada suku ke-n adalah barisan tersebut dapat menjadi 6 suku 18000000146758595a0000022c00000014625 000006c77747074000001 Un= (n + 3)3 000000000000000000000 01f000000014626b70740000020400000 4n + 8595a0000024000000014646d6e6400000254 berikutnya. f000000014626b7074000 Jadi, rumus suku ke-n barisan bilangan 000000000000000000000 00000070646d6464000002c40000008876756 0147258595a0000021800000014675859 1. 2n +1 3 15. Jawaban: 00204000000147258595a . Jadi, rumus suku ke-n adalahn U = tersebut adalahnU = (n + 3) . 5640000034c0000008676696577000003d400 000000000000000000000 5a0000022c000000146258595a0000024 4n 1 + 0000246c756d69000003f8000000146d656173 c Untuk 000002180000001467585 n =2n 30, 31, dan 000000000000000000001 +1 000000014646d6e640000025400000070 2. a. Barisan 1, 8, 27, 0000040c00000024746563680000043000000 U 95a0000022c0000001462 32: +1)=4n +1 2(12) +1 24 163707274000001500000 646d6464000002c400000088767565640 n 00c725452430000043c0000080c67545243000 U30 30(30 = = 465 64, · · · U1= 1 =13 U2= 8 =23 58595a000002400000001 2 003364657363000001840 0043c0000080c625452430000043c0000080c7 +1 25 000034c0000008676696577000003d400 U === 31(31 4646d6e64000002540000 U3 = 27= 33 U4 = 64= 43 465787400000000436f7079726967687420286 12 000006c77747074000001 0000246c756d69000003f8000000146d65 U31 = = 496 +1) 4(12) 3292031393938204865776c6574742d506163 0070646d6464000002c40 2 +1 48 +1 49 Jadi, f000000014626b7074000 61730000040c000000247465636800000 25 6b61726420436f6d70616e7900006465736300 32(32 . 000008876756564000003 49 00204000000147258595a 4300000000c725452430000043c000008 nilai U adalah U32 = =12528 0000000000001273524742204945433631393 +1) # 2 4c0000008676696577000 6362d322e3100000000000000000000001273 0c675452430000043c0000080c6254524 Jadi, 000002180000001467585 bilangan segitiga terbesar yang Un= n3 003d4000000246c756d69 5247422049454336313936362d322e3100000 95a0000022c0000001462 30000043c0000080c7465787400000000 3 kurang dari 500 adalah 496. Rumus suku ke-n: U = n 0000000000000000000000000000000000000 000003f8000000146d656 n 58595a000002400000001 436f70797269676874202863292031393 3. 000000000000 U12Jadi, = nilai 123 = 1.728 1730000040c0000002474 U12 adalah 4646d6e64000002540000 938204865776c6574742d5061636b6172 a. B. 656368000004300000000 0070646d6464000002c40 6420436f6d70616e79000064657363000 1.728. U n = 2( n + c725452430000043c0000 1) 3 U10 = 2(10 + 1)3 = 2 * 113 = 2 * 1.331 = 2.662. b. Barisan 2, 7, 12, 000008876756564000003 000000000001273524742204945433631 1. a. Barisan bilangan 4, 8, 16, 32, Uraian 080c675452430000043c0 17, · · · U1= 2 = 1 × 5 – 3 1 3936362d322e310000000000000000000 64, · · · 4c0000008676696577000 U 1= 4 = 2 × 2 = 2 × 2 U2= 8 = 000080c62545243000004 003d4000000246c756d69 00012735247422049454336313936362d U2= 7 = 2 × 5 – 3 U3 = 3c0000080c74657874000 2 × 4U= 2 22 U3= 16= 2 × 8 = 2 × 23 Jadi, nilai 10000003f8000000146d656 =× 2.662. 322e31000000000000000000000000000 12= 3 × 5 – 3 U4 = 17= 4 00000436f707972696768 1730000040c0000002474 U4 = 32= 0000000000000000000000000002, 2 × 16= 2 × 24 4, 7, 742028632920313939382 × 5# –3 b. 656368000004300000000 11, 16, 22, 29, 37 # 3 04865776c6574742d5061 = 343 U n = 2(n + (k 1) 3+U1) k == 686 ?2(k + 1)3 = 686 ? (k + 1)3 = 686 2 Un= n × 5 – 3 = 5n – 3 c725452430000043c0000 U= 2 × 2n =3321 × 2n 636b61726420436f6d706 n 3 080c675452430000043c0 Rumus suku ke-n: nU = 5n – 3 (k + 1) = 7 1 + n 16e790000646573630000 2 = 2n + 1 = 2n + 1 000080c62545243000004 U1 = 5n – 3 k + 1rumus =7 =suku Jadi, ke-n adalahn U. 000000000012735247422 k3c0000080c74657874000 =7–1 = 5(12) – 3 = 57 Jadi, 2 049454336313936362d32 00000436f707972696768 k= 6 Jadi, nilai k adalah nilai U12 adalah 57. 2e3100000000000000000 742028632920313939382 6. 000001273524742204945 04865776c6574742d5061 4336313936362d322e310 636b61726420436f6d706 108 Kunci Jawaban dan 000000000000000000000 16e790000646573630000 Pembahasan 000000000000000000000 000000000012735247422 00000000000 049454336313936362d32 2e3100000000000000000 000001273524742204945 4336313936362d322e310 000000000000000000000 000000000000000000000 00000000000 +2 +2 +2

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 4. a. Jika pola di atas dilanjutkan akan diperoleh 62d322e31000000000000000000000000000000000000000000000000000000

pola ke-4 dan ke-5 sebagai berikut.

A. Pilihan Ganda 1. Jawaban: d 1) 1, 2, 4, 7, 11, 16, · · · Selisih dua suku yang berurutan nilainya tidak tetap sehingga bukan merupakan barisan aritmetika. 2) 1, 3, 9, 27, 81, 243, · · · + 1 + 2+ 3 + 4+ 5

Pola ke-4

Pola ke-5

b. Banyak titik yang terlihat pada setiap pola sebagai berikut. Pola ke-1 = 5 × 3 + 22 = (2 ×1 + 3)(1 + 2) + (1 + 1)2 Pola ke-2 = 7 ×4 +32 = (2 × 2+ 3)(2+ 2)+ (2+ 1)2 Pola ke-3 = 9 ×5 +42 = (2×3 + 3) (3 + 2) + (3 + 1)2 Pola ke-4 = 11 ×6 +52= (2×4 + 3)(4 + 2) + (4 + 1)2 Pola ke-5 = 13 ×7 +62= (2×5 + 3)(5 + 2) + (5 + 1)2 Pola 2 ke-n= (2n + = 3)(n 2)7n + (n + 1) (2n+ + + 6) + 2(n2 + 2n + 1) = 3n2 + 9n + 7 Pola ke-8= 3(8)2 + 9(8) + 7 = 3(64) + 9(8) + 7 = 271 Jadi, banyak titik yang terlihat pada pola ke8 adalah 271.

Selisih dua suku yang berurutan nilainya tidak tetap sehingga bukan merupakan barisan aritmetika. 3) 1, 4, 5, 8, 9, 12, · · · + 2 + 6 +18 + 54 + 128

Selisih dua suku yang berurutan nilainya tidak tetap sehingga bukan merupakan barisan aritmetika. + 3 + 1+ 3 + 1+ 3

4)

Selisih dua suku yang berurutan nilainya tetap sehingga merupakan barisan aritmetika. Jadi, barisan bilangan pada pilihan d barisan aritmetika. merupakan +6+6+6+6

5. −21

t −1 +1t1

1 t1 =t22=

=+= 21

1 3

t −1 +1t2

t3= 2 =

1

−1

1 3

+1

3

t4= =

1 −+1

I. 17, 19, 21, 23, 25,· · · Jawaban: c + 2 + 2 + 2 + 2 Barisan di atas mempunyai beda = 2 (positif). 21, 17, 13, 9, 5,· · · + (–4) + (–4) + (–4) + (–4)

3t −1 +1t3 1 −−1

2.

II

1

= –2

1, 7, 13,19, 25, · · ·

Barisan di atas mempunyai beda = –4 (negatif). III. 32, 27, 22, 17, 12, · · · + (–5) + (–5) + (–5) + (–5)

2

Barisan di atas mempunyai beda = –5 IV. (negatif). 37, 44, 51, 58, 65,· · ·

2

= –3 t −1 4 +1

+7

t5= 4t

+7+7+7

Barisan di atas mempunyai beda = 7 (positif).

−− = =31 2 −+ 31

Barisan aritmetika turun adalah barisan aritmetika

yang mempunyai beda negatif. Perhatikan bahwa t5 = t1. Berarti, rumus di atas Barisan II dan III mempunyai beda negatif. berulang Jadi, barisan aritmetika turun adalah II dan III. tersebut dengan periode 4. Dengan kata lain, rumus = t ditulis: t , b = 1, 2, 3, 4, n (4 × a + b) dan a = 1, 2, 3,=4,t 5, · · · . b 1 t1999 = t(4 × 499 + 3)

= t3 = –

2

110 Kunci Jawaban dan Pembahasan

Matematika Kelas IX

3. Jawaban: d –5, 4, 13, 22, 31, · · · Dari barisan tersebut diketahui: Suku pertama = a = –5 dan beda b =4 – (–5) = 4 + 5 = 9 U = a + (n – 1)b

S20 =

20

= (2(27) + 2 19(–3)) 20 = (– 2= 3) 10(–3) = –30 Jadi, jumlah 20 suku pertama deret tersebut adalah –30.

n

= –5 + (n – 1)9 = –5 + 9n – 9 = 9n Jadi, U= 9n – 14. – 14rumus suku ke-n adalah n

4. Jawaban: b Rumus suku ke-n: U = 6n + 8 n U1 = 6(1) + 8 = 14 U2 = 6(2) + 8 = 20 U3 = 6(3) + 8 = 26 U4 = 6(4) + 8 = 32 U5 = 6(5) + 8 = 38 Jadi, barisan aritmetika yang mempunyai rumus suku ke-n = U = 6n + 8 adalah 14, 20, 26, 32, 38, · · · .

n

5. Jawaban: c 2, 5, 8, 11, 14, · · · ffd8ffe000104a464946000102 + 3 + 3 + 3 + 3 Suku 0100c800c80000ffe20c5849434 35f50524f46494c450001010000 pertama = a = 2 Beda 0c484c696e6f021000006d6e74 = b=3 Rumus umum suku ke-n barisan 725247422058595a2007ce0002 aritmetika: U = a + (n – 1)b 0009000600310000616373704d n U = a + 49b 53465400000000494543207352 5 47420000000000000000000000 =2+ 0 000000f6d6000100000000d32d 49(3) = 2 48502020000000000000000000 + 147 = 00000000000000000000000000 Jadi, suku ke-50 adalah 149. 149 00000000000000000000000000 6. Jawaban: c Deret aritmetika 2 + 6 + 00000000000000000000000011 10 + 14 + · · · . Diketahui a = 2 dan b = 6 – 2 63707274000001500000003364 = 4.657363000001840000006c7774 Rumus jumlah n suku pertama barisan aritmetika: 7074000001f000000014626b70 n 7400000204000000147258595a S n= (2a + (n – 2 00000218000000146758595a00 1)b) 00022c000000146258595a0000 n = (2(2) + (n – 024000000014646d6e64000002 2 5400000070646d6464000002c4 1)4) n = (4 + 4n – 00000088767565640000034c00 2 00008676696577000003d40000 4) n = (4n) = 00246c756d69000003f8000000 2 146d6561730000040c00000024 2 Jadi, jumlah n suku pertama deret 2nrumus 74656368000004300000000c72 tersebut S = 2n2. n 5452430000043c0000080c6754 7. 52430000043c0000080c625452 Jawaban: b 27 + 24 + 21 + · · · merupakan deret aritmetika dengan a = 27 430000043c0000080c74657874 dan00000000436f70797269676874 b = 24 – 27 = –3. Rumus jumlah n suku pertama: 20286329203139393820486577 6c6574742d5061636b61726420 n S = (2a + (n – 1)b) n 2 436f6d70616e79000064657363 00000000000000127352474220 49454336313936362d322e3100 00000000000000000000127352 47422049454336313936362d32 2e310000000000000000000000 00000000000000000000000000 000000

20 2 (2a + 19b)

8. Rumus jumlah n suku pertama deret Jawaban: b aritmetika: n S = (U nn 1 2

+ U ) 15 S15 = 2 (U + 1 U1515 ) = (4 + 2 144) =1 2

= 51.110 (148) Jadi, jumlah 15 suku pertama deret tersebut adalah 1.110. 9. Jawaban: a –7, –11, –15, –19, · · · , – 67 merupakan barisan aritmetika dengan a = –7 dan b = –11 – (–7) = –11 + 7 = –4. Rumus suku ke-n: U = a + (n – 1)b n

Besar suku terakhir (suku ke-n) adalah –67. U = –67 ⇒ a + (n – 1)b = –67 n ⇔ –7 + (n – 1)(–4) = – 67 ⇔ –4n – 3 =–67 ⇔ – 4n = –64 ⇔ n=

−64 −

⇔ n= 16 Diperoleh –67 merupakan suku 4 ke-16 (U16). Jadi, banyak suku pada barisan aritmetika –7, –11, –15, –19, · · · , –67 adalah 16. 10. Jawaban: c 2, 9, 16, 23, · · · merupakan barisan aritmetika dengan a = 2 dan b = 9 – 2 = 7. Misal: 177 merupakan suku ke-n U = 177 ⇒ a + (n – 1)b= 177 n n 2 + (n – 1)7= 177 2 + 7n – 7 =177 7n – 5 =177 7 7n = 177 + 5 7 7n = 182 1 182 n n= n= 26 Jadi, bilangan 177 merupakan suku ke-26.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 3 + (n – 1)3 = 81 11. Jawaban: d Rumus suku ke62d322e31000000000000000000000000000000000000000000000000000000

n barisan aritmetika: U = a + (n – 1)b n U1 = 3 ⇒ a= 3 U8 =

24 ⇒ a + 7b = 24 2 3 + 7b = 24 7b = 21 ⇔ b= 3 Jadi, beda barisan aritmetika tersebut adalah 3.

12. n

U1 = 4(1) Jawaban: a U– 5 = = 4n–1– U 5 = 4(2) – 5 2 = 3 beda = U2 = 3 – (–1) = 4 Jadi, beda dari –barisan U1 tersebut adalah 4.

3 + 3n – 3 = 81 3n = 81 3 n= 27 Jumlah n suku pertama deret aritmetika: n S nn = (a + U) 2

Jumlah 27 suku pertama deret aritmetika: 27

S27= (3 + 2 81) =2 2

7 = 1.134 Jadi, jumlah bilangan di antara 0 dan(84) 82 yang habis dibagi 3 adalah 1.134.

16. Jawaban: c Dari deret tersebut diketahui a = 8 dan b = 17 – 8 = 9 serta S = 13. Jawaban: b Barisan tersebut 690 merupakan barisan aritmetika dengan a = 400 n S = 690 dan b = 380 – 400 = –20. U = a + (n – 1)b = nn 120 nn ⇒ (2a + (n – 1)b) = 690 2 400 + (n – 1)(–20) = 120 4 400 – 20n + 20 = 120 ⇔ –20n =120 – 20 – 400 ⇔ n (2(8) + (n – 1)9) 2 ⇔ –20n = –300 = 690 n= 15 Diperoleh n = 15 sehingga barisan n ((16 + n 9n – 9) = 690 tersebut mempunyai 15 suku. 2

n(9n + 7) = 690 × 2 n S nn

n

= (a 2

9n2 + 7n – 1.380 = 0 Mencari nilai n yang memenuhi 9n2 + 7n – 1.380 = 0

+ U) 15 S15 = 2 (400 + 120) 15 = 2 = 3.900 Jadi, jumlah barisan (520) tersebut 3.900. 14. Jawaban: d Pada deret aritmetika berlaku U = S nn – Sn – 1 ⇒ – S 24 U25 = S25 S

=n3n2 – n S25 = 3(25)2 – 25 = 1.875 – 25 = 1.850 S24 = 3(24)2 – 24 = 1.728 – 24 = 1.704 U2 = S25 – = S24 – 1.704 = 146 Jadi, 5 1.850 besar suku ke-25 adalah 146.

15. Jawaban: c Penjumlahan bilangan di antara 0 dan 82 yang habis dibagi 3 adalah 3 + 6 + 9 + 12 + · · · + 81. 3 + 6 + 9 + 12 + · · · + 81 membentuk deret aritmetika dengan a = 3, b = 3, dan U = 81. n Unn= 81 ⇒ a + (n – 1)b = 81

7

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 18. Jawaban: c Tinggi tumpukan kursi 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 membentuk barisan aritmetika berikut 90, 96, 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 102, · · · . Diperoleh a = 90 dan b = 6. Tinggi 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 tumpukan 10 kursi: 0000000000000000000000000000000000000000000000 U1 = a + 9b = 90 + 9 × 6 = 144 Jadi, tinggi 0 tumpukan 10 kursi adalah 144 cm. 19. Jawaban: b Bulan Januari = U1 = a = 7.500 Bulan Februari = U2 = 10.000 Bulan Maret = U3 = 12.500 # Bulan Desember = U12 Besar uang yang ditabung membentuk barisan aritmetika dengan a = 7.500 dan beda = b = 2.500. Jumlah uang sampai bulan Desember = S12: n Sn= (2a + (n – 2

1)b) 12 S12 2=12 (2a + 11b) = (2(7.500) + 2

11(2.500)) = 6(15.000 + 27.500) = 6(42.500) = Jadi,255.000 tabungan Angga sampai bulan Desember pada tahun yang sama berjumlah Rp255.000,00. 20. Jawaban: a Pada waktu menyebutkan bilangan urutan ke-k, bilangan yang diucapkan Fachmy sama dengan bilangan yang diucapkan Zeldy. Misal bilangan tersebut Uk. Bilangan urutan ke-k yang diucapkan Fachmy: Uk = a + (k – 1)b ⇒ Uk = 1.008 + (k – 1)8 ⇔ Uk = 1.008 + ⇔ Uk = 8k + 1.000 Bilangan urutan ke-k 8k – 8 yang diucapkan Zeldy: Uk = a + (k – 1)b ⇒ Uk = 2.004 + (k – 1)(–4) ⇔ Uk = 2.004 – 4k + 4 4 Uk = –4k + 2.008 Bilangan urutan ke-k yang diucapkan Fachmy dan Zeldy sama. Uk = 8k + 1.000 = –4k + 2.008 1 12k = 1.008 1 k= 84 Uk = 8(84) + 1.000 = 1.672 Uk = – 4(84) + 2.008 = 1.672 Jadi, bilangan yang diucapkan Zeldy dan Fachmy adalah 1.672.

Untuk n = 10: 9(102) + 7(10) – 1.380 = –410 ≠ 0 Untuk n = 11: 9(112) + 7(11) – 1.380 = –214 ≠ 0 Untuk n = 12: 9(122) + 7(12) – 1.380 = 0 Untuk n = 13: 9(132) + 7(13) – 1.380 = 232 ≠ 0 Nilai n yang memenuhi adalah 12. Jadi, banyak bilangan dari deret tersebut adalah 12. 17. Jawaban: b Banyak koin yang dikumpulkan Desta setiap bulan membentuk barisan aritmetika 5, 11, 17, . . . dengan a = 5 dan b = 6. Banyak koin yang dikumpulkan Desta pada bulan ke-15:

U1 = a + 14b = 5 + 14 × 6 5 = 89 Jadi, banyak koin yang dikumpulkan Desta pada bulan ke-15 adalah 89 koin.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 0 0000000000000000000000000000000000000000000

a + (n – 1)b = 415 7 + (n – 1)8 = 415 7 + 8n – 8 =415 8n = 415 – 7 + 8 8 8n = 416 8 n= 52 Deret tersebut mempunyai 52 suku. Jumlah 52 suku deret aritmetika tersebut: n

Sn= (2a + (n – 2

1)b) 52 S52 2=52 (2a + 51b) = (2(7) + 2 52 51(8)) = (14 + 408) =

B.

Deret yang baru mempunyai suku pertama: a = 2 dan beda:n b = 2 dan U 1. a. 42. Deret 5 +baru: 7 + 92 + · mempunyai Deret yang + ·4· + 6 + 8 + 10 = Uraian nilai a = 5 dan b = 2. U = a + (n + 12 + 14 + 16 + 18 + · ·–· 1)b + 42 b. ⇒nnn U = = 42a + 29b U30 = 5 + 29(2) a + (n – 1)b = 42 5+ 2 + (n – = 1)2 = 58 42 = 63 2 + 2n –n2 = 42 2n = 42 S = (2a + (n – 2 n 2 aritmetika yang baru n= 21 Jadi, deret 30 1)b) mempunyai m 21 suku. ⇒ S30 = 2 30 (2a + Jumlah deret yang baru: n 29b) = (2(5) + 2

= 29(2)) (10 + 30

2 30 58)

2

10.972 Jadi, jumlah deret tersebut 10.972. b. 222 + 214 + 206 + 198 + · · · + 30 Diperoleh a = 222 dan b = 214 – 222 = –8 dan U = 30 n Un n= 30 a + (n – 1)b = 30 222 + (n – 1)(–8) = 30 2 222 – 8n + 8 =30 ⇔ –8n = 30 – 8 – 222 ⇔ –8n = –200 −200

⇔ n= = 25 Deret tersebut mempunyai 25 suku. Jumlah 25 suku deret aritmetika tersebut: n −8

Sn= (2a + (n – 2

1)b) 25 S25 = (2a + 24b) 25 = (2(222) + 2 2 24(–8)) 2= (444 – 192) = 5 2

Jadi, jumlah deret tersebut 3.150. 3.150 3. a. Deret aritmetika: 2 + 10 + 18 + 26 + 34 + 42 Di antara dua suku yang berurutan disisipi 3 bilangan sehingga membentuk deret aritmetika baru. Misalkan barisan tersebut digambarkan sebagai berikut. 2 + + + 10 U = a + + (n – 1)b n U1 = 2 ⇔ a= 2 U5 = 10 ⇔ a + 4b = 10 2 + 4b = 10 4b = 8 4 b= 2 Diperoleh a = 2 dan b = 2.

= × 68 1.020 Jadi, besar U30 = 63 dan S30 = 1.020. 2=

b. Deret 100 + 75 + 50 + · · · mempunyai nilai a = 100 dan b = –25. U = a + (nn – 1)b ⇒ U30 ==100 a ++29b 29(– 25) = 100 + (–725) = – 625 n Sn= (2a + (n – 2

30 1)b) ⇒ S30 = 2 (2a + 29b) 30 = (2(100) + 29(– 2 25)) 30 = (– 2

= –7.875 Jadi, besar U30 = –625 dan 525) S30 = –7.875. c. Deret (–24) + (–21) + (–18) + · · · mempunyai nilai a = –24 dan b = 3. U = a + (n –n 1)b U30 = = –24 a ++ 29b 29(3) = –24 + 87 = 63 n

Sn= (2a + (n – 2

1)b) 30 S30 2=30 (2a + 29b) = (2(–24) + 2

30 29(3)) = × (39) = 2Jadi, besar U = 63 dan S = 30 30 585 585. 2. a. 7 + 15 + 23 + 31 + · · · + 415 Diperoleh a = 7 dan b = 15 – 7 = 8 dan U = 415. n Un= 415

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 n Rumus umum: S = (2a + (n – 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 n8 2 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 S8 = 37.600 ⇒1)b) (2(4.000) + 7b) = 2 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 4(8.000 4 + 7b) = 37.600 3 37.600 0000000000000000000000000000000000000000000000 8.000 + 7b = 9.400 8 7b = 1.400 7 b = 200 Jadi, penambahan produksi setiap tahun 200 kompor.

Sn = (2a + (n – 2

1)b) 21 S21= (2a + 20b) =

2 21

(2(2) +

2 21 20(2))

= (4 + 40) = Jadi, jumlah deret yang 462 baru 462. 2

4. a.

Sn= 5n2 + 2n 2 2 Sn – 1 = 5(n = 5(n – 1) – –2n1)+ + 1)2(n + 2(n 2 – 1) = 5n – 10n + 5 + 2n – 2 = 5n2 – 8n + 3 U= S – Sn – 1 = (5n2 + 2n) – (5n2 – 8n + 3) = 5n2 + 2n – 5n2 + 8n – 3 = 10n – 3 Jadi, rumus suku ke-n: U = 10n – 3. nn

n

b.

n

U = 10n – 3

U1 = 10(1) – 3 = 10 – 3 = 7 U2 = 10(2) – 3 = 20 – 3 = 17 U3 = 10(3) – 3 = 30 – 3 = 27 U4 = 10(4) – 3 = 40 – 3 = 37 Jadi, deret aritmetika tersebut 7 + 17 + 27

+ 37 + . . . . 5. a.

Misal b = penambahan banyak kompor yang diproduksi setiap tahun Banyak kompor yang diproduksi setiap tahun membentuk deret aritmetika dengan a 4.000. Deret aritmetika = yang terbentuk 4.000+(4.000 + b) + (4.000 + 2b) + · · · + (4.000 + 7b) ↓ ↓↓ ↓ Tahun ke-1 ke-2 ke-3 ke-8

S8 = total produksi kompor sampai tahun kedelapan

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0104a464 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 94600010 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 20100c80 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0c80000ff 0000000000000000000000000000000000000000000

95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 e20c5849 6. 2. 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 43435f50 Jumlah 8 suku pertama: 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 524f4649 Barisan geometri 128, 64, 32, 16, 8, a. Barisan geometri 2, 6, 18, 8 8 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 4c450001 Jawaban: Jawaban: a(r −1(( 3) · · · . 64 1c 54, · · · 6b−1) 18 54 S8 =rasio =r1− a = 128 dan r = = = = −1) =−− =3 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0100000c 31 1282 n1 268Besar rasio barisan 2, 6, 18, 54, · · · 0000000000000000000000000000000000000000000000 − 484c696e adalah = 1(6.561 ⎛ ⎛ 1 −1)−4 6f021000 Un = arn – 1 = 1282 3. 1 006d6e74 6.560 ⎛⎛ b. Barisan 9 = −geometri 7, 14, 28, 11 72524742 U = 128 ⎛⎛ = 128 · = 56, · · · 1428 56 4 10 ⎛⎛ 2 5124 1 rasio = ===2 2058595a 71428Besar rasio barisan 7, 14, 28, 56, = –1.640 Jadi, besar suku ke-10 2007ce00 adalah. · · · adalah 2. b. Rumus umum: U = a4 + (n – 1)b 02000900 Jadi, jumlah 8 suku pertama n c. Barisan geometri 27, barisan 9, 3, 1, · · · U 06003100 = a + 9b = 4.000 + 9(200) = 5.800 7. Jawaban: a 17 1717 tersebut adalah –1.640. 10 Jadi, pada tahun kesepuluh 9 31 diproduksi 5.800 00616373 rasio = 2793 10. Jawaban: b == Barisan geometri –17, 704d5346 ,– ,,···. kompor. 28 17 4 S = 2 · 3n + 1 – 16+ 1 1 54000000 nS = 2·3 – 6 = 2 · 32 – 6 17121 . 1 Besar rasio barisan 27, 9, 3, 1, · · · adalah 3 a = –17 dan r =00494543 =×=– 2+1 −172−172 n1 = 12 S = 2 · 3 – 6 = 2 · 20735247 2Barisan geometri 64, 32, 16, 8, · · · d. − 3 42000000 3 – 6 = 48 U1 =S1 = 12 S12 = n – 1 ⎛− 32168 Un = ar = (–17) ⎛⎛ rasio = = = = 2  8 00000000 A. Pilihan 1⎛ U1 + U2 ⇔U2 = S2 – 6432162 U1 = 48 – 00000000 Ganda  1 1 12 = 36Besar U2 36 = 1. U9 = ar8 000000f6 rasio barisan 64, 32, 16, 8, ·.· · 2  1

d6000100 I. Barisan bilangan 3, 12, 48, (–17) ⎛− ⎛ 000000d3 Jawaban: 192, · · ·b 12(–17)( 48 192 = == =4 2d485020 31248Oleh setiap ) 17karena perbandingan 256 =– 20000000 dua suku besarnya sama 256berurutan 17 00000000 12, ke-9 48, 192, · · · –merupakan . Jadi,maka besar3, suku adalah 256 barisan geometri. 00000000 00000000 8. II. Jawaban: Rumus 6, suku Barisan b bilangan 18, 54, 00000000 n–1 ke-n barisan geometri: U = ar 216, · · · 18 54=216 00000000 = 3 akan tetapi = 4 6

1854

n 00000000 U1 =Oleh 15 ⇔ a= 15 perbandingan U3 = 135 karena setiap dua 00000000 2 ⇔arsuku = 135 Substitusi a = berurutan besarnya tidak 00000000 sama (yaitu atau ar 4)2maka 15 ke ar2 =3 135: =00000000 1356, 18, 54, 216, · 2 00000000 · · bukan merupakan barisan geometri. ⇔15 r = 135 1 135 00000000 2 1 r = = bilangan 9, 3, III. Barisan 1, 15 11637072 3 2 19 r = = 11 74000001 , · · ·3 3= = ⇔ r2= ± 50000000 9313 Oleh karena perbandingan setiap 33646573 dua suku 1 63000001 Untuk r = 3, U5besarnya = ar4 = 15 berurutan sama maka (yaitu 84000000 4 · 3 =)15 · 81 = 1.215 1 6c777470 3 maka 9, 3, 1, , ·43·74000001 · merupakan Untuk r = –3, U5 = ar barisan f0000000 geometri. = 15(–3)4 = 15 14626b70 · 81 = 1.215 1 1 74000002 IV. Barisan bilangan 10, 5, 21.215. ,,·· Jadi, nilai suku ke-5 adalah 2 1151 1 2 04000000 2 2 · tetapi 9. 14725859 2 = = akan = 1 1052 5 5a000002 2 Barisan geometri 1, –3, 9, –27, 2 · Jawaban: b 18000000 ·a·= . −3 Oleh perbandingan setiap dua 1 dankarena r = = –3 14675859 1 Rumus n suku suku jumlah berurutan besarnya tidak sama 5a000002 n pertama: 1) 11 maka a(r 10,−5, 2c000000 2 , , · · · bukan merupakan 2 Snr1 2 = 14625859 − Jadi, barisan I dan III merupakan barisan geometri. 5a000002 40000000 barisan geometri. 14646d6e 64000002 54000000 70646d64 64000002 c4000000 88767565 64000003 4c000000 86766965 77000003 d4000000

2

adalah rasio: r = = =geometri 3 U1 Jadi, barisan yang mempunyai 12 rasio 2 adalah barisan 7, 14, 28, 56, · · · . Jadi, rasio deret geometri tersebut adalah 3. 3. Jawaban: b Pada barisan geometri 11. Jawaban: berlaku perbandingan setiap dua suku yang aberurutan Rumus jumlah n suku besarnya sama. pertama a(rn −1) Sm = 16 410 nr1= = − a(r 256 m16

−1)

⇔ = S1 r1 10 164 16 − a(2 = ⇔0 m= × 16 m 16 4−1) ⇔ 1.023 = − 21 ⇔ m= 64 Diperoleh barisan geometri a(1.024 256, 64, 16, 4, · · · . Jadi, nilai m adalah 1.023 = 1 − −1) 64. 1.023 1 = a(1.023) a 4. a= a= 1 U1 geometri = a = 1 Jadi, pertama Barisan 3, p,suku 12, 4p, 48, deret tersebut adalah 1. Jawaban:·pb· · 12 4p 48 rasio = === 3 p 12 4p p 12 = 3p

⇔ p2 = 36 ⇔ p = ±6

6p Untuk p = 6, diperoleh rasio 33 = −6 = = 2. p Untuk p = –6, diperoleh rasio = = = – 33 Oleh karena rasio positif maka nilai p 2. yang memenuhi adalah 6. Jadi, nilai p yang memenuhi adalah 6. 5. Jawaban: a Barisan geometri 12, 36, 108, 324, · · · diperoleh 36

a = 12 dan r = = 3. 12 = arn – 1 = 12 × 3n – 1 = 3 × = 4 × 3n Jadi, rumus suku ke-n adalah U = 4 × 3n. n 3n U n 12

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0104a464 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 94600010 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 20100c80 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0c80000ff 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 e20c5849 13. Jawaban: b Misal U1 = panjang B. 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 43435f50 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 524f4649 potongan tali terpendek dan U4 = panjang 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 4c450001 1. I. Deret geometri 2 + 6 + 18 + Uraian potongan tali terpanjang. 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0100000c 54 + · ·a· = 6 2 dan r = = 3 U1 = a = 25 0000000000000000000000000000000000000000000000 484c696e 3 2 U = arn – 1 cm U4 = ar = 200 6f021000 n U4ar 2003 006d6e74 cm U = ar6 = 2(3)6 = 2 × 729 = = = Ua25 7 72524742 1 2058595a 1.458 a(rn −1) Sn= ⇔ r3 = 8 ⇔r = r1 7 7 2007ce00 − a(r −1) 2(3 Jumlah keempat potongan 02000900 S7= =−1) tali: a(rn −1) − 31 2(2.187 06003100 r1− Sn= − r1 00616373 −1) = = 2.186 a(r4 −1) 704d5346 ⇔S4= 2 Jadi, U = 1.458 dan S = 7 7 − r1 54000000 2.186. 25(24 −1) 00494543 = −= 25(16 – 1) = 375 II. Deret geometri 1 – 2 + 4 – 8 + · 21 20735247 Jadi, panjang tali mula-mula 375 cm. · −2 42000000 a = 1 dan r· = = –2 n–1 1 U = ar 00000000 14. Jawaban: a Tinggi tanaman setiap hari n membentuk barisan 00000000 U7 = ar6 = 1(–2)6 = 000000f6 1 n geometri dengan ad6000100 = U1 = 64 a(r −1) 2 1 dan U2 Sn= r1 − a(r7 −1) − 000000d3 =U2. 24 2d485020 2 2 1(( 2)7 . Diperoleh r = 1= = = 1 33 S = = U −1) 7 20000000 2 −− 21 1( 128 −1) r1−

00000000 48 − = 2 × 00000000 U3 = U2 × r =12 3 −3 00000000 = 8 4 32 = −129 = 43 U43= U3 × r = × 00000000 −3 Jadi, U = 64 dan S 3 3 7 7 00000000 = 32 4128 20 = 43. U59= U4 × r =932727 × 00000000 ==4 00000000 27· · · III Deret geometri 2 + 3 +9 + + 2 4 Jadi, tinggi tanaman pada pengamatan 00000000 hari ke-5 20 00000000 3 a = 2 dan r= n – 11 adalah 427cm. 2 U = ar 00000000 n ⎛⎛ 3 00000000 15. Jawaban: d Jumlah penduduk setiap 6 729 729 1.023 11637072 tahun mengikuti deret geometri. U7 = ar6n= 2 ⎛⎛= 2 × = 1.023  a(r −1) 2 6432 74000001 50000000 U1 = Sn= U2a = 64 r1 33646573 − 37 2(( ) 7 = 96 63000001 a(r −1) 2 S = = 3 U2 96 3 7 r1 − −1 84000000 2 rasio = U=1 642 = 12. −1) ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f5 6c777470 2.187 128 40 menit 128 60 2 menit ... n–1 −1283 jam = 180 menit 0524f46494c4500010100000c484c696e6f021000006d6e74725247 Rumus suku ke-n: U 74000001 n= ar = 1 422058595a2007ce00020009000600310000616373704d53465400 Jawaban: 2a 20 4 8 Jumlah penduduk pada tahun f0000000 00000049454320735247420000000000000000000000000000f6d6 2  2.059  2.059 5 menit 14626b70 777 7 2012: U6 = ar 000100000000d32d48502020000000000000000000000000000000 = 4 U 1U⎛⎛ 2U3 = U 9 000000000000000000000000000000000000000000000000000000 74000002 5 ⎛⎛ 3 128 32 729 2.059 000000000011637072740000015000000033646573630000018400 04000000 = 64 ⎛⎛ a = U1 = 00006c77747074000001f000000014626b707400000204000000147  2 Jadi, U7 = dan S7 = . 14725859 258595a00000218000000146758595a0000022c000000146258595 32 32 2 U4 2 5a000002 ⎛243 ⎛ a0000024000000014646d6e640000025400000070646d646400000 r= ==2 = 6432 ⎛⎛ 2. Rumus suku ke-n: U =n n – 1 U2 1 2c400000088767565640000034c0000008676696577000003d4000 18000000  n – ar 1 2= 50 . . . (1) U = 1.250 ar 000246c756d69000003f8000000146d6561730000040c0000002474 U3==250 14675859 U · 2⇒ 5 = 486 n 656368000004300000000c725452430000043c0000080c67545243 4 2 2 5a000002 ⇒ ar = 1.250 . . . (2) ⇔ (ar Jadi, jumlah penduduk pada tahun 2012 sebanyak Banyak Amoeba setelah 3 jam:)r = 0000043c0000080c625452430000043c0000080c746578740000000 2 2c000000 0436f70797269676874202863292031393938204865776c6574742d = 1.250 486 jiwa. U9 = ar81.250 = 2 · ⇔ 50r = 5061636b61726420436f6d70616e790000646573630000000000000 14625859 8 2 012735247422049454336313936362d322e3100000000000000000 2 = 2 · 256r == 25 5a000002 0000012735247422049454336313936362d322e310000000000000 512 40000000 r= ±5 00000000000000000000000000000000000000000 Jadi, banyak Amoeba setelah 3 jam adalah 512. 14646d6e 64000002 54000000 70646d64 64000002 c4000000 88767565 64000003 4c000000 86766965 77000003 d4000000

(

)

ffd8ffe0001 ffd8ffe00 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe 04a4649460 0104a464 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000104a 001020100c 94600010 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 4649460 800c80000ff 20100c80 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 0010201 e20c5849434 0c80000ff 35f50524f46 e20c58494 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 00c800c 494c450001 3435f5052 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 80000ffe 0100000c48 4f46494c4 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 20c5849 5. Oleh karena rasio positif maka r 4c696e6f021 50001010 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 43435f5 000006d6e7 0000c484c = 5. Substitusi r = 5 ke ar2 = 4725247422 696e6f021 ffd8ffe000104a464946 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0524f46 2 2 ffd8ffe000104 50: ar = 50 ⇔ a · 5 = 50 058595a200 000006d6 a46494600010 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 8 m494c450 0001020100c800c80000 7ce00020009 e7472524 50 20100c800c80 ⇔ a= 0000000000000000000000000000000000000000000000 0010100 0006003100 74220585 ffe20c584943435f50524f 000ffe20c5849 25 000c484 0061637370 95a2007ce 43435f50524f4 ⇔ a= 2 Diperoleh a = 2 6494c4500010 dan r 46494c4500010100000c 4d53465400 00020009 c696e6f 0000004945 00060031 = 5. U4 = ar3 = 2 · 53 = 100000c484c6 2· 0210000 Pantulan 484c696e6f021000006d ke-1 ke-2 ke-3 ke-4 ke-5 96e6f0210000 4320735247 00006163 5 6e74725247422058595a 06d6e74 125 = 250 U6 = ar = 2 06d6e7472524 ·5= 4200000000 73704d53 a. Perhatikan ketinggian bola saat jatuh: 7422058595a2 7252474 0000000000 46540000 2007ce00020009000600 2 · 3.125 = 6.250 U4 + U007ce0002000 = 6 82205859 m, 6,8 m, 5,78 m, · 0000000000f 00004945 310000616373704d5346 900060031000 6d60001000 43207352 250 + 6.250 = 6.500 Jadi, nilai · · 6,8 5,78 5a2007c 0616373704d5 r = 86,8 = = 0,85 00000d32d4 47420000 54000000004945432073 U + U6 = 6.500. 346540000000 e000200 8502020000 00000000 3. a. 4 Misalkan deret tersebut digambarkan 049454320735 Jadi, ketinggian bola membentuk barisan 52474200000000000000 0900060 0000000000 00000000 247420000000 sebagai berikut. 15 + + geometri dengan rasio = 0,85. 0000000000 00000000f 00000000000000f6d600 0310000 + suku000000000000 + 480 Rumus ke-n = + 0000000000 6d600010 000000000f6d 6163737 b. Perhatikan ketinggian pantulan bola: 0100000000d32d485020 0000000000 0000000d U = arn – 1 n600010000000 04d5346 6,8 m, 5,78 0000000000 m, 4,913 32d48502 m, . . . Tinggi pantulan 20000000000000000000 U1 = 15 ⇔ a=0d32d4850202 0000000000 02000000 5400000 5000000000000 ke-3 = 5,78 × 0,8500000000000000000000 = 4,913. Tinggi pantulan U ⇔ ar 000000000000 = 15 0000000000 00000000 6 = 480 0004945 5 ke-4 = 4,913 × 0,85 = 4,17605 ≈4,176. 0000000000 00000000 15(r )= 480 480 4 000000000000 480 4 00000000000000000000 4320735 0000000000 00000000 000000000000 Tinggi pantulan ke-5 = 4,17605 × 0,85 = 5 r = = = 32 00000000000000000000 2474200 000000000000 0116370727 00000000 3,5496425 ≈3,550 Panjang lintasan yang 15 000000000000 4000001500 00000000 0000000 ⇔ r= 5 00000000000000001163 dilalui bola 0000033646 = 8 + 200000000 × (6,8 + 5,78 + 4,913 + 000000000000 0000000 70727400000150000000 000000000001 5736300000 00000000 4,176)0000000 + 3,550 163707274000 1840000006 00000000 33646573630000018400 001500000003 00000f6 c777470740 00000000 b. Oleh karena terdiri atas 6 suku 364657363000 = 8 + 43,338 + 00006c77747074000001 00001f00000 00000000 d600010 001840000006 maka n = 6 a(rn −1) 3,550 = 54,888 0014626b70 00000000 Sn= 0000000 f000000014626b707400 c77747074000 r1 − 15(26 01163707 ≈55 m7400000204 001f00000001 d32d485 00020400000014725859 0000001472 27400000 4626b7074000 − 1) Jadi, panjang lintasan yang dilalui bola 55 m. 0202000 S6= 58595a0000 15000000 5a000002180000001467 002040000001 21 − 0218000000 03364657 0000000 47258595a000 58595a0000022c000000 146758595a 36300000 002180000001 0000000 = 15(64 0000022c00 18400000 46758595a000 1 = 15 × 63 = 945 Jadi, 146258595a0000024000 0000000 −1) 0000146258 06c77747 0022c0000001 jumlah deret tersebut 945. 0000000 000014646d6e64000002 595a000002 07400000 46258595a000 002400000001 4000000014 1f0000000 0000000 5400000070646d646400 4646d6e64000 646d6e6400 14626b70 A. Pilihan 0000000 0002c400000088767565 002540000007 1+··· 0002540000 74000002 4. 30 + 20 + 13 Ganda 0000000 3 0646d6464000 0070646d64 04000000 1. Jawaban: 640000034c0000008676 002c40000008 0000000 64000002c4 14725859 Jarak perjalanan unta Baktrian c 0000000 876756564000 696577000003d4000000 0000008876 5a000002 0034c0000008 membentuk suatu 2 7565640000 18000000 0000000 246c756d69000003f800 deret geometri dengan a676696577000 = 30 dan 034c000000 14675859 003d40000002 0000000 n 0000146d656173000004 8676696577 5a000002 r = . a(r −1) 46c756d69000 3 0000000 S= 000003d400 2c000000 n r1 0c000000247465636800 003f80000001 − 0000246c756 14625859 0000000 Pola ke-n 46d656173000 0004300000000c725452 d69000003f8 5a000002 ⎛⎛2⎛ − 32 Banyak Bunga 0040c0000002 0116370 30⎛  30 −1 ⎛ 3 000000146d  ⎛243 Pola474656368000 ke-1 Pola ke-2 Pola ke-3 Pola ke-4 Pola7274000 ke-5 Pola ke-640000000 Pola ke-7 Pola ke-8 430000043c0000080c67 S5= 2= 1 1 1 004300000000 + 2 1 + 2 + 3 1 + 2 + 3 + 4 1 + 2 + 3 + 4 + 6561730000 5 1 + 2 14646d6e +3+4+5+61+2+3+4+5+6+71+ −1 − 0015000 5452430000043c000008 1 33 040c000000 64000002 c72545243000 0043c0000080 2474656368 54000000 0000336 0c625452430000043c00  211⎛ 0000043000 70646d64 = 30 ⎛−c67545243000 ⎛× (– 4657363 Jadi, banyak bunga pada pola ke-8 adalah 00080c74657874000000 0043c0000080 243  00000c7254 64000002 0000018 3) c62545243000 1 + 2 + 5243000004 3 + 4 c4000000 + 5 + 6 + 7 + 8 = 36 00436f70797269676874 0043c0000080 4000000 bunga. 3c0000080c6 88767565 2.110 = 6.330 =c74657874000 = 784 2. Jawaban: 20286329203139393820 7545243000 64000003 6c77747 812727 Jadi, jarak yang 00000436f707 ditempuh unta Pola ke0043c00000 4c000000 b 0740000 4865776c6574742d5061 972696768742 Banyak Lingkaran 80c62545243 86766965 Baktrian selama 4 028632920313 01f0000 1 636b61726420436f6d70 0000043c00 77000003 939382048657 1d4000000 *2 0001462 5 hari adalah 78 km. 00080c7465 616e7900006465736300 76c6574742d5 2 27 7874000000 246c756d 6b70740 061636b61726 200000000000012735247 *3 00436f70797 69000003f 420436f6d706 0000204 3 2696768742 80000001 16e790000646 0000001 342204945433631393636 *4 0286329203 46d65617 573630000000 4 4725859 2d322e31000000000000 1393938204 30000040 000000012735 4c0000002 *5 865776c6574 247422049454 5a00000 00000000001273524742 # 336313936362 742d506163 47465636 2180000 # 2049454336313936362d d322e3100000 6b61726420 80000043 0014675 20 000000000000 436f6d70616 00000000 20322e3100000000000000 * 21 000001273524 8595a00 e790000646 c7254524 742204945433 00000000000000000000 5736300000 30000043 00022c0 6313936362d3 0000000001 c0000080c 0000014 00000000000000000000 22e310000000 2735247422 67545243 000000000000 6258595 0494543363 0000043c 000000000000 a000002 13936362d3 0000080c 000000000000 22e3100000 62545243 4000000 000000000003 0000000000 014646d 0000043c 2 = 2 Jadi, 0000000127 0000080c 6e64000 3524742204 74657874 rasio deret 0025400 9454336313 00000000 tersebut 2. 5

()

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 000000000000000000000000000000000000000000000000000000000000000000000000000000000 –8, 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 –1, 0, ke-20 = 20 × Banyak lingkaran pada pola 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 1, 21 = 420 Jadi, banyak lingkaran pada pola 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 8, ke-20 adalah 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 27, 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 x, 420. 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 y 3. 7 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 7 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 Jawaban: d 7 00000000000000000000000000000000000000000000000 7 7 7 7

7 (–2)3 (–1)3 03 13 23 33 43 53

ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe0 104a46494 104a46494 104a46494 104a46494 00104a4 600010201 600010201 600010201 600010201 6494600 00c800c80 00c800c80 00c800c80 00c800c80 0102010 000ffe20c5 000ffe20c5 000ffe20c5 000ffe20c5 0c800c8 84943435f 84943435f 84943435f 84943435f 13. Jawaban: b 50524f464 Barisan aritmetika 50524f464 50524f464 50524f464 0000ffe2 memiliki beda yang tetap. Pada barisan –2, 0, 94c450001 94c450001 94c450001 94c450001 0c58494 2, 4, 6, · ·0100000c4 · berlaku 00100000c4 – (–2) = 2 – 0 = 4 – 2 = 6 0100000c4 0100000c4 – 4 = 2. Jadi, –2, 84c696e6f 0, 2,84c696e6f 4, 3435f505 6, · · · merupakan 84c696e6f 84c696e6f 24f46494 barisan aritmetika. 021000006 021000006 021000006 021000006 c450001 d6e747252 d6e747252 d6e747252 d6e747252 474220585 474220585 474220585 474220585 0100000 14. Jawaban: a Pada barisan bilangan di 95a2007ce 95a2007ce 95a2007ce 95a2007ce c484c69 atas mulai suku ke-4 dan seterusnya 000200090 000200090 000200090 000200090 6e6f0210 merupakan penjumlahan 3 suku di depannya. 006003100 006003100 006003100 006003100 3 00006d6 U1 =x1=U42006163737 ==2006163737 = 3006163737 U = 6 =y 1=+532= +53× U55=× 4U×3006163737 4× 4 4= 64 5 = 125 Jadi, dua bilangan selanjutnya e747252 04d534654 04d534654 04d534654 04d534654 11 = 2 + 3 + 6 U6= 20 = 3 + 6 + 11 U7= x = 000000004 000000004 000000004 000000004 adalah 64 dan 125. 4742205 6 + 11 + 20 = 37 U8= 68 = 11 + 20 + 37 Jadi, 945432073 945432073 945432073 945432073 4. Jawaban: a 50, 45, 39, 8595a20 isian yang tepat adalah 37. 524742000 524742000 524742000 32, U5 U6524742000 07ce000 000000000 000000000 000000000 000000000 2000900 000000000 –5 000000000 –6 –7 –8 000000000 –9 000000000 0600310 0000000f6 0000000f6 0000000f6 0000000f6 Selisih dua bilangan berurutan d60001000 d60001000 d60001000 d60001000 0006163 ffd8ffe000104a4649460 membentuk barisan bilangan bulat 00000d32d 00000d32d 00000d32d 00000d32d 001020100c800c80000ff 73704d5 berurutan. U = 32 – 8 = 24 U6 = U5 – 9 = 5 485020200 485020200 485020200 485020200 e20c584943435f50524f4 3465400 24 – 9000000000 = 000000000 15 Jadi, dua suku berikutnya 000000000 000000000 15. Jawaban: 6494c4500010100000c4 0000004 000000000 24 000000000 dan 000000000 15. 000000000 d adalah I.84c696e6f021000006d6e 3, 000000000 000000000 000000000 000000000 9454320 5. Jawaban: 74725247422058595a20 + 2+ 2+ 3+ 3+ 4 000000000 000000000 000000000 000000000 c ke-Segitiga 7352474 07ce0002000900060031 Baris Pascal 11 →1 = 20 2berurutan 11 →2 = 21 3 121 →4 = Selisih dua suku yang tidak 000000000 22 4 1331 →000000000 8 = 000000000 23 5 000000000 14641 →2000000 16 = 24 6 1 510 10 5 1 →32 = 0000616373704d534654 tetap sehingga bukan barisan 000000000 000000000 000000000 25 7 1 6152015 6 1→ 26 8000000000 1 7 21 35 35 21 7 1 →27 9 1 8 00000000494543207352 0000000 aritmetika. 2856705628 8 1 → 2 8 10 1 9 36 84126126 84 36 9 1 →29 000000000 000000000 000000000 000000000 II.47420000000000000000 0, 0000000 000000000 000000000 000000000 000000000 000000000000f6d60001 + 3+ 3+ 3+ 3 +3 000000000 000000000 000000000 000000000 0000000 00000000d32d48502020 Selisih dua suku yang berurutan selalu 000000000 000000000 000000000 000000000 0f6d6000 barisan 00000000000000000000 tetap sehingga merupakan 000116370 000116370 000116370 000116370 00000000000000000000 1000000 aritmetika. 727400000 727400000 727400000 727400000 III. 5, 00000000000000000000 00d32d4 150000000 150000000 150000000 150000000 00000000000000000000 + 1+ 2+ 3+ 4+ 5 8502020 336465736 336465736 336465736 336465736 00000000000000116370 ffd8ffe000104a4649460001020100 Jumlah bilangan padayang barisberurutan ke-10 = 1tidak +9 Selisih dua suku 300000184 300000184 300000184 300000184 0000000 c800c80000ffe20c584943435f50524 +72740000015000000033 36tetap + 84 + 126 + 126 + 84 + 36 + 9 +1 sehingga bukan barisan 0000006c7 0000006c7 0000006c7 0000006c7 0000000 f46494c4500010100000c484c696e6 =64657363000001840000 512 Jadi, jumlah bilangan pada baris kearitmetika. 774707400 774707400 IV.774707400 33,774707400 28,0000000 23, 18, 13, 8, · · · 006c77747074000001f00 f021000006d6e74725247422058595 10 segitiga Pascal adalah 512. 0001f0000 0001f0000 0001f0000 0001f0000 0000014626b707400000 a2007ce0002000900060031000061 0000000 00014626b +00014626b (–5)00014626b + (–5) +00014626b (–5) + (–5) + (–5) Selisih dua 204000000147258595a0 6373704d534654000000004945432 0000000 707400000 707400000 707400000 6. 00002180000001467585 Jawaban: c707400000 5, 8, 13,20, x, y, z, suku yang berurutan selalu tetap 07352474200000000000000000000 204000000 204000000 204000000 204000000 0000000 sehingga merupakan barisan aritmetika. 95a0000022c000000146 00000000f6d6000100000000d32d48 147258595 147258595 147258595 147258595 0000000 258595a0000024000000 50202000000000000000000000000 Jadi, barisan bilangan aritmetika yaitu II + 3a00000218 + 5 a00000218 + 7 a00000218 + 9 +11 + 13 a00000218 014646d6e64000002540 00000000000000000000000000000 0000000 16. dan Jawaban: b Barisan IV. 000000146 x0000070646d646400000 = 20 +000000146 9 =000000146 29000000146 y = x +aritmetika 11 = 29 +turun 11 = 00000000000000000000000000000 0000000 mempunyai beda negatif. Perhatikan barisan 758595a00 758595a00 40 z= y 758595a00 + 13 = 758595a00 40 + 13 = 53 Jadi, tiga 2c400000088767565640 00000000000001163707274000001 padabilangan pilihan b. 30, 26, 22, 18, · ·29, · 40, dan 53. 0000000 00022c000 00022c000 00022c000 00022c000 berikutnya adalah 000034c0000008676696 50000000336465736300000184000 000146258 000146258 000146258 000146258 0000000 577000003d4000000246 0006c77747074000001f0000000146 595a00000 595a00000 595a00000 595a00000 0000000 c756d69000003f8000000 26b70740000020400000014725859 + (–4)240000000 + (–4) +240000000 (–4)240000000 Selisih dua suku 240000000 146d6561730000040c00 5a00000218000000146758595a000 0000000 7. berurutan Jawaban: c Barisan bilangan 1, 3, (beda) dari barisan tersebut 14646d6e6 14646d6e6 14646d6e6 14646d6e6 00002474656368000004 0022c000000146258595a00000240 0001163 5, 7, 9, 11, 13, 15, 17, 19, 21, 23 selalu tetap yaitu (–4). Oleh karena beda 400000254 400000254 400000254 400000254 300000000c7254524300 00000014646d6e640000025400000 barisan negatif maka barisan 30, 26, 22, 7072740 000000706 000000706 000000706 000000706 00043c0000080c675452 070646d6464000002c40000008876 18, · ·46d646400 · merupakan barisan aritmetika ↓↓ ↓46d646400 46d646400 46d646400 0000150 430000043c0000080c62 7565640000034c000000867669657 turun. U1 U20002c4000 U120002c4000 .barisan . .0002c4000 0002c4000 Jadi, bilangan di atas 0000003 5452430000043c000008 7000003d4000000246c756d690000 000887675 000887675 000887675 000887675 0c746578740000000043 03f8000000146d6561730000040c00 terdiri atas 12 suku.3646573 656400000 656400000 656400000 656400000 6f707972696768742028 00002474656368000004300000000 6300000 34c000000 34c000000 34c000000 34c000000 63292031393938204865 c725452430000043c0000080c6754 1840000 867669657 867669657 867669657 867669657 776c6574742d5061636b 52430000043c0000080c625452430 7000003d4 7000003d4 7000003d4 7000003d4 006c777 61726420436f6d70616e 000043c0000080c74657874000000 000000246 000000246 000000246 000000246 4707400 79000064657363000000 00436f707972696768742028632920 c756d6900 c756d6900 c756d6900 c756d6900 00000000127352474220 31393938204865776c6574742d506 0001f000 0003f8000 0003f8000 0003f8000 0003f8000 49454336313936362d32 1636b61726420436f6d70616e79000 0000146 000146d65 000146d65 000146d65 000146d65 2e310000000000000000 06465736300000000000000127352 26b7074 617300000 617300000 617300000 617300000 00000012735247422049 47422049454336313936362d322e3 40c000000 40c000000 40c000000 40c000000 0000020 454336313936362d322e 10000000000000000000000127352 247465636 247465636 247465636 247465636 4000000 31000000000000000000 47422049454336313936362d322e3 800000430 800000430 800000430 800000430 00000000000000000000 10000000000000000000000000000 1472585 0000000c7 0000000c7 0000000c7 0000000c7 0000000000000000 00000000000000000000000000

8. 17. Jawaban: d Diketahui barisan aritmetika –18, I.–22, –26, · · 5, sehingga diperoleh: a = – 1, –30, 1, 2,·3, ··· 18 b = –22 –a(–18) 777 Jawaban: 1+11+22+3 merupakan barisan = –4Fibonacci U = a + II. 1, 3, 6, 10, 15, · · · (nn –1)b 7 777 7 U20 = a + 19b 1 1+1 1+2+3 1+2+3+4 = –18 + 1+2+3+4+5 merupakan barisan 19(–4) = –94 = bilangan segitiga a + 21b III. 1, 6, 15, 20, 15, 6, · · · merupakan U22 = –18 + 21(–4) = – bilanganbilangan pada baris ke-7 segitiga 102 Nilai U20 + U22 = –94 Pascal IV. 2, 3, 5, 7, 11, · · · merupakan barisan + (–102) = –196 bilangan prima Jadi, barisan I merupakan barisan Fibonacci. 9. 18. Jawaban: d Barisan 127, 119, 111, 103, 95, · · · merupakan barisan aritmetika n 3 U1= a13=–b 1= 1 – 1b=0 U2 =– 2127 – 1= 8 – 1 =7 Jawaban: dengan 127 dan = 119 = –8. 3 3 3 U =Un= –31– 1= 27 – 1 = 26 U = 4 – 1= 64 – 3 4 3 Rumus suku ke1 = 63 U = 5 – 1= 125 – 1 =124 dan 5 n:n U = a + (n – seterusnya. Jadi, barisan bilangan tersebut 1)b= 127 + (n – 1) adalah 0, 7, 26, 63, 124, · · · . (–8) = 127 – 8n + 8 = 135 – 8n Un = 23 b Rumus suku 10. Misal: Jawaban: ⇔ 135 – 8n= 23 ⇔ ke-n: U = 3 × 2n – 1. –8n = 23 – 5135 ⇔– n U5 = 3 × 2 – 1 = 3 × 24 = 3 × 16 = 48 U8 8n = –112 = 3 × 28 – 1 =−112 3 × 27 = 3 × 128 = 384 ⇔ n= = 14 Hasil penjumlahan sukusuku ke-5 dan ke-8: U5 −8 Jadi, 23 merupakan ke-14. + U8 = 48 + 384 = 432 Jadi, hasil 19 penjumlahan Jawaban: suku b Rumus ke- barisan ke-5 suku dan ke-8 n barisan aritmetika: U = a + (n – 432.b U = 4 11. itu adalah Jawaban: 1 1)b n = 2U × 1=×5 (1 + 1) U = ⇔ a + 2b2 = 12 5 3 = 2U × 2=×13 (2⇔ + a1)+U6b = 24 3 = 7 = 2 13 × 3 × (3 + –––––––––– 1) U4= 40 – ––4b = –8 = 2 b= × 42×Substitusi (4 + 1) U5b==60 2 ke a + 2b = 2= × 5 × (5 + 1) # 5: a a + Un 2(2)= = 25× n × (n + a= 5 – 4 = 1 U = a + (n – 1)b 1) Jadi, rumus suku ke-n barisan tersebut U = 2n(n + 1). n 12. Jawaban: d 2 12 +1 U1= = 11 5 +1 22

U2= = U3= U4=

22 10 32 +1 = 33 17 42 +1 = 44 n2+1 Un = n

n2+1 n

Jadi, rumus suku ke-n adalah U=. n

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe0001 ffd8ffe000104a ffd8ffe00010 ffd8ffe00010 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 04a4649460 4649460001020 4a464946000 4a464946000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 001020100c 100c800c80000f 1020100c800 1020100c800 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 800c80000ff fe20c584943435 c80000ffe20c c80000ffe20c 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 e20c584943 f50524f46494c4 584943435f5 584943435f5 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 435f50524f4 500010100000c 0524f46494c 0524f46494c4 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 6494c45000 484c696e6f0210 45000101000 5  5  50001010000 8 m + 2n = 4−  + 2 ⎛⎛ = (2(8) + 7(– 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 10100000c4 00006d6e74725 00c484c696e 0c484c696e6f 32 2 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 84c696e6f02 247422058595a 6f021000006 021000006d6 6)) = 4(16 – 55 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 1000006d6e 2007ce0002000 d6e74725247 e7472524742 =−+ 4 16 42) = 4(– 0000000000000000000000000000000000000000000000 7472524742 9000600310000 422058595a2 2058595a200 26) = – −20 + 5 15 2058595a20 616373704d534 007ce000200 = 16 16 = – 7ce00020009 Jadi, jumlah deret tersebut –104. 104 07ce000200 6540000000049 09000600310 00060031000 4543207352474 00061637370 0616373704d 21. Jawaban: c Barisan 25. 0900060031 0000616373 2000000000000 4d534654000 53465400000 bilangan ganjil: 1, 3, 5, 7, 9, · · · , Dari barisan geometri tersebut 704d534654 0000000000000 00000494543 00049454320 U25 Diperoleh a = 1 dan b = 3 – 1 Jawaban: c 50 1 diperoleh a0000000049 = 100 dan r = =73524742000 000f6d60001000 20735247420 1002 =2 4543207352 00000d32d4850 00000000000 00000000000 n Rumus suku keSn = (2a + (n 4742000000 2020000000000 00000000000 00000000000 2 n: = arn – 1 U 0000000000 0000000000000 00000f6d600 000f6d60001 n – 1)b)25 1n– 0000000000 0100000000d 00000000d32 = 0000000000000 100( S25 = 2 (2a + ) 00f6d60001 0000000000000 32d48502020 d4850202000 24b) 1 26. a Dari deret tersebut 2Jawaban: 25 00000000d3 0000000000000 00000000000 = (2(1) + diperoleh a = 2, r = 2,00000000000 dan U = 2.048. 2d48502020 0000000000000 00000000000 00000000000 2 24(2)) n 0000000000 0000000000000 00000000000 00000000000 2 = (2 + 48) Pada deret geometri berlaku: 2 0000000000 0000000116370 00000000000 00000000000 5 Jadi,=jumlah 625 25 bilangan ganjil yang U= 0000000000 00000000000 00000000000 t 1n 7274000001500 pertama adalah 625. 0000000000 0000033646573 00000000000 00000000000 0000000000 6300000184000 00000000000 00000000000 = 2 × 2.048 = 22. 0000000000 0006c77747074 00000000000 00000000000 n 0000000000 000001f0000000 00000011637 00001163707 27. 0000000000 S1 = 12d+S1 Jawaban: 14626b7074000 07274000001 27400000150 Ketinggian anak tangga dari 0000000000 0020400000014 50000000336 00000033646 = n2=+2nS = 22 2 Jawaban: d 0000116370 7258595a00000 46573630000 57363000001 lantai: 25, + 2 = 6 S1 = 7274000001 2180000001467 01840000006 840000006c7 + 20 + 20 U1 = 2 S2 = 5000000033 58595a0000022 c7774707400 77470740000 ⇔ U2 = S2 – U1 = 6 – 2 = 4 6465736300 c000000146258 0001f000000 01f00000001 Ketinggian anak tangga dari lantai U1 + U2 Diperoleh U1 = 2 dan U2 = 0001840000 595a000002400 014626b7074 4626b707400 membentuk barisan aritmetika dengan a 006c777470 00000204000 4. Beda = U2 – U1 = 25 0000014646d6e dan beda = 00020400000 b = 20. Rumus suku ke74000001f0 6400000254000 00014725859 0147258595a n: U = a + (n – 1)b n = 4 – 2 = 2 Jadi, beda deret 0000001462 00070646d6464 5a000002180 00000218000 Untuk n = 15 tersebut adalah 2. 6b70740000 000002c400000 00000146758 U maka: =a 14b 00014675859 15 = 25 ++ 14(20) = 305 Jadi, ketinggian 23. Jawaban: c U1, U2, U3, · · ·, U disebut 0204000000 0887675656400 595a0000022 5a0000022c0 anak tangga ke-15 dari lantai adalah 305 147258595a 00034c0000008 c0000001462 00000146258 barisan geometri jika cm = 3,05 meter. nUUU U 0000021800 6766965770000 58595a00000 595a0000024 23 4n 0000146758 03d4000000246 24000000014 00000001464 = = = · · · = . UU U n1 12 3U − 595a000002 c756d69000003f 646d6e64000 6d6e6400000 Pilihan c merupakan barisan geometri 28. 2c00000014 Jawaban: c Banyak potongan kertas: 8000000146d65 00254000000 25400000070 karena berlaku: 1, 2, 6258595a00 4, 8,61730000040c0 .70646d64640 . . Banyak 646d6464000 potongan kertas membentuk barisan 002c4000000 1 11 0002400000 0000024746563 00002c40000 −− 6 24 96 1 = = = –. 0014646d6e 6800000430000 00887675656 88767565640 21 2 b. n 4 3 624 − 6400000254 0000c72545243 40000034c00 000034c0000 Banyak pada gambar = 1lingkaran + (n – 2 membentuk geometri dengan 00867669657 a = 1 dan r = 0000007064 0000043c00000 00008676696 1)2 = 1 + 1 24. Jawaban: a 5 6d64640000 80c6754524300 577000003d4 7000003d400 = 2. 2n – 2 = 2n Misal: a = U = kertas mula-mula U2 = pola bilangan berikut. 55 1 02c4000000 00043c0000080c 000000246c7 0000246c756 10, –5, , m, , – , n, · · · merupakan Jadi, rumus suku ke-n barisan aritmetika – 1 16 2 8 banyak kertas setelah pemotongan 8876756564 6254524300000 56d69000003 d69000003f8 tersebut U2=+2n −51 Pola 4 – 1. n ke-1 ?0 barisan 0000034c00 43c0000080c746 f8000000146 000000146d6 geometri dengan a = 10 dan r = = U3 = banyak kertas setelah 10 20. pertama 0000867669 5787400000000 d6561730000 56173000004 n–1 − . U =3 ar sehingga diperoleh: pemotongan U4 = banyak Pola ke-2 ?12 +kedua 4 dengan Deret aritmetika n = 8, akertas = 8, dan 6577000003 436f7079726967 040c0000002 0c000000247 n2 1   1 10 5 b = –6. n Jawaban: apemotongan 6874202863292 47465636800 setelah ke-3 m = U4 = 10 × −  = 10 × −  = − = − Sd400000024 (2a + (n – 46563680000 # n=ke-3 6 20313939382048 Pola ?22 + 4 04300000000 6c756d6900 00043000000  2  8 84  1  110 5 U 8 banyak kertas setelah 1)b) 65776c6574742 00c72545243 c7254524300 S0003f80000 = 11 (2apemotongan + n = U7 = 10 × −  = 10 × = = ke-10 8 =2d5061636b6172  2 646432 00146d6561 0000043c000 Pola ke-4 ?32 + 4 00043c00000 7b) 730000040c 6420436f6d7061 0080c675452 80c67545243 0000002474 6e79000064657 430000043c0 0000043c000 # 6563680000 3630000000000 000080c6254 0080c625452 0430000000 0000127352474 52430000043 430000043c0 Pola ke-n ?(n – 1)2 + 4 0c72545243 2204945433631 c0000080c74 000080c7465 0000043c00 3936362d322e3 65787400000 78740000000 Banyak lingkaran pada pola ke-25 yaitu 00080c6754 1000000000000 000436f7079 0436f707972 5243000004 0000000000127 72696768742 69676874202 = 3c0000080c (25 –3524742204945 1)2 + 4 02863292031 86329203139 6254524300 4336313936362 39393820486 39382048657 d322e31000000 5776c657474 76c6574742d = 00043c0000 242 + 4 080c746578 0000000000000 2d5061636b6 5061636b617

2. a. 24, 6, 1 1 2 ,

38

, E E E merupakan barisan geometri

dengan a = 24 dan r = 6 24 =

1, 2, 4,8, . . . ↑ ↑ ↑↑ ↓ ↓ ↓↓ U1 U2 U3U4 =

1 21

6

=

14

.

U n = arn – 1

U arn – 1 10 10 n U = ar = 1 · 2 = 210 = 1.024 Jadi, 11 jumlah potongan kertas pada pemotongan ke-10 adalah 1.024. 29. Jawaban: b Total panjang terowongan yang tergali: 4, 10, 16, 22, . . . , 238 Barisan 4, 10, 16, 22, . . . , 238 merupakan barisan aritmetika dengan a = 4, b = 6, dan U = 238.

= 24 n11 4 ??? ????

= 24 n1 4 ?? ???? E

11

4 ??? ????

= 24 n1 4 ?? ????(4)

n

Un= a + (n – 1)b Unn= 238 ⇒ a + (n – 1)b = 238 4 + (n – 1)6 = 238 4 + 6n – 6 =238 6n – 2 =238 6 6n = 240 ⇔ n= 40 Jadi, penggalian terowongan akan selesai dalam 40 hari. 30. Jawaban: a Harga televisi mula-mula = 5.000.000 Harga televisi akhir tahun ke-1

= 96 n1 4 ?? ????

Jadi, rumus suku ke-n adalah Un = 96

n1

4 ?? ????

b. 18, 11, 4, –3, . . . merupakan barisan aritmetika

dengan a = 18 dan b = 11 – 18 = 4 – 11 = –7.

10

= 5.000.000 – × 5.000.000 100= 4.500.000 Harga televisi akhir tahun ke-2 10

= 4.500.000 – × 4.500.000 100= 4.050.000 Harga televisi akhir tahun ke-3 10

= 4.050.000 – × 4.050.000 100= 3.645.000 Jadi, harga televisi pada akhir tahun ke-3 adalah Rp3.645.000,00.

U n = a + (n – 1)b

= 18 + (n – 1)(–7)

= 18 – 7n + 7

= 25 – 7n

Jadi, rumus suku ke-n adalah Un = 25 – 7n.

B. 1. a. Banyak segi empat pada gambar Uraian

membentuk pola bilangan berikut. Pola ke-1 →4 × (1 – 1) = 0 Pola ke-2 →4 × (2 – 1) = 4 Pola ke-3 →4 × (3 – 1) = 8 Pola ke-4 →4 × (4 – 1) = 12

# Pola ke-n →4 × (n – 1) Jadi, banyaknya segi empat pada pola ke20 yaitu 4 × (20 – 1) = 4 × 19 = 76.

1357

c. 1,

8

,

27

,

64

,EEE

Barisan bilangan pembilang: 1, 3, 5, 7, E E E

merupakan barisan aritmetika dengan a = 1

dan b = 2.

U n = a + (n – 1)b ?U n = 1 + (n – 1) * 2

= 1 + 2n – 2

.

= 2n – 1

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d5346540 000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000 Pembilang pada suku ke-n adalah 2n – 1. 000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000001163707274000001500000003364657363000001840000006c77747074000001f00000 3. a. 200 + 196 + 192 + · · · + 8 + 4 0014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 Barisan bilangan penyebut. merupakan deret aritmetika dengan a = 200 95a0000024000000014646d6e640000025400000070646d6464000002c40000008876756564000003 dan b = –4. U = a + (n – 1)b 4c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002 n 1, 8, 27, 64, E E E U n 474656368000004300000000c725452430000043c0000080c675452430000043c0000080c62545243 U = 4 ⇒ 200 + (n – 1)(–4)= 4 n 0000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c657 ⇔ 200 – 4n + 4 =4 ⇔ –4n = 4 4742d5061636b61726420436f6d70616e790000646573630000000000000012735247422049454336 –4– 77777 313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000 200 ⇔ –4n = –200 00000000000000000000000000000000000000000000000 ⇔ n= 50 Deret tersebut mempunyai 50 suku. n Sn = (2a + (n

13 23 33 43 n3

2

– 1)b)50 S50 = 2 (2a + 49b) = 50 2

(2(200) +

49(–4))– 196) = 25(204) = = 25(400 5.100 Jadi, jumlah deret tersebut 5.100. b. 2 + 6 + 18 + . . . + 1.458 merupakan deret geometri dengan a = 2 dan r = 3. U = arn – 1 n–1 n U = 1.458 ⇒ 2 × 3 = n 1.458

⇔ 3n – 1 = ⇔ 729 3n – 1 = 36 n – 1=6 n= 7 Deret tersebut mempunyai 7 suku. a(rn − 1) Sn r1 = 7 S

7

− 2(3 − 1) 2(2.187 −1) = = = 2.186 − 2 31

Jadi, jumlah deret tersebut 2.186. c. 3 + 6 + 12 + · · · + 768 + 1.536 merupakan deret geometri dengan a = 3 dan r = 2. U = arn – 1 n–1 n U = 1.536 ⇒ 3 × 2 = n 1.536

⇔ 2n – 1 = ⇔ 512 2n – 1 = 29 n – 1=9 n= 10

Penyebut pada suku ke-n adalah n3 .

Jadi, rumus suku ke-n adalah Un = 3 2n 1 n ? .

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Jadi, pada bulan ke-30 Pak Dodi d. 100 + 94 + 88 + · · · + (–68) + (– 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 menabung Rp352.000,00. 74) merupakan deret aritmetika dengan a 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 = 100 dan b = –6. U = a + (n – 1)b b. 3 tahun = 36 bulan 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 n Jumlah uang yang ditabung selama 36 0000000000000000000000000000000000000000000000 U = –74 ⇒ 100 + (n – 1)(–6) = –74 n bulan: n Sn = (2a + (n ⇔ 100 – 6n + 6 =–74 ⇔ –6n= 2 –74 – 6 – – 1)b)36 S36 = (2a + 2 = 18(2(120.000) + 35(8.000)) = 100 ⇔ –6n = –180 35b) 18(520.000) = 9.360.000 Jadi, jumlah ⇔ n= 30 uang yang ditabung Pak Dodi selama 3 Deret tersebut mempunyai 30 tahun pertama adalah Rp9.360.000,00. suku. n Sn = (2a + 2

6.

30 (n – 1)b S30= (2a +

Misalkan: U = usia anak ke-n n U1 = usia anak paling muda = 3 U5 =

2

29b) 30 = (2(100) + 2 29(–6))– 174) = 15(26) = 390 = 15(200 Jadi, jumlah deret tersebut 390.

usia anak paling tua = 23 U = a + (n – n1)b U1 = a = 3 U5 = a + 4b = 23 Substitusi a = 3 ke a + 4b = 23: a + 4b = 23 ⇒ 3 + 4b = 23 ⇔ 4b = 20 ⇔ b= 5 Diperoleh a = 3 dan b = 5. U1 = a = 3 U2= a + b = 3 + 5 = 8 U3= a + 2b = 3

4. a. Deret geometri denganS= n+2 n 3 – 32 Rumus suku ke– Sn –S1 U = S – S n: nn U = 6 6 5 6+2 8 = 63= 6.561 – 32–=93= –6.552 32 S

+ 2 × 5 =13 U4= a + 3b = 3 + 3 × 5 =18 U5= a + 4b = 3 + 4 × 5 =23 Jadi, usia

= 35 + 2 – 32 = 37 – 32 S5= 2.187 – 9 = 2.178 U6 = S6 – S5= 6.552 – 2.178 = 4.374 Jadi, besar suku ke-6 adalah 4.374. = 3n + 2 – 32

kelima anak tersebut 3 tahun, 8 tahun, 13 tahun, 18 tahun, dan 23 tahun.

7. S n= panjang besi mula-mula 2,4 m = 240 cm U1 = a = panjang potongan besi terpendek = 24 cm U = panjang potongan besi terpanjang = 72 cm. nn S = (a 2

nnn

+ U) 240 = (24 + 2 72) n n 240= (96) (

b.

n S 2 S1= =27 31 –+ 29 –=318 ==333–2 32 +2

– 32 = 34 – 32

S2= 81 – 9 = 72 S1 = U1 = 18 S2 = U1 + U2 ⇔ 72 = 18 + U2

2

240= 48n 240 2 n= = 5

⇔ U2 = 72 ⇔ U2 =– 54 18U2 54

Deret tersebut mempunyai 10 suku. a(rn − 1) Sn r1 = 10 − 3(2 1) S10 = − 21 − 3(1.024 = −1) = 3.069 1 Jadi, jumlah deret

3.069.

tersebut

rasio = U= = 3 1 18

Jadi, rasio deret tersebut adalah 3. 5. 120.000 + 128.000 + 136.000 + 144.000 + · · · Uang yang ditabung Pak Dodi membentuk deret aritmetika dengan a = 120.000 =(n 8.000. a. U ndan = ab+ – 1)b U30 = a + 29b = 120.000 + 29(8.000) = 352.000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 Latihan Ujian U1, U2, U3, · · · , U10 merupakan deret 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 Sekolah 8 geometri 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 dengan r = dan a = U1 = 30 Pilihan 7 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 n Ganda a(r − 1) cm. 0000000000000000000000000000000000000000000000 S n r1 = 1. Jawaban: c − 24 : (–6) × 8 – 20 : 8 10 30(( ) − 1) 8 (–5) = (–4) × 8 – 20 7 − 1 S10= 7 : (–5) = (–32) – (–4) = (–32) + 4 = –28 30(3,8 − 1) = 1 ≈ 588 cm = 5,88 m 7

2. Jawaban: a Nilai Dian = 21 × 4 + 4 × (–1) + 5 × 0 = 84 – 4 = 80 3. Jawaban: d Pakan Ikan 12 20 x Perbandingan 20 banyak + 10

ikan hias dan banyak pakan yang diperlukan merupakan perbandingan senilai, sehingga: 12 20 = x 20 +10 12

⇔

20 x

=

30

1 ×30 x 12 x= x = 18 kantong Jadi, banyak pakan yang diperlukan 18 kantong per bulan.

Jadi, panjang bambu tersebut 5,88 m. 9. Populasi kijang setiap lima tahun bertambah menjadi dua kali lebih banyak dari 5 tahun sebelumnya atau dapat dikatakan populasi kijang 5 tahun1 sebelumnya adalah 2dari populasi kijang 5 tahun berikutnya. Populasi kijang pada: Tahun 2009 = 1.224 1 Tahun 2004 = (1.224) = 2 612 1 Tahun 1999 = (612) = 2

306 1 Tahun 1994 =

2

(306) =

153 Jadi, banyak kijang pada tahun 1994 adalah 153 ekor. 10. Tinggi kecambah setiap hari sebagai berikut. Hari pertama = 1 60

Hari kedua = 1 +

100

(1)

60 = 1,6 Hari ketiga = 1,6 + 100 = 1,6 (1,6) + 0,96 = 2,56

60

Hari keempat = 2,56 + (2,56)

48Jadi,

banyak potongan besi tersebut ada 5. 8. Misal U = panjang ruas ke-n dari pangkal n bambu. 7 8 U = U1 = 8U⇔ U 2 7 21 7 88

U2 = 8U77 ⇔ U3 = U2 = ( )2 U1 3 # dan seterusnya.

100

= 2,56 + 1,536 = 60 4,096 Hari kelima = 4,096 + 100 = 4,096 + 2,4576 = 6,5536 Jadi, tinggi (4,096) kecambah pada hari kelima 6,5536 cm.

ffd8ffe0001 ffd8ffe0001ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe00 04a4649460 04a4649460 104a46494 104a46494 104a46494 104a46494 104a46494 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 0104a464 001020100c 001020100c600010201 600010201 600010201 600010201 600010201 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 94600010 800c80000ff 800c80000ff 00c800c800 00c800c800 00c800c800 00c800c800 00c800c800 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 20100c80 e20c584943 e20c58494300ffe20c58 00ffe20c58 00ffe20c58 00ffe20c58 00ffe20c58 435f50524f4 435f50524f4 4943435f50 4943435f50 4943435f50 4943435f50 4943435f50 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 0c80000ff 6494c45000 6494c45000524f46494c 524f46494c 524f46494c 524f46494c 524f46494c 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 e20c5849 10100000c4 10100000c4450001010 450001010 450001010 450001010 450001010 10. Jawaban: 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 43435f50 6. ((105Jawaban: Modal – 70) + 70c + (82 –yang 70) + x = 150 84c696e6f02 84c696e6f02 0000c484c6 0000c484c6 0000c484c6 0000c484c6 0000c484c6 ffd8ffe000104 a 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 524f4649 1000006d6e 1000006d6e 96e6f02100 96e6f02100 96e6f02100 96e6f02100 96e6f02100 35 3 + 70 + 12 + x =150 dikeluarkan = harga beli + biaya a46494600010 x2+ 3x − 10 (x − 2)(x + 5) x2 − 7472524742 7472524742 0006d6e74 0006d6e74 0006d6e74 0006d6e74 0006d6e74 = = 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 4c450001 117 + x==150 1 perbaikan 900.000 + 100.000 20100c800c80 − 5)(x x22058595a20 − 25 (x + −5) x5 2058595a20 725247422 725247422 725247422 725247422 725247422 000ffe20c5849 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 0100000c x = 150 – 117 = 33 Jadi, yang memilih = Rp1.000.000,00 Harga jual siswa = 07ce000200 07ce000200058595a20 058595a20 058595a20 058595a20 058595a20 43435f50524f4 0000000000000000000000000000000000000000000000 484c696e kegiatan lain ada 33 orang. modal + keuntungan 0900060031 07ce000200 07ce000200 07ce000200 07ce000200 11. 0900060031 Jawaban:07ce000200 6494c4500010 6f021000 0000616373 0000616373 090006003 090006003 090006003 090006003 090006003 100000c484c6 d 35 704d534654 704d534654 100006163 100006163 x – 6=100006163 x100006163 –100006163 7 96e6f0210000 006d6e74 ffd8ffe000104a = modal + 15% × 46 0000000049 0000000049 73704d534 73704d534 73704d534 73704d534 73704d534 15. Jawaban: 4649460001020 06d6e7472524 72524742 100 15 4543207352 4543207352 654000000 654000000 654000000 654000000 654000000 ag g(x)= 5x + = 4 modal 100c800c80000 7422058595a2 35 × modal + × modal 2058595a 004945432 004945432 004945432 004945432 004945432 ⇔4742000000 x4742000000 – x= 6–7 ffe20c58494343 007ce0002000 100100115 g(a)= 5a + 4 g 46 2007ce00 0000000000 0000000000 073524742 073524742 073524742 073524742 073524742 5f50524f46494c 900060031000 44= 5a +=4 4 0000000000 0000000000 000000000 000000000 000000000 000000000 000000000 × modal 02000900 4500010100000 0616373704d5 9x − 10x 100 00f6d600010 000000000 000000000 000000000 000000000 000000000 ⇔00f6d600010 = –1 5a = 40 5 c484c696e6f02 346540000000 06003100 12 0000000d32 0000000d32 000000000 000000000 000000000 000000000 000000000 1000006d6e747 049454320735 a= 8 a = 115 × 1.000.000 00616373 d485020200 d485020200 0f6d600010 0f6d600010 0f6d600010 0f6d600010 0f6d600010 2524742205859 247420000000 −x a 3= 8 – 3100 = 5 Jadi, 0000000000 0000000d3 704d5346 0000000d3 0000000d3 0000000d3 0000000d3 ⇔0000000000 = –1 = –Rp1.150.000,00 Jadi, harga jual telepon 5a2007ce00020 000000000000 120000000000 ⇔ –x = –12 0000000000 2d4850202 2d4850202 2d4850202 2d4850202 2d4850202 a – 3 = 5. 54000000 0090006003100 000000000f6d seluler tersebut Rp1.150.000,00. 0000000000 0000000000 000000000 000000000 000000000 000000000 000000000 x= 12 00616373704d5 600010000000 00494543 0000000000 0000000000 000000000 000000000 000000000 000000000 000000000 16. Jawaban: c 3465400000000 0d32d4850202 20735247 0000000000 0000000000 000000000 000000000 000000000 000000000 000000000 Diketahui h: y – 2x c = Jumlah 7. 4945432073524 000000000000 7. Jawaban: 0000000000 000000000 000000000 000000000 000000000 42000000 7420000000000 000000000000 12. 0000000000 Jawaban:000000000 b 2x −3 y – 2x = 7 pinjaman selama 10 bulan = 0000000000 0000000000 000000000 000000000 000000000 000000000 000000000 00000000 0000000000000 000000000000 3x ⇔ y = 2x ++ 7 Gradien 0000000000 0000000000 000000000 000000000 000000000 000000000 000000000 pinjaman awal bunga = 00000f6d60001 000000000000 +000000000 6> + 5 00000000 0000000000 0000000000 000000000 000000000 000000000 000000000 garis h = m 2. 452x −3000000f6 3x00000000d32d4 000000000000 h = 1.000.000 + 10 × 1%Garis × 1.000.000 0001163707 0001163707 000000000 000000000 000000000 000000000 000000000 8502020000000 000000000000 g ⊥ garis + h sehingga: = 1.000.000 100.000 = ⇔2740000015 – > –1000000000 2740000015 000000000 000000000 000000000 000000000 d6000100 0000000000000 000000000000 mg· mh= – 45 2x −3 0000000336 0000000336 000000000 000000000 000000000 000000000 000000000 1.100.000 000000d3 0000000000000 000000000001 4657363000 4657363000 000001163 000001163 000001163 000001163 000001163 ⇔1 mg1· 2=–1 3 ))> 20 163707274000 2d485020 20( – 3x ×0000000000000 (–1) Besar angsuran setiap bulan 0018400000 0018400000 707274000 707274000 707274000 707274000 707274000 4 5 0000000000000 001500000003 20000000 06c7774707 06c7774707001500000 001500000 001500000 001500000 001500000 ⇔ m =g– 0000000000000 364657363000 5(2x 5 – 3) – 4(3x) > –20 1.100.000 2 g melalui titik (4, 3) 4000001f000 4000001f000 003364657 003364657 003364657 003364657 003364657 00000000 =Garis 0000000000000 001840000006 ⇔000014626b 10x –363000001 15 –363000001 12x 10 = 110.000 angsuran setiap 000014626b 363000001 363000001 363000001 sehingga: y= Jadi, mx +besar c 00000000 ffd8ffe000104a464946 0000000000116 c77747074000 7074000002 7074000002 840000006c 840000006c 840000006c 840000006c 840000006c > –20 bulan yang harus dibayar Rp110.000,00. 0001020100c800c8000 ⇔ –2x > 00000000 –53707274000001 5 001f00000001 5 g 0400000014 0400000014 777470740 777470740 777470740 777470740 777470740 0ffe20c584943435f5052 5000000033646 4626b7074000 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c48 x< 00000000 7258595a00 7258595a00 00001f0000 00001f0000 00001f0000 00001f0000 00001f0000 1+ c 4f46494c450001010000 5736300000184 002040000001 3= – × 41 3 4c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d53465400 0002180000 0002180000 00014626b 00014626b 00014626b 00014626b 00014626b 00000000 2 0c484c696e6f02100000 0000006c77747 47258595a000 25 3= 3 –2 + c 0014675859 0014675859 707400000 707400000 707400000 707400000 707400000 00000049454320735247420000000000000000000000000000f6d6000100000000d32d48502020000 00000000 6d6e747252474220585 074000001f000 002180000001 8. Jawaban: a Suku ke20 5a0000022c 5a0000022c 204000000 204000000 204000000 204000000 204000000 Jadi, nilai x yang memenuhi {x | x c= 5 1 95a2007ce0002000900 000014626b707 46758595a000 000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000 n barisan aritmetika: U = a 0000001462 0000001462 147258595 147258595 147258595 147258595 147258595 0600310000616373704 < 4000002040000 }. 0022c0000001 00000000001163707274000001500000003364657363000001840000006c77747074000001f0000000 00000000 58595a0000 a00000218 a00000218 a00000218 a00000218 a00000218 2 + (n n– 1)b 13. 58595a0000 Jawaban: d M =46258595a000 {x | x < 12, x d534654000000004945 00147258595a0 Diperoleh persamaan g: y 2= – x + 5. 14626b707400000204000000147258595a00000218000000146758595a0000022c000000146258595 0240000000 0240000000 000000146 000000146 000000146 000000146 000000146 00000000 4320735247420000000 0000218000000 002400000001 U3 + U7= (a + 2b) + (agaris + 6b) bilangan asli} = 758595a00 {1, 2, 3,758595a00 4, 5, 6, 7, 8, 9, 10, 14646d6e64 14646d6e64 758595a00 758595a00 758595a00 a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c 00000000 0000000000000000000 146758595a000 4646d6e64000 11} N0000025400 =0000025400 {x | xd ≤00022c0000 12, x bilangan cacah 00022c0000 00022c0000 00022c0000 00022c0000 4. Jawaban: 17. ⇔–16 Jawaban: b ⇔–16 = 2(a + = 2a + 8b 00f6d6000100000000d3 0022c00000014 002540000007 0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474 00000000 000070646d 000070646d 001462585 001462585 001462585 001462585 001462585 kelipatan 3} –3 –26258595a00000 5 0646d6464000 2 –3 3 – Persamaan P 4b) ⇔ –8 garis = a + y = 4b3x ⇔+ –86= U5 2d48502020000000000 A6464000002 ×6464000002 B = (3 m n95a000002 ) ×95a000002 (6 k m n 656368000004300000000c725452430000043c0000080c675452430000043c0000080c625452430000 11637072 95a000002 95a000002 95a000002 0000000000000000000 2400000001464 002c40000008 2 = {3, 6, 9, 12} )c400000088 3 – 2400000001 Titik potong dengan sumbu Yx= 043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d c400000088 400000001 400000001 400000001 74000001 Jadi, suku ke-5 barisan tersebut = 3–3 ×{3, 62 400000001 × m × n5 – 2 × k– 0000000000000000000 6d6e640000025 876756564000 M7675656400 ∩7675656400 N =3 –3 4646d6e64 2 4646d6e64 24646d6e643 5061636b61726420436f6d70616e79000064657363000000000000001273524742204945433631393 0 → y = 3 × 0 + 6 = 6 Titik = ×6,50000000 24646d6e64 ×4646d6e64 30034c0000008 ×m×n –8. 0000000000000000000 400000070646d 3 00034c0000 00034c0000 000002540 000002540 000002540 000002540 000002540 3 9} 2 6464000002c40 –1 0000000000000000000 6362d322e31000000000000000000000012735247422049454336313936362d322e31000000000000 potong (0, 6) p 33646573 14. 0086766965 Jawaban: c Misal: Am = himpunan × k–3 =000007064 2 ×000007064 3676696577000 × ×n 0086766965 000007064 000007064 000007064 0000000001163707274 0000088767565 003d40000002 3 000000000000000000000000000000000000000000 AB –3olahraga 63000001 siswa77000003d4 memilih B = himpunan TitikJawaban: potong dengan sumbu X y = 9. 77000003d4 6d6464000 6d6464000 6d6464000 6d6464000 6d6464000 × k 0000015000000033646 640000034c000 46c756d69000 4mn = 384000000 000000246c 002c400000 002c400000 002c400000 002c400000 002c400000 siswa000000246c memilih seni n(S) = 150 n(A) = 105 0 0 → 3x +6=0 5736300000184000000 0008676696577 003f80000001 b U2 2 1 3k 088767565 756d690000 756d690000 088767565 088767565 088767565 088767565 r= = = 3x = –6 6 6c777470 6c77747074000001f000 000003d400000 46d656173000 n(B) = 82 n(A ∩ B) = 70 x = banyak siswa 6 U1 63 U1 5. Jawaban: b 527 03f80000001 03f80000001 640000034c 640000034c 640000034c 640000034c 640000034c ffd8ffe000104a4649460001020100c8 000014626b707400000 0246c756d6900 0040c0000002 74000001 yang 46d6561730 memilih kegiatan lain x = – = –2 46d6561730 000000867 000000867 000000867 000000867 000000867 00c80000ffe20c584943435f50524f464 204000000147258595a 0003f80000001 474656368000 S = − f0000000 000040c000 000040c000 669657700 669657700 669657700 669657700 669657700 ∞ 1r = 94c4500010100000c484c696e6f02100 0000021800000014675 46d6561730000 004300000000 0002474656 0002474656 0003d4000 0003d4000 0003d4000 0003d4000 0003d4000 14626b70 758595a0000022c000000 0006d6e74725247422058595a2007ce 040c000000247 c72545243000 6 3680000043 3680000043 000246c756 000246c756 000246c756 000246c756 000246c756 74000002 00020009000600310000616373704d5 146258595a000002400 4656368000004 0043c0000080 = 1 × 00000000c7 00000000c7 d69000003f d69000003f d69000003f d69000003f d69000003f = 1 −3 3465400000000494543207352474200 0000014646d6e640000 300000000c725 c67545243000 04000000 53 2545243000 2545243000 800000014 800000014 800000014 800000014 800000014 00000000000000000000000000f6d60 025400000070646d646 452430000043c 0043c0000080 6 14725859 0043c00000 0043c00000 6d6561730 6d6561730 6d6561730 6d6561730 6d6561730 00100000000d32d4850202000000000 4000002c40000008876 0000080c67545 c62545243000 = 2 = 80c6754524 80c6754524 000040c000 000040c000 000040c000 000040c000 000040c000 5a000002 0000000000000000000000000000000 7565640000034c00000 0043c0000080 3 53 2430000043c00 30000043c0 30000043c0 000247465 000247465 000247465 000247465 000247465 18000000 0000000000000000000000000000000 08676696577000003d4 00080c6254524 c74657874000 000080c625 000080c625 636800000 636800000 636800000 636800000 = 6 ×3 0000000000000000000000001163707 000000246c756d69000 30000043c0000 00000436f707 = 636800000 14675859 2 = 9 Jadi, jumlah semua suku barisan 4524300000 4524300000 430000000 430000000 430000000 430000000 430000000 2740000015000000033646573630000 003f8000000146d65617 3 18080c746578740 972696768742 5a000002 43c0000080c 43c0000080c 0c72545243 0c72545243 0c72545243 0c72545243 0c72545243 geometri 9. 01840000006c77747074000001f00000 30000040c0000002474 0000000436f70 028632920313 7465787400 7465787400 0000043c00 0000043c00 0000043c00 0000043c00 0000043c00 = 2c000000 0014626b70740000020400000014725 6563680000043000000 7972696768742 939382048657 x 000000436f7 000000436f7 00080c6754 00080c6754 00080c6754 00080c6754 00080c6754 3 14625859 8595a00000218000000146758595a00 00c725452430000043c 0286329203139 76c6574742d5 0797269676 0797269676 524300000 524300000 524300000 524300000 524300000 00022c000000146258595a000002400 0000080c67545243000 3938204865776 061636b61726 = 43c0000080 65a000002 8742028632 8742028632 43c0000080 43c0000080 43c0000080 43c0000080 0000014646d6e640000025400000070 0043c0000080c625452 c6574742d5061 420436f6d706 40000000 9203139393 9203139393 c625452430 c625452430 c625452430 c625452430 c625452430 646d6464000002c4000000887675656 430000043c0000080c7 636b617264204 16e790000646 8204865776 8204865776 000043c000 000043c000 000043c000 000043c000 000043c000 14646d6e 40000034c0000008676696577000003 465787400000000436f7 36f6d70616e79 573630000000 c6574742d5 c6574742d5 0080c74657 0080c74657 0080c74657 0080c74657 0080c74657 64000002 d4000000246c756d69000003f8000000 0797269676874202863 0000646573630 000000012735 061636b617 061636b617 874000000 874000000 874000000 874000000 874000000 146d6561730000040c0000002474656 2920313939382048657 0000000000000 247422049454 54000000 26420436f6d 26420436f6d 00436f7079 00436f7079 00436f7079 00436f7079 00436f7079 368000004300000000c725452430000 76c6574742d5061636b 1273524742204 336313936362 70646d64 70616e7900 70616e7900 726967687 726967687 726967687 726967687 726967687 043c0000080c675452430000043c000 61726420436f6d70616e 9454336313936 d322e3100000 0064657363 0064657363 420286329 420286329 420286329 420286329 420286329 64000002 0080c625452430000043c0000080c74 7900006465736300000 362d322e31000 000000000000 0000000000 0000000000 203139393 203139393 203139393 203139393 203139393 c4000000 65787400000000436f7079726967687 0000000001273524742 0000000000000 000001273524 0000127352 0000127352 820486577 820486577 820486577 820486577 820486577 4202863292031393938204865776c65 2049454336313936362 0000001273524 742204945433 88767565 4742204945 4742204945 6c6574742d 6c6574742d 6c6574742d 6c6574742d 6c6574742d 74742d5061636b61726420436f6d706 d322e31000000000000 7422049454336 6313936362d3 64000003 4336313936 4336313936 5061636b6 5061636b6 5061636b6 5061636b6 5061636b6 16e7900006465736300000000000000 0000000000127352474 313936362d322 22e310000000 362d322e31 362d322e31 172642043 172642043 172642043 172642043 172642043 4c000000 12735247422049454336313936362d3 2204945433631393636 e310000000000 000000000000 0000000000 0000000000 6f6d70616e 6f6d70616e 6f6d70616e 6f6d70616e 6f6d70616e 86766965 22e3100000000000000000000001273 2d322e3100000000000 0000000000000 000000000000 0000000000 0000000000 790000646 790000646 790000646 790000646 790000646 5247422049454336313936362d322e3 0000000000000000000 0000000000000 000000000000 77000003 0012735247 0012735247 573630000 573630000 573630000 573630000 573630000 1000000000000000000000000000000 0000000000000000000 0000000000000 000000000004 d4000000 4220494543 4220494543 000000000 000000000 000000000 000000000 000000000 000000000000000000000000 0000012 593 0000025 ×3×× ×3 53

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 104a464946 ffd8ffe000104a4 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 0001020100 64946000102010 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 c800c80000 0c800c80000ffe2 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 ffe20c58494 0c584943435f505 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 3435f50524f 24f46494c450001 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 46494c4500 18. Jawaban: c Garis g mempunyai 22. Jawaban: b AD = BC = 4 cm E 0100000c484c69 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 010100000c persamaan 2x – y – 11 = 0 sehingga: 2x – y – 6e6f021000006d 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 484c696e6f 11= 0 ffd8ffe000104a464 22 DE= 6e747252474220 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 021000006d 9460001020100c80 58595a2007ce00 0000000000000000000000000000000000000000000000 0c80000ffe20c5849 ⇔ y =6e74725247 2x – 11 Gradien 2 = 02000900060031 3 +42D garis g422058595a = m1 = 2 h ⊥g 43435f50524f46494 0000616373704d 2007ce0002 c4500010100000c4 + 916 = 53465400000000 sehingga gradiennya: 0009000600 84c696e6f0210000 m1 × m2 = – 49454320735247 A 6 cmB CE = DE = 5 cm = 25 = 5 cm 3100006163 06d6e74725247422 42000000000000 1 73704d5346 1 058595a2007ce000 00000000000000 2 × 5400000000 m2 = –1 2000900060031000 00f6d600010000 2 1 4945432073 0616373704d53465 Keliling bangun = AB + BC + CE + DE 0000d32d485020 m2gradien =– Jadi, garis h = – 5247420000 4000000004945432 + AD 20000000000000 =6+4+5+5+4= 24 cm 1 . 0000000000 0735247420000000 00000000000000 19. 2 Jawaban: b 0000000000 ffd8ffe000104a4 00000000000000 Jadi, keliling bangun 24 cm.0000000000000000 Misal x = harga buku 649460001020100 0000f6d600 00000f6d60001000 00000000000000 23. Jawaban: dc800c80000ffe20c ∠CAB, ∠ABC, ∠ACD tulis y = harga 0100000000 00000d32d4850202 00000000000000 584943435f50524 merupakan sudut dalam ∆ABC. ∠0000000000000000 ACD dan ∠BCE bolpoin d32d485020 f46494c45000101 00000000000000 merupakan sudut luar ∆ABC. ∠BCE = ∠CAB + 00000c484c696e6 0000000000000000 Sistem2000000000 persamaan linear: 3x + 5y = 00000000000011 f021000006d6e74 ∠ ABC 0000000000000000 43.0000000000000 . . . (1) 4x + 2y = 34.000 ⇔2x + y 63707274000001 725247422058595 0000000000 0000000000000000 = 17.000 . . . (2) 50000000336465 a2007ce00020009 0000000000 = 30° + 0000000000000000 000600310000616 73630000018400 Eliminasi y dari (1) dan 0000000000 21° = 51° 373704d53465400 0000000000000001 00006c77747074 3x + 5y = 43.000 × 1 2x + y = (2): 3x0000000000 + 5y = 43.000 000000494543207 000001f0000000 24. Jawaban: a Besar ∠A2 1637072740000015 = 65° 17.0000000000000 ×10x 5 + 5y = 85.000 352474200000000 0000000336465736 14626b70740000 000000000000000 –––––––––––––––– – (sehadap) ∠ A berpelurus dengan ∠A2 A ffd8ffe000104a46494600 0000000000 1 3000001840000006 00000f6d6000100 02040000001472 ––7x = –42.000 01020100c800c80000ffe20 0000000000 000000d32d48502 c77747074000001f x = 6.000 Substitusi nilai x = 6.000 ke (2): 58595a00000218 c584943435f50524f46494c 020000000000000 0000001163 000000014626b707 2 × 6.000 + y = 17.000 4500010100000c484c696e 00000014675859 000000000000000 ∠A1 +5a0000022c0000 65°= 180° 6 65° 4000002040000001 12.0007072740000 1 + y = 17.000 6f021000006d6e74725247 000000000000000 422058595a2007ce000200 0150000000 47258595a0000021 y = 17.000 – 12.000 = 5.000 Nilai 5x + 7y 000000000000000 ∠A1 = =00146258595a00 180° – 65° 090006003100006163737 3364657363 000000000000000 8000000146758595 00024000000014 04d534654000000004945 000000000000000 ∠A1 =646d6e64000002 115° Jadi, besar ∠A1a0000022c0000001 = 115°. 0000018400 432073524742000000000 000000011637072 00006c7774 46258595a0000024 0000000000000000000f6d 5400000070646d 740000015000000 25. Jawaban: c y 6000100000000d32d4850 = 5 × 7074000001 6.000 + 7 × 5.000 = 30.000 + 000000014646d6e6 033646573630000 6464000002c400 202000000000000000000 f000000014 4000002540000007 35.000 = 65.000 Jadi, harga 5 buku tulis x : y =00008876756564 1 : 2 ⇒01840000006c777 x= 000000000000000000000 47074000001f000 2 626b707400 0646d6464000002c 000000000000000000000 0000034c000000 dan 7 0002040000 bolpoin Rp65.000,00. 000014626b70740 y : z = 4 : 3 ⇒z =3 y 000000000000000000000 4000000887675656 86766965770000 4 000020400000014 20. Jawaban: a ∆ABC dan ∆ACD siku000000000000001163707 0014725859 40000034c0000008 7258595a0000021 03d4000000246c 2 2 2 2 274000001500000003364 siku di B dan C AC = AB + BC = 4 + x, y, dan z saling berpelurus 5a00000218 800000014675859 676696577000003d 756d69000003f8 657363000001840000006c 32 = 16 + 90000001467 = 25 AC2 + CD2 = AD2 CD2 sehingga: x + 5a0000022c00000 y + z =180° y3 4000000246c756d6 77747074000001f0000000 000000146d6561 2 2 0146258595a0000 = AD – AC58595a0000 14626b707400000204000 9000003f80000001 ⇔ + 730000040c0000 y + y= 024000000014646 180° 000147258595a00000218 2 4 022c000000 46d6561730000040 d6e640000025400 00247465636800 000000146758595a00000 146258595a 2y0004300000000c ++ 4y 3y000070646d64640 c000000247465636 22c000000146258595a000 ⇔= 180° 00002c400000088 0000024000 8000004300000000 4 4 0024000000014646d6e64 72545243000004 2 767565640000034 = 000014646d 13 – 25 = 0000025400000070646d6 c725452430000043 3 9y = 720° 9 3c0000080c6754 c00000086766965 464000002c400000088767 6e64000002 169 – 25 = Titik potong (–2, 0). Grafik garis y = c0000080c6754524 77000003d400000 52430000043c00 y = 80° y 565640000034c000000867 5400000070 144 0246c756d690000 3x00080c62545243 + 6 seperti di bawah30000043c0000080 ini. Y CD= 6696577000003d4000000 03f8000000146d6 646d646400 c625452430000043 –2 246c756d69000003f80000 0000043c000008 6 561730000040c00 0002c40000 00146d6561730000040c00 c0000080c7465787 0c746578740000 000024746563680 000024746563680000043 21. 0088767565 400000000436f707 00004300000000c 0000436f707972 00000000c7254524300000 640000034c 725452430000043 L= Lsetengah + 9726967687420286 69676874202863 lingkaran 43c0000080c67545243000 c0000080c675452 0000008676 3292031393938204 Jawaban: 0043c0000080c625452430 29203139393820 11 c Ltrapesium 430000043c00000 = 6965770000 × π× r2 + (AB + CD) × 000043c0000080c7465787 865776c6574742d5 4865776c657474 80c625452430000 2 2 400000000436f707972696 03d4000000 X 061636b617264204 043c0000080c746 2d5061636b6172 tinggi 768742028632920313939 1 246c756d69 22 1 578740000000043 36f6d70616e79000 6420436f6d7061 = × × 142 + (20 + 28) 38204865776c6574742d50 6f7079726967687 2 000003f800 7 2 61636b61726420436f6d70 6e790000646573 Jadi, grafik420286329203139 garis y = 3x0646573630000000 + 6 adalah × 308 250000146d65 = + 600 0000000127352474 63000000000000 gambar pada 3938204865776c6 pilihan b. 616e79000064657363000 2 000000000001273524742 = 908 cm 6173000004 2204945433631393 574742d5061636b 00127352474220 2049454336313936362d3 0c00000024 61726420436f6d7 6362d322e3100000 49454336313936 22e310000000000000000 0616e7900006465 7465636800 0000000000000000 000000127352474220494 362d322e310000 736300000000000 0004300000 54336313936362d322e31 0127352474220494 00000000000000 000127352474220 000000000000000000000 000c725452 54336313936362d3 494543363139363 00001273524742 000000000000000000000 430000043c 62d322e31000000 22e3100000000000 20494543363139 00000000000012 ∠A + 000000000000000 1 0000080c67 0000000000000000 36362d322e3100 012735247422049 ∠A2 = 180° 4 3 3 5452430000 0000000000000000 00000000000000 454336313936362 043c000008 00000000000 d322e3100000000 00000000000000 0c62545243 000000000000000 00000000000000 000000000000000 0000043c00 0000000000DF 000000000000000 00080c7465 +EF 00

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000104 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 a46494600010 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 20100c800c80 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 000ffe20c5849 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 43435f50524f4 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 30. Jawaban: b Panjang AB = CD = 3 6494c4500010 BD garis bagi sudut B 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 cm Panjang AD = BC = 4 cm 100000c484c6 sehingga: ∠ABD = ∠1DBC = × ∠ABC = 60° 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 2 96e6f0210000 ffd8ffe000104a464946000 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 ∠BDC = 180° – (∠DBC + 06d6e7472524 1020100c800c80000ffe20c 0000000000000000000000000000000000000000000000 ∠BCA) = 180° – (60° + 25°) 7422058595a2 584943435f50524f46494c4 = 180° – 85° = 95° 500010100000c484c696e6f 007ce0002000 021000006d6e7472524742 A 900060031000 2058595a2007ce00020009 27. ffd8ffe000104a4649 0616373704d5 000600310000616373704d B 460001020100c800c α 5346540000000049454320 ∆ABD dan ∆BCD kongruen. 2 346540000000 L =b360 ×π×r Jawaban: jurin80000ffe20c5849434 7352474200000000000000 049454320735 g ° D B 35f50524f46494c450 00000000000000f6d600010 60° 22 ffd8ffe000104a464 247420000000 0000000d32d48502020000 0010100000c484c69 = 360°7 × × 21 × 21 9460001020100c80 6e6f021000006d6e74 000000000000 4 cm 4 cm 4 cm 0000000000000000000000 0c80000ffe20c5849 1 22 725247422058595a2 43435f50524f46494 000000000f6d 0000000000000000000000 4 cm = × × 21 × 21 c4500010100000c4 007ce000200090006 67= 231 600010000000 0000000000000000000000 D 84c696e6f02100000 A 00310000616373704 cm2 6d6e747252474220 0d32d4850202 0000000000000000000000 d5346540000000049 58595a2007ce0002 0001163707274000001500 28. Jawaban: b Panjang garis singgung 000000000000 0000033646573630000018 45432073524742000 0009000600310000 persekutuan luar = d. R = 8 cm, r = 2 cm, 616373704d534654 00000000000000000 000000000000 40000006c7774707400000 AD = BD = 4 cm dan BD = BC = PQ = 10 cm 0000000049454320 00000000f6d6000100 ffd8ffe000104a46494600 1f000000014626b70740000 4000000000000 cm Jadi, panjang BC = 47352474200000000 cm. 22 d= 000000d32d4850202 01020100c800c80000ffe2 0000000000000000 0204000000147258595a00 000000000000 31. 0c584943435f50524f4649 Jawaban: a Prisma segi enam 00000000000000000 0000f6d6000100000 000218000000146758595a 22 000000000000 mempunyai 18 rusuk dan 8 sisi. 000d32d485020200 00000000000000000 4c4500010100000c484c6 = 10 − (8 − 0000022c00000014625859 0000000000000000 000000000000 5a0000024000000014646d 00000000000000000 96e6f021000006d6e74725 0000000000000000 2) 00000000000000000 = 100 − 36 247422058595a2007ce00 000000000000 6e64000002540000007064 0000000000000000 64 0000000000000000 020009000600310000616 = 00000000000000000 000000000001 6d6464000002c400000088 0000000000000000 373704d53465400000000 = 8 00000000001163707 cm Jadi, panjang garis singgung 163707274000 767565640000034c000000 0000000000000116 27400000150000000 494543207352474200000 persekutuan luar dua lingkaran tersebut 8 8676696577000003d40000 3707274000001500 001500000003 000000000000000000000 0000033646573630 cm. 33646573630000018 00246c756d69000003f8000 364657363000 000146d6561730000040c0 00001840000006c7 40000006c77747074 00f6d6000100000000d32 29. 7747074000001f000 000001f00000001462 d48502020000000000000 001840000006 0000024746563680000043 000014626b707400 6b707400000204000ffd8ffe000104a464946000 000000000000000000000 c77747074000 00000000c7254524300000 0002040000001472 Jawaban: 160 000147258595a0000 cmb→ 2 meter x m 1020100c800c80000ffe20c5 →5 000000000000000000000 58595a0000021800 43c0000080c67545243000 001f00000001 Cara 1 02180000001467585 0000146758595a00 000000000000000000000 meter Menggunakan 84943435f50524f46494c45 0043c0000080c625452430 00010100000c484c696e6f0 4626b7074000 00022c0000001462 95a0000022c000000 000000000000000000011 1 000043c0000080c7465787 perbandingan senilai 16021000006d6e74725247422 2 58595a0000024000 32. Jawaban: 002040000001 146258595a0000024 637072740000015000000 058595a2007ce0002000900 400000000436f7079726967 000014646d6e6400 b = 0600310000616373704d53 000000014646d6e64 033646573630000018400 47258595a000 6874202863292031393938 0002540000007064 x5 4654000000004945432073 6d6464000002c400 00000254000000706 00006c77747074000001f0 204865776c6574742d5061 002180000001 5247420000000000000000 5 5 0000887675656400 46d6464000002c400 00000014626b707400000 x = × 160 636b61726420436f6d70616 000000000000f6d60001000 2 46758595a000 00034c0000008676 2 204000000147258595a00 00000d32d4850202000000 3 x = 00008876756564000 400 cm e790000646573630000000 696577000003d400 0022c0000001 0000000000000000000000 0034c000000867669 000218000000146758595 5 0000246c756d6900 0000000127352474220494 0000000000000000000000 6 46258595a000 0003f8000000146d6 a0000022c000000146258 = 46577000003d400000 m Jadi, tinggi tiang0000000000000000000000 54336313936362d322e310 561730000040c000 0246c756d69000003f 595a00000240000000146 002400000001 0000000000000000000001 sebenarnya 4 m. 0000000000000000000000 0002474656368000 4 8000000146d656173 46d6e6400000254000000 0116370727400000150000 4646d6e64000 2735247422049454336313 004300000000c725 Cara 0003364657363000001840 0000040c000000247 70646d6464000002c4000 452430000043c000 002540000007 936362d322e31000000000 2 46563680000043000 ∆ADE sebangun dengan000006c77747074000001f0 ∆ABC 0080c67545243000 00088767565640000034c 0000000000000000000000 C 00000014626b7074000002 x000000867669657700000 = = 40° 0646d6464000 0043c0000080c625 00000c72545243000 2 1,6 04000000147258595a0000 ADDE AE 0000000000000000000000 2 452430000043c000 = = 002c40000008 ABBCAC 0043c0000080c6754 3d4000000246c756d6900 D 0218000000146758595a00 3y = 0080c74657874000 z = 0 00022c000000146258595a0 52430000043c00000 0003f8000000146d65617 876756564000 4 00000436f70797269 000024000000014646d6e6 ⇔ =80c62545243000004 Jadi, besar sudut x, y, dan6768742028632920 z berturut-turut 30000040c000000247465 60° 0034c0000008 5 BC 40000025400000070646d6 3139393820486577 3c0000080c7465787 6368000004300000000c7 40°, 80°, dan 60°. 464000002c4000000887675 676696577000 6c6574742d506163 × 51,6 400000000436f70797 25452430000043c000008 ⇔ BC= A65640000034c00000086766 2 mD B 6b61726420436f6d7 26. Jawaban: 003d40000002 2 96577000003d4000000246c 26967687420286329 0c675452430000043c000 0616e79000064657 5 ∠ABC = 180° – ( ∠ BAC + d 756d69000003f8000000146 46c756d69000 3630000000000000 0080c625452430000043c = 4 20313939382048657 Jadi, tinggi tiang m d6561730000040c00000024 ∠BCA) = 180° – (35° + 25°) 0127352474220494 76c6574742d506163 0000080c7465787400000 003f80000001 sebenarnya 4 m. 7465636800000430000000 54336313936362d3 = 180° – 60° = 120° 6b61726420436f6d70 000436f70797269676874 0c725452430000043c00000 46d656173000 22e3100000000000 80c675452430000043c0000 616e7900006465736 202863292031393938204 0000000000012735 0040c0000002 080c625452430000043c000 2474220494543363 30000000000000012 865776c6574742d506163 0080c746578740000000043 474656368000 13936362d322e310 73524742204945433 6b61726420436f6d70616e 6f707972696768742028632 0000000000000000 004300000000 6313936362d322e31 790000646573630000000 92031393938204865776c65 0000000000000000 74742d5061636b61726420 00000000000000000 000000012735247422049 c72545243000 0000000000000000 436f6d70616e79000064657 00000127352474220 454336313936362d322e3 00000 3 cm BC 0043c0000080 3630000000000000012735 49454336313936362 100000000000000000000 2474220494543363139363 c67545243000 d322e310000000000 001273524742204945433 62d322e310000000000000 0043c0000080 0000000001273524742204 00000000000000000 6313936362d322e310000 9454336313936362d322e3 c62545243000 00000000000000000 000000000000000000000 1000000000000000000000 0000000000PQ − (R0000000000000000000000 000000000000000000000 0043c0000080 − r) 00000000 00000000000E

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe000104 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 a46494600010 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 20100c800c80 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 000ffe20c5849 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d ffd8ffe000104a4 43435f50524f4 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 2 649460001020100 Luas ABCD Satu topi membutuhkan 816,4 cm karton sehingga sepuluh ffd8ffe000104a4 6494c4500010 62d322e31000000000000000000000000000000000000000000000000000000 ffd8ffe000104a4 2 H c800c80000ffe20c 649460001020100 topi membutuhkan karton = 10 × 816,4 = 8.164 cm 100000c484c6 1 649460001020100 ffd8ffe000104a464946 584943435f50524 c800c80000ffe20c 2 c800c80000ffe20c 0001020100c800c8000 96e6f02100000 f46494c45000101 584943435f50524 0ffe20c584943435f505 584943435f50524 6d6e74725247 1 00000c484c696e6 t 24f46494c4500010100 f46494c45000101 = 2 f46494c45000101 × 16 × 12 36. Jawaban: c Perhatikan 000c484c696e6f02100 f021000006d6e74 422058595a20 00000c484c696e6 00000c484c696e6 0006d6e747252474220 ∆DEF. FD 725247422058595 = AC = 7 cm ∆DEF = 96 2 f021000006d6e74 07ce00020009 f021000006d6e74 C 58595a2007ce0002000 A siku-siku di D sehingga: cm a2007ce00020009 725247422058595 9000600310000616373 725247422058595 B 000600310000 Luas ABFE 000600310000616 704d534654000000004 a2007ce00020009 = AB a2007ce00020009 × 616373704d53 9454320735247420000 373704d53465400 000600310000616 DE= 000600310000616 BF = 10 0000000000000000000 465400000000 000000494543207 373704d53465400 00000f6d60001000000 × t = 373704d53465400 10t 352474200000000 494543207352 000000494543207 00d32d4850202000000 000000494543207 Luas permukaan prisma =2× 000000000000000 0000000000000000000 352474200000000 474200000000 352474200000000 lluas ABCD + 4 × luas0000000000000000000 ABFE 00000f6d6000100 000000000000000 000000000000000 000000000000 0000000000000000000 792 = 2 × 96 + 4 × 10t 7 000000d32d48502 576 = 00000f6d6000100 0000000000000000000 00000f6d6000100 00000000f6d6 792 = 192 + 40t 7 020000000000000 24000000d32d48502 cm 0000000000000116370 000000d32d48502 ffd8ffe000104a464946 000100000000 40t = 792 – 192 4 7274000001500000003 000000000000000 1 020000000000000 020000000000000 0001020100c800c8000 L∆DEF = × DEd32d48502020 ×FD 3646573630000018400 000000000000000 40t =000000000000000 600 6 600 000000000000000 0ffe20c584943435f5052 00006c7774707400000 000000000000000 t= = 15 000000000000000 000000000000 1f000000014626b7074 000000000000000 = × 24 × 7 4f46494c450001010000 000000000000000 0000020400000014725 40 21 2 = 84000000000000000 cm2 Luas permukaan = 2 × L∆DEF + 000000000000000 0c484c696e6f02100000 000000000000 8595a00000218000000 0000000000000006d6e747252474220585 000000000000000 000000000000000 000000000000 Diperoleh tinggi prisma 15 cm. 32 × (DE + EF + FD) = 2 × 84 + 32 × (24 146758595a0000022c0 000000011637072 000000000000000 000000000000000 Volume prisma = luas00000146258595a0000 ABCD + 25 + 7) = 168 + 32 ×95a2007ce0002000900 56 = 168 + 1.792 000000000000 740000015000000 000000011637072 024000000014646d6e6 000000011637072 0600310000616373704 3 2 × t = 96 × 15 = 1.440 cm = 1.960 cm Jadi, luas permukaan prisma 033646573630000 000000000000 4000002540000007064 740000015000000d534654000000004945 740000015000000 2 6d6464000002c400000 01840000006c777 ABC.DEF 1.960 cm . 033646573630000 000000000000 033646573630000 4320735247420000000 088767565640000034c 47074000001f0000000000000000000000 Jadi, volume prisma 1.440 cm3. 01840000006c777 01840000006c777 000000000000 0000008676696577000 000014626b7074000f6d6000100000000d3 47074000001f000 003d4000000246c756d 47074000001f000 34. Jawaban: 000000000011 000020400000014 69000003f8000000146 000014626b70740 000014626b70740 2d48502020000000000 c Volume = Vtabung + d6561730000040c0000 637072740000 7258595a0000021 000020400000014 000020400000014 0000000000000000000 0024746563680000043 Vsetengah bola 14 800000014675859 015000000033 7258595a0000021 7258595a0000021 0000000000000000000 = π × r2 2×00000000c7254524300 t + ×3 × π × r3 5a0000022c00000 37. 800000014675859 646573630000 00043c0000080c67545 800000014675859 Nilai 0000000000000000000 0146258595a0000 2430000043c0000080c 5a0000022c00000 22 1422 5a0000022c00000 0000000000000000000 01840000006c Frekuensi 2 = × 7 × 8 + 625452430000043c000 ×7237 ××7×7×7 024000000014646 0146258595a0000 5 0146258595a0000 0000000001163707274 Jawaban: d 777470740000 0080c74657874000000 d6e640000025400 3 024000000014646 024000000014646 0000015000000033646 × ×× 22277 00436f7079726967687 01f000000014 6 000070646d64640 = 22 × 7 × 8 4202863292031393938 + d6e640000025400 d6e640000025400 5736300000184000000 3 8 00002c400000088 626b70740000 204865776c6574742d5 000070646d64640 2.156 000070646d64640 6c77747074000001f000 7 = 1.232 + 061636b61726420436f 767565640000034 00002c400000088 020400000014 10 00002c400000088 000014626b707400000 3 6d70616e79000064657 c00000086766965 8 767565640000034 ffd8ffe000104a4649460001020100c80 767565640000034 204000000147258595a 7258595a0000 3630000000000000012 11 2 77000003d400000 = 1.232 + 718 c00000086766965 7352474220494543363 0c80000ffe20c584943435f50524f46494c c00000086766965 0000021800000014675 021800000014 9 3 0246c756d690000 13936362d322e310000 77000003d400000 4500010100000c484c696e6f021000006 77000003d400000 8595a0000022c000000 6 6758595a0000 0000000000000000001 03f8000000146d6 0246c756d690000 10 d6e74725247422058595a2007ce000200 0246c756d690000 146258595a000002400 = 1.9502 cm3 2735247422049454336 561730000040c00 022c00000014 2 03f8000000146d6 3 09000600310000616373704d53465400 03f8000000146d6 0000014646d6e640000 313936362d322e31000 000024746563680 561730000040c00 6258595a0000 0000000000000000000 00000049454320735247420000000000 561730000040c00 025400000070646d646 35. Jawaban: 00004300000000c 0000000000000000000 B 000024746563680 Jika persegi nomor 1 menjadi sisi alas 000024746563680 4000002c40000008876 024000000014 d 000000000000000000f6d600010000000 22 725452430000043 AB = 00004300000000c 0000000000000G = × 0d32d485020200000000000000000000 7565640000034c00000 maka00004300000000c persegi nomor 4 menjadi sisi atas 646d6e640000 c0000080c67545208676696577000003d4 AC × BD E 725452430000043 00000000000000000000000000000000 725452430000043 22 =c0000080c675452 10 + 24 00000000000000000000000000000000 00000000000116370727400000150000 =430000043c00000 100 + 576 0003364657363000001840000006c7774 80c625452430000 =043c0000080c746 676 7074000001f000000014626b707400000 =578740000000043 26 204000000147258595a0000021800000 cm 0146758595a0000022c00000014625859 6f7079726967687 5a0000024000000014646d6e64000002 420286329203139 = AB = 26 cm Luas 5400000070646d6464000002c40000008 3938204865776c6 selimut kerucut = πrs = 8767565640000034c0000008676696577 574742d5061636b 3,14 × 10 × 26 = 816,4 000003d4000000246c756d69000003f80 2 61726420436f6d7 cm 00000146d6561730000040c0000002474 0616e7900006465 656368000004300000000c72545243000 736300000000000 0043c0000080c675452430000043c0000 000127352474220 080c625452430000043c0000080c74657 494543363139363 87400000000436f707972696768742028 62d322e31000000 63292031393938204865776c6574742d5 000000000000000 061636b61726420436f6d70616e790000 012735247422049 64657363000000000000001273524742 454336313936362 2049454336313936362d322e31000000 d322e3100000000 00000000000000001273524742204945 000000000000000 4336313936362d322e31000000000000 000000000000000 00000000000000000000000000000000 000000000000000 0000000000 0AO + BO A Garis pelukis = s

kubus. Jadi, sisi atas kubus adalah persegi 025400000070 430000043c00000 c0000080c675452 000000246c756d69000 nomor 4. 33. Jawaban: d D646d64640000 80c625452430000 430000043c00000 003f8000000146d65617 043c0000080c746 02c400000088 30000040c0000002474 AB= 80c625452430000 5787400000000436563680000043000000 043c0000080c746 767565640000 6f7079726967687 2 00c725452430000043c = 578740000000043 8 + 62 034c00000086 420286329203139 6f7079726967687 A= 100 = 10 cm 0000080c67545243000 766965770000 3938204865776c6 420286329203139 0043c0000080c625452 574742d5061636b 03d400000024 3938204865776c6 430000043c0000080c7 61726420436f6d7 B 574742d5061636b 6c756d690000 465787400000000436f7 0616e7900006465 61726420436f6d7 0797269676874202863 03f800000014 7363000000000002920313939382048657 0616e7900006465 6d6561730000 00012735247422076c6574742d5061636b 736300000000000 040c00000024 494543363139363 000127352474220 61726420436f6d70616e 62d322e31000000 746563680000 494543363139363 7900006465736300000 000000000000000 62d322e31000000 04300000000c 0000000001273524742 0127352474220492049454336313936362 000000000000000 725452430000 454336313936362d322e31000000000000 012735247422049 043c0000080c d322e3100000000 454336313936362 0000000000127352474 675452430000 000000000000000 d322e3100000000 2204945433631393636 000000000000000 043c0000080c 000000000000000 2d322e3100000000000 000000000000000 000000000000000 625452430000 0000000000000000000 2 2 2 000000000000000 0EF − FD 043c0000080c = 25 0000000000000000000 22 2 0AO −7 = + 625 BO − 49 =00000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 38. Jawaban: c Misal rata-rata tinggi 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 kedua orang yang baru 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 masuk = 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 x2 0000000000000000000000000000000000000000000000 Rata-rata mula-mula = x1 Rata-rata delapan pemain = x x

6x×+ ×122x

=

8

6172×+ 2

×x 173= 2

8 8

8 × 173 = 6 × 172 + 2 · x22 1.384 = 1.032 + 2 x22 2 x2 = = 1.384 – 1.032 2 x2 = = 352 x2=

3 352

x2 = 176 Jadi, rata-rata tinggi kedua pemain 176 cm.

Jumlah nilai: = 3 × 5 + 8 × 6 + 10 × 7 + 11 × 8 + 6 × 9 + 2 × 10 = 15 + 48 + 70 + 88 + 54 + 20 = 295 Banyak data = 3 + 8 + 10 + 11 + 6 + 2 = 40 Jumlah nilai 295

Nilai rata-rata =Banyak = = 7,375 data40

Nilai di atas rata-rata yaitu 8, 9, dan 10. Banyak siswa yang mendapat nilai 8, 9, dan 10 = 11 + 6 + 2 = 19 Jadi, ada 19 anak yang mendapat nilai di atas ratarata.

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 ffd8ffe00 ffd8ffe00 ffd8ffe ffd8ff 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 0104a464 0104a464 000104 e00010 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 94600010 94600010 a46494 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 4a4649 20100c80 20100c80 600010 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 460001 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 0c80000ff 0c80000ff 20100c8 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 020100 e20c5849 e20c5849 00c8000 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 43435f50 43435f50 0ffe20c 8. Jawaban: c Barisan ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000 c800c8 62d322e31000000000000000000000000000000000000000000000000000000 5.

Latihan Ujian

n–1 524f4649 524f4649 584943U = ar 006d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000 geometri: 0000ff n ffd8ffe000104a4649460001020100c800c800 Nasional 4c450001 4c450001 435f505 000000000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000 Panjang pipa terpendek: U1 = a = 20 Pilihan 00ffe20c584943435f50524f46494c450001010 e20c58 0100000c 0100000c 24f4649 0000000000000000000000000000000000000000000000000000116370727400000150000000336465736300000184 Jawaban: d Ganda 0000c484c696e6f021000006d6e74725247422 cm Panjang pipa terpanjang: U5 = ar4 = ===== 20 · 3 + 20484c696e 494343 484c696e 0000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a000002 1. 4c45000 058595a2007ce0002000900060031000061637 = 32020r cm = 20 ke ar4 = 4 Substitusi 5f5052 6f021000 6f021000 101000 2c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000 = 320 = Diitentukan: menanga→ 3704d53465400000000494543207352474200 320: 006d6e74 006d6e74 00c484c 4 034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000 4f4649 rnilai = 16 = 10aseri → nilai 5 kalah 00000000000000000000000000f6d600010000Jawaban: 72524742 72524742 696e6f0 004300000000c725452430000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400 4 4 4 0000d32d48502020000000000000000000000 nilai r→ = 2 –2 4c4500 2058595a 2058595a 210000 000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e7900006 Sebuah kesebelasan menang 6 kali, 0000000000000000000000000000000000000 010100 r= 2 Diperoleh a = 20 dan r = 2 U2kalah = ar 3 = 20 × 2007ce00 2007ce00 06d6e7 46573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494 0000000000000000000000000000000000001 kali dan seri 10 – 6 – 3 = 1 kali. Nilai 6 kali 000c48 2 2 02000900 02000900 472524 54336313936362d322e31000000000000000000000000000000000000000000000000000000 2 = 40 cm U3 = ar = 20 × 2 = 20 × 4 = 80 1637072740000015000000033646573630000 menang, 3 kali kalah, dan 1 kali seri: = 6 × 06003100 06003100 742205 3 4c696e 01840000006c77747074000001f00000001462 cm+U43 = ar =+20 ×(5) 23 = = 60 20 –×68+=5 160 10 1× = 59cm 00616373 00616373 8595a2× (–2)6f0210 6. 6b707400000204000000147258595a0000021 Jawaban: d Biaya produksi 10 lembar Jadi, nilai yang diperoleh kesebelasan Panjang pipa yang lain 40 cm, 80 cm, 007ce00ketiga 00006d tembikar = 10 ×704d5346 Rp2.800,00704d5346 = Rp28.000,00 8000000146758595a0000022c0000001462585 tersebut dan 160 cm. Jadi, 120 cm bukan panjang dari 54000000 54000000 020009 Sebanyak 8 tembikar laku dengan harga 95a0000024000000014646d6e640000025400 6e7472 59. salah satu pipa tersebut. 00494543 00494543 000600 Rp5.000,00 setiap lembar dan 2 tembikar laku 000070646d6464000002c4000000887675656 524742 20735247 20735247 2. 310000 Jawaban: a1 40000034c0000008676696577000003d40000 dengan Rp4.500,00 setiap lembar. 205859 42000000 42000000 616373 00246c756d69000003f8000000146d65617300 halamanb = 45 : = kursi 45 × penonton 3 = 135 halaman 00000000 00000000 704d53 9. Jumlah Jawaban: Banyak 5a2007 3 Sisa buku yang belum 00040c0000002474656368000004300000000c 2 00000000 00000000 46540021 barisan membentuk aritmetika 32, 36, 40, 44, dibaca ce0002 725452430000043c0000080c67545243000004 Harga penjualan 10 tembikar 000000f6 000000f6 . . . 000000 U12. a = 32, b = 4, dan n = 12 Jumlah n 3c0000080c625452430000043c0000080c7465 = 1 – ( + ) 000900 = 8 × Rp5.000,00 + 2 ×d6000100 d6000100 494543 5 3deret aritmetika pertama 787400000000436f7079726967687420286329 suku 1565 4 060031 000000d3 207352 Rp4.500,00 000000d3 = Rp40.000,00 + ffd8ffe000104 = – ( + ) = bagian ffd8ffe000104 2031393938204865776c6574742d5061636b61 151515 154 000061 a46494600010 a46494600010 39. Rp9.000,00 Jawaban: Sudut daerah 2d485020 2d485020 474200 =a Rp49.000,00 n 20100c800c80 20100c800c80 726420436f6d70616e79000064657363000000 S =halaman (2a + kegiatan sosial = 360° – (70° + 60° + 000ffe20c5849 000ffe20c5849 20000000 20000000 000000 = n × 135 = 36 halaman 637370 Keuntungan penjualan = 2 43435f50524f4 0000000012735247422049454336313936362 15Jadi, sisa halaman43435f50524f4 yang belum dibaca 60° Harga + 50° + 40°) = 360° – 280° = 00000000 00000000 000000 6494c4500010 6494c4500010 (n – 1)b) penjualan – biaya 4d5346 d322e31000000000000000000000012735247 100000c484c6 100000c484c6 berjumlah Jumlah 12 suku pertama 80° produksi = Rp49.000,00 00000000 00000000 000000 96e6f0210000 – 36. 1296e6f0210000 422049454336313936362d322e31000000000 540000 06d6e7472524 06d6e7472524 S000000 00000000 00000000 = (2a + 7422058595a2 12 2 12 7422058595a2 12 Rp28.000,00 = Rp21.000,00 0000000000000000000000000000000000000 007ce0002000 007ce0002000 000049 3. 00f6d60 Jawaban: b 4 bulan → 30 pekerja x 00000000 00000000 11b)900060031000 900060031000 Banyak responden yang melakukan 00000000 Persentase 00000000 00000000 0616373704d5 0616373704d5 = (2(32) + 11(4)) = (108) = 001000 454320 bulan → (30 + 18) pekerja Permasalahan kegiatan sosial 346540000000 346540000000 2 2 keuntungan 00000000 00000000 00000d 049454320735 049454320735 735247 = Sudut × 100% tersebut merupakan 247420000000 perbandingan 247420000000 daerah kegiatan sosial Keuntungan 00000000 648 32d485 Jadi, jumlah kursi penonton 000000000000 000000000000 = Biaya ×00000000 banyak 420000 360 000000000f6d 000000000f6d 10. 020200600010000000 00000000 600010000000 = produksiRp × 100% ° 00000000 responden 648 buah. 000000 0d32d4850202 0d32d4850202 Rp28.000, berbalik nilai, 00000000 00000000 000000 80° 21.000,00 x2− 3x −000000000000 10 (x −sehingga 5)(x + 000000000000 2) berlaku: = = 003 × 180 42 x − a 30 +18 000000 000000000000 000000000000 360 00000000 00000000 000000 Jawaban: 8x + 15 (x − 5)(x − 3) = × 100% = 75% 000000000000 000000000000 = ° x30 4Jadi, keuntungan yang diperoleh 000000000000 000000000000 00000000 00000000Pak 000000 000000 + x2 = 40 orang 000000000000 000000000000 =48 − 4 000000000000 11637072 11637072 000000 Sanusi adalah 75%. 000000000000 x3 000000 000000000000 000000000000 ⇔ = x 000000000001 30 Jadi, ada 4074000001 orang responden yang 74000001 000000 f6d600 11. Jawaban: d 2x 000000000001 163707274000 163707274000 430× 120 50000000 50000000 000000001500000003 kegiatan pada waktu 001500000003 7. melakukan Jawaban: c Besarsosial pinjaman: M= 010000 − 5 4x − 5 ⇒ x= = = 2,5 364657363000 33646573 33646573 000000 48 48Jadi, pekerjaan = 364657363000 tersebut akan selesai 001840000006 001840000006 Rp2.000.000,00 Persentase bunga: b= 40. luang. Jawaban: c S = {1, 2, 3, 4, 5, 6} n(S) = 4 43 0000d3 c77747074000 c77747074000 63000001 63000001 000000 dalam waktu 2,5 bulan. 3(2x – 5) 3 = 4(4x – 5)001f00000001 001f00000001 18% per tahun pinjaman: = 10dadu 6 Misalkan A = Lama kejadian muncul nmata 4626b7074000 4626b7074000 84000000 84000000 000000 2d4850 6x 6 – 15002040000001 = 16x – 20 002040000001 bulan Jumlah pinjaman dan bunga setelah 6c777470 6c777470 00000047258595a000 202000 6x – 16x = – 47258595a000 10 bulan 4. ⇔ 002180000001 002180000001 74000001 74000001 000000 −2 −2 46758595a000 −23 000000 20 +−2315 ⇔ –10x = y– ⎛ 46758595a000 ⎛ 48x y40022c0000001 ⎛ ⎛ 23x n 3 = {3, 6} kelipatan 0022c0000001 f0000000 f0000000 00000046258595a000 = M+ × 46258595a000 = ⎛ 5⎛−5 − n(A) =12 2 × b14626b70 002400000001 002400000001 36y ⎛d ⎛ 23 y000000 4–2 14626b70 Jawaban: 000000 = (2 · 31 – 2 · x–2 · M 4646d6e64000 4646d6e64000 10 x= 000000 002540000007 002540000007 74000002 74000002 000000 = Rp2.000.000,00 + × 18% × 0646d6464000 y3 – 5)–20646d6464000 12= Rp2.000.000,00 +04000000 n(A) 204000000 1 000000002c40000008 000000 Rp2.000.000,00 = (22 · 002c40000008 3–1 · x–2 · y–2)–2 876756564000 876756564000 P(A)= = = Rp300.000,00 = n(S)6314725859 14725859 0001160034c0000008 0034c0000008 000000 = 2–4 · 676696577000 32 · x4 · y4 Rp2.300.000,00 5a000002 5a000002 370727676696577000 Jadi, peluang muncul mata dadu kelipatan 003d40000002 003d40000002 000000 244 44 46c756d69000 3xy 9xy 18000000 Rp2.300.000,00 18000000 40000046c756d69000 3 1 003f800000014 003f80000001 Angsuran = = 2 adalah . setiap 000000 10 16 46d656173000 46d656173000 14675859 14675859 1500000040c0000002 bulan =3 0040c0000002 = Rp230.000,00 Jadi, besar angsuran 000000 474656368000 5a000002 5a000002 000336474656368000 004300000000 004300000000 tiap bulan Rp230.000,00. 2c000000 2c000000 465736c72545243000 000000 c72545243000 0043c0000080 0043c0000080 14625859 14625859 300000c67545243000 c67545243000 000000 0043c0000080 5a000002 5a000002 1840000043c0000080 c62545243000 c62545243000 000000 0043c0000080 0043c0000080 40000000 40000000 0006c77 c74657874000 c74657874000 000000 00000436f707 14646d6e 14646d6e 74707400000436f707 972696768742 972696768742 000000 64000002 64000002 000001f 028632920313 028632920313 939382048657 939382048657 54000000 54000000 00000076c6574742d5 000000 76c6574742d5 061636b61726 061636b61726 70646d64 70646d64 014626420436f6d706 420436f6d706 000000 16e790000646 16e790000646 64000002 64000002 b70740573630000000 573630000000 001163 000000012735 000000012735 c4000000 c4000000 000020247422049454 247422049454 707274 336313936362 88767565 88767565 400000336313936362 d322e3100000 d322e3100000 000001 64000003 64000003 014725000000000000 000000000000 000001273524 000001273524 4c000000 4c000000 8595a0742204945433 500000 742204945433 6313936362d3 86766965 86766965 0000216313936362d3 22e310000000 22e310000000 003364 000000000000 77000003 77000003 800000000000000000 000000000000 000000000000 657363 000000000000 d4000000 d4000000 014675000000000000 00000000000

00000000000

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2 007ce00020009000600310000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d485020 20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011637072740000015000000 03364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c00000 0146258595a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d40000 00246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452430000043c0000 080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e310000000000000000000000127352474220494543363139363 ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d53465400000000 12. Jawaban: d 3(x Gradien garis n: 62d322e31000000000000000000000000000000000000000000000000000000 49454320735247420000000000000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011 63707274000001500000003364657363000001840000006c77747074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014 ∆y63 – 2) ≤ 4(x – 5) + 6

m===

646d6e640000025400000070646d6464000002c400000088767565640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c7 25452430000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d70616e n ∆4 2 x 790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e310000000000000000000000000000000000000000000000 00000000 2

3 – 6 ≤ 4x – 20 + 6 3x 3 – 6 ≤ 4x – 14 3x 3x – 4x ≤ –14 + 6 ⇔ –x ≤ –8 3 x ≥ 8 ← (dikali –1) Jadi, penyelesaiannya x ≥ 8.

13. Jawaban: b P = {2, 3, 5, 7, 11} Q = {1, 2, 3, 4, 6, 12} P ∩ Q = {2, 3} 14. Jawaban: a M = {siswa suka Matematika} → n(M) = 25 B = {siswa suka Bahasa Inggris} → n(B) = 23 M ∩ B = {siswa suka Matematika dan Bahasa Inggris} n(M ∩ B) = 15 n(M ∪ B) = n(M) + n(B) – n(M ∩ B) = 25 + 23 – 15 = 33 Jumlah siswa = n(M ∪ B) + n((M ∪ B)C) = 33 + 7 = 40 anak. 15. Jawaban: c Fungsi f: x → 3x – k Rumus fungsi: f(x) = 3x – k Diketahui f(2) =2 3(2) – k = 2 3 6 – k= 2 ⇔ –k = 2 – 6 ⇔ ––k = –4 k= 4 Diperoleh f(x) = 3x –k f(x) = 3x – 4 Bayangan dari 5: f(5) =3(5) – 4

= 15 – 4 = 11 Jadi, bayangan dari 5 adalah 11. 16. Jawaban: b Gradien garis k: ∆y −63

mk = = = – x ∆42 Gradien garis A: ∆y −42

mA =∆63 ==– x

Gradien garis m: mm

∆y 42

===

∆63 x

Jadi, garis yang bergradien – adalah 3 garis A. 17. Jawaban: c Garis 3x – 5y + 15 = 0 memotong sumbu X jika y = 0. y = 0 ⇒ 3x – 5(0) + 15 = 0 5 3x = –15 3 x= –5 → (–5, 0) Garis memotong sumbu X di (–5, 0).

Garis 3x – 5y + 15 = 0 memotong sumbu Y jika x = 0. x = 0 ⇒ 3(0) – 5y + 15 = 0 ⇔ –5y = –15 ⇔ y= 3 → (0, 3) Garis memotong sumbu Y di (0, 3). Grafik 3x – 5y + 15 = 0 diperoleh dengan menghubungkan titik (–5, 0) dan (0, 3). ffd8ffe000104a4649

Y 460001020100c800c 80000ffe20c5849434 35f50524f46494c450 0010100000c484c69 6e6f021000006d6e74 725247422058595a2 007ce000200090006 00310000616373704 Jadi, grafik yang sesuai pada d5346540000000049 45432073524742000 c. 18. pilihan Jawaban: b Garis h melalui 00000000000000000 titik 00000000f6d600010 (0, 6) dan (4, 0). Gradien garis h: 0000000d32d485020 20000000000000000 2− yy 061 − mh = ⇔ mh= 00000000000000000 x2 − x1 − 40 00000000000000000 ⇔m = 00000000000000000 h 00000000000000000 −6 4 00000000000116370 ⇔ mh = –3 2 72740000015000000 03364657363000001 Garis g tegak lurus garis h, maka 840000006c7774707 berlaku mg × mh = –1 ⎛ 3 ⎛ 4000001f000000014 626b7074000002040 ⇔m× ⎛− ⎛ = –1 g 00000147258595a00 ⎛2⎛2 00021800000014675 ⇔ m= 8595a0000022c0000 g 3 00146258595a00000 Garis g melalui (2, 3) dan 24000000014646d6e 2 3 64000002540000007 bergradien . −1 0646d6464000002c4 2 y m(x – 0x11) ⇒ y – 3= (x – 2) 3 00000088767565640 3 ⇔– y1 =x= 000034c0000008676 3(y – 3) = 2(x –2 2) 3 2x − 5 4x − 5 696577000003d4000 3y Jadi, 3 – 9 =penyelesaian 2x – 4 persamaan = 000246c756d690000 43 3y – 2x – 9 + 4 = 0 3 03f8000000146d656 1 1730000040c000000 adalah . 0 Jadi, persamaan garis g adalah 3y – 2x x– = 5= 2 24746563680000043 3y – 2x – 5 = 0. 00000000c72545243 0000043c0000080c6 75452430000043c00 00080c62545243000 0043c0000080c7465 787400000000436f7 07972696768742028 63292031393938204 865776c6574742d50 61636b61726420436 f6d70616e790000646 57363000000000000 00127352474220494 54336313936362d32 2e310000000000000 00000000012735247 42204945433631393 6362d322e31000000 00000000000000000 00000000000000000 00000000000000X

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 ffd8ffe000 ffd8ffe000 ffd8ffe000 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 104a46494 104a46494 104a464946 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 600010201 600010201 0001020100 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 00c800c80 00c800c80 c800c80000 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 000ffe20c5 000ffe20c5 ffe20c58494 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 84943435f 84943435f 3435f50524f 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 50524f464 50524f464 46494c4500 22. Jawaban: b Misal: b = 19. Jawaban: ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 94c450001 010100000c 1 baju a harga 94c450001 f021000006d6e74725247422058595a2007ce00020009000600310000616373704d534654000000004945432 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 0100000c4 0100000c4 484c696e6f 0735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 84c696e6f0 84c696e6f0 021000006d ffd8ffe000104a 000000000000000000000000000000000000000000000000000000000000000000000001163707274000001 Harga dua baju dan satu kaos Rp170.000,00 0000000000000000000000000000000000000000000000 21000006d 21000006d 6e74725247 4649460001020 500000003364657363000001840000006c77747074000001f000000014626b70740000020400000014725859 diperoleh 2b + k = 170.000 . . . (1) 8 m Harga satu baju100c800c80000f 6e7472524 6e7472524 422058595a 5a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e6400000254000000 6m dan tiga 742205859 kaos Rp185.000,00 742205859 1 2007ce0002 fe20c584943435 70646d6464000002c400000088767565640000034c0000008676696577000003d4000000246c756d69000003 D f50524f46494c4 F 8 diperoleh b + 3k = 5a2007ce0 5a2007ce0 0009000600 500010100000c f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c675452 m 185.000 . . . (2) Terbentuk 002000900 002000900 3100006163 484c696e6f0210 430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031 10 2b 060031000 + k =060031000 SPLDV: S 73704d5346 00006d6e74725 393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735247 247422058595a m 170.000 b + 3k061637370 061637370 5400000000 2007ce0002000 422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e310 B 20 m G6 m C = 185.000 4d5346540 4d5346540 4945432073 00000000000000000000000000000000000000000000000000000Denah lahan k 9000600310000 = harga 1 kaos A 20 m E Eliminasi b: ∆ DEF5247420000 siku-siku di F,616373704d534 sehingga berlaku: 000000049 000000049 2b + k = 170.000 ×2b 1 + k = 170.000 ffd8ffe000104a46494 2 2 2 454320735 454320735 DE = EF 0000000000 + FD = 6540000000049 b + 3k = 185.000 × 2 2b + 6k = 370.000 60001020100c800c80 4543207352474 2 247420000 247420000 82 + 60000000000 = 64 + 36 2000000000000 –––––––––––––––––– – 000ffe20c584943435f5 000000000 000000000 = 1000000f6d600 0000000000000 –5k = –200.000 0524f46494c45000101 000000000 000000000 −200.000 DE = 0100000000 000f6d60001000 00000c484c696e6f021 ⇔k= −5 000000f6d 000000f6d d32d485020 00000d32d4850 000006d6e747252474 2020000000000 Keliling lahan = AB + BC + CD + DE + ⇔ k = 40.000 600010000 600010000 2000000000 0000000000000 22058595a2007ce000 AE = (18 + 26 + 10 + 10 + 20) m = 0000d32d4 0000d32d4 0000000000 0000000000000 200090006003100006 Substitusi k = 40.000 ke b + 3k = 84 m 0000000000000 850202000 850202000 0000000000 16373704d534654000 185.000: b + 3k = 185.000 ⇒b + 0000000000000 000000000 000000000 0000000000 Biaya memagari lahan = 84 × 000004945432073524 b + 120.000 = 185.000 3(40.000) = 185.000 3 0000000000000 000000000 000000000 0000000000 Rp150.000,00 = Rp12.600.000,00 0000000000000 742000000000000000 b = 65.000 000000000 000000000 0000000000 0000000116370 0000000000000f6d600 23. Jawaban: 000000000 0000000000 Diperoleh harga 1000000000 baju (b) = Rp65.000,00 HG Rusuk-rusuk yang7274000001500 saling bersilangan 0100000000d32d4850 d 0000033646573 000000000 000000000 0000000000 tidak seE bidang. Misalnya: Rusuk BC dan harga 1 kaos (k) = Rp40.000,00. 202000000000000000 6300000184000 000000000 000000000 0000000000 bersilangan 0006c77747074 000000000000000000 Harga tiga baju dan dua kaos 000000000 0000001163 000001f0000000 = 3 × 000000000 Rp65.000,00 +2× 000000000000000000 000000000 000000000 + 7072740000 14626b7074000 Rp40.000,00 = Rp195.000,00 000000000000000000 000000000 000000000 0150000000 0020400000014 = Rp275.000,00 Rp80.000,00 7258595a00000 dengan rusuk EF dan 000000000000000000 000000000 000000000 3364657363 Jadi, harga tiga baju dan dua kaos Rp275.000,00. 2180000001467 B rusuk GH. A000000001163707274 001163707 001163707 0000018400 58595a0000022 000001500000003364 20. Jawaban: b ∆ABC siku-siku di C, Adapun hubungan rusuk-rusuk yang lain c000000146258 2740000012274000001 00006c7774 657363000001840000 595a000002400 sehingga berlaku: AC 500000003 = AB2 + BC2 = sebagai s berikut. 500000003 7074000001 006c77747074000001f 2 2 0000014646d6e 8 + 10 364657363 364657363 BC dengan AD saling6400000254000 sejajar dan terletak pada f000000014 000000014626b70740 satu bidang s yaitu ABCD. 000001840 000001840 626b707400 00070646d6464 000020400000014725 BC dengan EH saling000002c400000 sejajar dan terletak pada = 64 +000006c77 100 = 164000006c77 ∆ACD siku-siku 0002040000 8595a0000021800000 0887675656400 satu bidang s yaitu BCHE. 747074000 747074000 0014725859 di D, sehingga berlaku: AC2 = AD2 0146758595a0000022 BC dengan FB saling 00034c0000008 berpotongan di B. 5a00000218 + CD2 001f00000 001f00000 6766965770000 c000000146258595a0 0014626b7 0000001467 ⇔ AD2 = AC2 0014626b7 – 03d4000000246 24. Jawaban: b ∠ABC +∠ CBE 000024000000014646 58595a0000 c756d69000003f CD2 = 074000002 164 – 112 =074000002 164 = 180° (berpelurus) d6e640000025400000 8000000146d65 040000001 040000001 022c000000 – 121 = 43 ⇔∠ABC = 180° – 070646d6464000002c 61730000040c0 ∠CBE146258595a = 180° – 104°0000024746563 = 400000088767565640 AD = 47258595a 47258595a 000002180 000002180 76° 0000024000 6800000430000 000034c00000086766 Jadi, panjang AD =000001467 0000c72545243 000001467 96577000003d400000 ∠ACF 000014646d merupakan sudut luar segitiga 0000043c00000 58595a000 58595a000 0246c756d69000003f8 ABC ∠6e64000002 A ACF = ∠ABC + ∠BAC 21. 80c6754524300 0022c0000 0022c0000 5400000070 00043c0000080 000000146d65617300 x = 76° + 52° Luas halaman = 6001462585 ×9= 001462585 646d646400 c625452430000 00040c000000247465 2 x = 128° Jadi, nilai x 043c0000080c7 adalah 128°. Jawaban: c 54 m membeli 54 95a000002 95a000002 0002c40000 Biaya rumput = × 636800000430000000 4 4657874000000 400000001 400000001 0088767565 0c725452430000043c0 00436f70797269 Rp25.000,00 = Rp337.500,00 4646d6e64 4646d6e64 640000034c 6768742028632 000080c67545243000 000002540 000002540 0000008676 9203139393820 0043c0000080c625452 000007064 000007064 6965770000 4865776c65747 430000043c0000080c7 6d6464000 6d6464000 03d4000000 42d5061636b61 465787400000000436f 726420436f6d70 002c40000 002c40000 246c756d69 616e790000646 707972696768742028 008876756 008876756 000003f800 5736300000000 632920313939382048 564000003 564000003 0000146d65 0000001273524 65776c6574742d5061 7422049454336 4c0000008 4c0000008 6173000004 636b61726420436f6d7 313936362d322 676696577 676696577 0c00000024 e310000000000 0616e7900006465736 000003d40 000003d40 7465636800 0000000000001 300000000000000127 00000246c 00000246c 0004300000 2735247422049 352474220494543363 756d69000 756d69000 000c725452 4543363139363 13936362d322e31000 62d322e310000 003f80000 003f80000 430000043c 0000000000000 000000000000000000 00146d656 00146d656 0000080c67 0000000000000 012735247422049454 173000004 173000004 5452430000 0000000000000 336313936362d322e3 0c0000002 0c0000002 043c000008 00000000000 100000000000000000 474656368 474656368 0c62545243 000000000000000000 000004300 000004300 0000043c00 000000000000000000 000000c72 000000c72 00080c7465 0

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006 d6e74725247422058595a2007ce00020009000600310000616373704d5346540000000049454320735247420000000000 000000000000000000f6d6000100000000d32d48502020000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c7774 7074000001f000000014626b707400000204000000147258595a00000218000000146758595a0000022c0000001462585 95a0000024000000014646d6e640000025400000070646d6464000002c400000088767565640000034c00000086766965 29. Jawaban: c D 77000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c72545243 25. Jawaban: a Sepasang sudut dalam 0000043c0000080c675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874 berseberangan besarnya sama, berarti ∠P1 = 202863292031393938204865776c6574742d5061636b61726420436f6d70616e790000646573630000000000000012735 ffd8ffe000104a46494600010201 132°. 247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000 00c800c80000ffe20c584943435f5 Sudut P1 dan P2 berpelurus, 0000000000000000000000000000000000000000000000 Kunci Jawaban0524f46494c4500010100000c484 dan Pembahasan ∠ P1 + ∠P2 = 180° ⇒∠P2 = 180° – berarti: c696e6f021000006d6e74725247 422058595a2007ce00020009000 ∠P1 = 180° – 132° = 48° Jadi, besar sudut P2 adalah 26. 48°.Jawaban: a AC merupakan garis tinggi ∆ABD maka AC tegak lurus BD. ∠ACB = ∠ACD = 90°. AC merupakan garis bagi sudut A maka ∠BAC = ∠CAD = 50°. Perhatikan ∆ACD. ∠ADC + ∠ACD + ∠CAD = 180° ⇒∠ADC = 180° – (∠ACD – ∠CAD) = 180° – (90° + 50°) = 40° Jadi, besar ∠ADC = 40°. 27. Jawaban: c Sudut keliling lingkaran yang menghadap busur sama besarnya sama. ∠ABE, ∠ACE, dan ∠ADE menghadap busur AE sehingga ∠ABE = ∠ACE = ∠ADE. ∠ABE A + ∠ACE + ∠ADE = 96° 3∠ABE = 96° 3 ∠ABE = 32° ∠ABE dan ∠AOE merupakan ffd8ffe000104a46494600010201 sudut keliling dan sudut pusat yang 00c800c80000ffe20c584943435f menghadap busur sama yaitu busur AE 50524f46494c4500010100000c4 sehingga: ∠AOE = 2 × ∠ABE 84c696e6f021000006d6e747252 47422058595a2007ce000200090 = 2 × 32° = 64° Jadi, 00600310000616373704d534654 besar ∠AOE adalah 64°. 000000004945432073524742000 28. 0000000000000000000000000f6 d6000100000000d32d485020200 000000000000000000000000000 Jawaban: d 000000000000000000000000000 000000000000000000000000000 000000000000116370727400000 150000000336465736300000184 0000006c77747074000001f0000 00014626b707400000204000000 R = 25 cm r = 147258595a00000218000000146 16 cm d = PQ 758595a0000022c000000146258 = 45 cm R + r 595a0000024000000014646d6e6 = 25 + 16 40000025400000070646d646400 = 41 cm Oleh karena berlaku d > R + r 0002c4000000887675656400000 maka kedua lingkaran terpisah. 34c0000008676696577000003d4 000000246c756d69000003f8000 000146d6561730000040c000000 2474656368000004300000000c7 25452430000043c0000080c6754 52430000043c0000080c6254524 30000043c0000080c7465787400 000000436f70797269676874202 863292031393938204865776c65 74742d5061636b61726420436f6 d70616e79000064657363000000 000000001273524742204945433 6313936362d322e310000000000 000000000000127352474220494 54336313936362d322e31000000 000000000000000000000000000 000000000000000000000

600310000616373704d53465400 A000000494543207352474200000 B 00000000000000000000000f6d6 ∠000100000000d32d48502020000 DEC = ∠AEB (sudut bertolak belakang) ∠000000000000000000000000000 CDE = ∠ABE (sudut dalam berseberangan) ∠000000000000000000000000000 DCE = ∠BAE (sudut dalam berseberangan) ffd8ffe000104a4 000000000000000000000000000 Diperoleh ∆ABE dengan ∆CDE 64946000102010 000000000011637072740000015 sebangun. 0c800c80000ffe2 000000033646573630000018400 AE AB ×AECD 86 × 0c584943435f505 = ⇔AB = = = 12 cm 00006c77747074000001f000000 CECD CEffd8ffe000104a4 4 Jadi, panjang AB = 12 24f46494c450001 014626b70740000020400000014 649460001020100 0100000c484c696 7258595a0000021800000014675 cm. c800c80000ffe20c e6f021000006d6e 8595a0000022c00000014625859 584943435f50524f 74725247422058 30. 5a0000024000000014646d6e640 Jawaban: c 46494c450001010 595a2007ce0002 000025400000070646d64640000 0000c484c696e6f 00090006003100 02c400000088767565640000034 021000006d6e747 00616373704d53 c0000008676696577000003d400 25247422058595a 46540000000049 0000246c756d69000003f800000 2007ce000200090 45432073524742 0146d6561730000040c00000024 674656368000004300000000c725 cm006003100006163 00000000000000 73704d534654000 00000000000000f 452430000043c0000080c675452 000004945432073 6d600010000000 430000043c0000080c625452430 524742000000000 0d32d485020200 000043c0000080c746578740000 A Q00000000000000 8 cm 000000000000000 0000436f7079726967687420286 0000f6d60001000 00000000000000 3292031393938204865776c6574 ∆ABC00000d32d485020 ≅∆PQR, berlaku hal-hal berikut. 00000000000000 742d5061636b61726420436f6d7 1)0616e7900006465736300000000 ∠A200000000000000 = ∠Q, ∠B = ∠R,00000000000000 ∠C = ∠P 2) 00000000000000 AC =000000000000000 PQ, AB = QR, BC = PR Jadi, 000000127352474220494543363 000000000000000 13936362d322e31000000000000 pernyataan yang benar00000000000000 pada pilihan c. 000000000000000 00000000011637 000000000012735247422049454 000000000000000 07274000001500 31. 336313936362d322e3100000000 Jawaban: b ∆TBE merupakan salah 000000000000000 00000336465736 000000000000000000000000000 satu bidang diagonal limas T.ABCDEF. Bidang 000000116370727 30000018400000 0000000000000000000C diagonal 400000150000000 yang lain adalah TAC, TAD, TAE, 06c77747074000 TBD, 001f00000001462 TBE, TBF,336465736300000 TCE, TCF, dan TDF. 1840000006c7774 6b707400000204 7074000001f0000 00000014725859 32. Jawaban: a ∆SUT siku-siku di 00014626b707400 5a000002180000 S sehingga berlaku: 000204000000147 00146758595a00 258595a00000218 00022c00000014 000000146758595 6258595a000002 a0000022c000000 4000000014646d 146258595a00000 6e640000025400 24000000014646d 000070646d6464 6e6400000254000 000002c4000000 00070646d646400 88767565640000 0002c4000000887 034c0000008676 67565640000034c 696577000003d4 000000867669657 000000246c756d 7000003d4000000 69000003f800000 246c756d6900000 0146d656173000 3f8000000146d65 0040c000000247 61730000040c000 46563680000043 000247465636800 00000000c72545 0004300000000c7 2430000043c000 25452430000043c 0080c675452430 0000080c6754524 000043c0000080c 30000043c000008 62545243000004 0c6254524300000 3c0000080c74657 43c0000080c7465 87400000000436f 787400000000436 70797269676874 f70797269676874 20286329203139 202863292031393 3938204865776c 938204865776c65 6574742d506163 74742d5061636b6 6b61726420436f6 1726420436f6d70 d70616e7900006 616e79000064657 46573630000000 363000000000000 00000001273524 001273524742204 74220494543363 945433631393636 13936362d322e3 2d322e310000000 10000000000000 000000000000000 00000000012735 127352474220494 24742204945433 54336313936362d 6313936362d322 322e31000000000 e3100000000000 000000000000000 00000000000000 000000000000000 00000000000000 000000000000000 00000000000000 B 0R

ffd8ffe000104 ffd8ffe0001 ffd8ffe00010 ffd8ffe000 ffd8ffe000104a a46494600010 04a4649460 4a464946000 104a46494 4649460001020 20100c800c80 001020100c 1020100c800 600010201 100c800c80000f 000ffe20c5849 800c80000ff c80000ffe20c 00c800c80 fe20c584943435 43435f50524f4 e20c584943 584943435f5 000ffe20c5 f50524f46494c4 6494c4500010 435f50524f4 0524f46494c4 84943435f 500010100000c 2 100000c484c6 6494c45000 50001010000 50524f464 484c696e6f0210 ffd8ffe000104a46 sehingga AT 2 96e6f0210000 2 10100000c4 0c484c696e6f 94c450001 00006d6e74725 = AE 49460001020100c + ET 2 2021000006d6 06d6e7472524 84c696e6f0 0100000c4 247422058595a 800c80000ffe20c5 ⇔ ET = AT – 2 2 2 7422058595a2 21000006d6 e7472524742 84c696e6f 2007ce0002000 AE = 84943435f50524f4 26 – 10 = 007ce0002000 e747252474 2058595a200 021000006 9000600310000 6494c4500010100 676 – 100 = 576 900060031000 22058595a2 7ce00020009 d6e747252 616373704d534 000c484c696e6f02 0616373704d5 007ce00020 00060031000 474220585 6540000000049 1000006d6e74725 ET = 346540000000 0090006003 0616373704d 95a2007ce 4543207352474 247422058595a20 049454320735 1000061637 53465400000 000200090 2000000000000 1 07ce00020009000 × × 247420000000 3704d53465 00049454320 006003100 0000000000000 600310000616373 3 tinggi 006163737 000000000000 4000000004 73524742000 000f6d60001000 704d53465400000 L 1alas AC × BD 000000000f6d 9454320735 00000000000 04d534654 00000d32d4850 = 32 × × ET 000494543207352 600010000000 2474200000 00000000000 000000004 2020000000000 474200000000000 1 20 × 20 0d32d4850202 0000000000 000f6d60001 945432073 0000000000000 000000000000000 = 32 × × 24 000000000000 0000000000 00000000d32 524742000 0000000000000 00f6d6000100000 000000000000 000f6d6000 d4850202000 000000000 0000000000000 000d32d48502020 1 = × 200 × 24 000000000000 100000000d 00000000000 000000000 0000000000000 000000000000000 3 3= 1.600 cm Jadi, volume limas 000000000000 32d4850202 00000000000 0000000f6 0000000000000 000000000000000 T.ABCD adalah 160 cm3. d60001000 000000000000 0000000000 00000000000 0000000000000 000000000000000 000000000000 0000000000 00000000000 00000d32d 0000000116370 000000000000000 000000000000 0000000000 00000000000 485020200 7274000001500 34. Jawaban: 000000000000000 a Volume 000000000001 0000000000 000000000 0000033646573 000000000000000 air = volume tabung 00000000000 163707274000 0000000000 00000000000 000000000 6300000184000 000011637072740 = 001500000003 0000000000 00000000000 000000000 0006c77747074 000015000000033 πr2t 2 364657363000 0000000000 00001163707 000000000 000001f0000000 = × 3 × 14 646573630000018 22 7 001840000006 0000000000 27400000150 000000000 14626b7074000 3 40000006c777470 = 396 ffd8ffe000104a46494 c77747074000 0000000000 00000033646 000000000 0020400000014 74000001f000000 cm 60001020100c800c800 Volume 21 001f00000001 0000011637 57363000001 000000000 7258595a00000 014626b70740000 00ffe20c584943435f50 4626b7074000 0727400000 000000000 2180000001467 kelereng = 21 × 840000006c7 524f46494c450001010 020400000014725 0000c484c696e6f0210 002040000001 1500000003 000000000 58595a0000022 8595a0000021800 volume bola 4 377470740000 = 21 ×47258595a000 πr 01f00000001 00006d6e74725247422 3646573630 000000000 c000000146258 0000146758595a0 3 4 22 058595a2007ce000200 002180000001 0000184000 4626b707400 000116370 595a000002400 000022c00000014 0900060031000061637 = 21 ×46758595a000 × × 13 00020400000 727400000 0000014646d6e 3704d53465400000000 6258595a0000024 37= 4 0006c77747 × 22 4945432073524742000 0022c0000001 074000001f 0147258595a 150000000 6400000254000 000000014646d6e 0000000000000000000 46258595a000 0000000146 00000218000 336465736 00070646d6464 3 640000025400000 = 88 cm Volume air yang tersisa = 000000f6d6000100000 002400000001 26b7074000 00014675859 300000184 000002c400000 070646d64640000 000d32d485020200000 volume tabung – volume 6 kelereng = 396 4646d6e64000 0020400000 5a0000022c0 0000006c7 0887675656400 – 88 =02c400000088767 308 Jadi, volume0000000000000000000 air yang tersisa 0000000000000000000 002540000007 0147258595 00000146258 774707400 00034c0000008 565640000034c00 dalam0646d6464000 tabung 308 cm3.0000000000000000000 a000002180 595a0000024 0001f0000 6766965770000 000086766965770 0000000000000000000 002c40000008 0000014675 00000001464 00014626b 03d4000000246 00003d400000024 0000000000000011637 876756564000 8595a00000 6d6e6400000 707400000 c756d69000003f 0727400000150000000 6c756d69000003f 3364657363000001840 0034c0000008 22c0000001 25400000070 204000000 8000000146d65 8000000146d6561 000006c777470740000 676696577000 46258595a0 646d6464000 147258595 61730000040c0 730000040c00000 35. Jawaban: 01f000000014626b707 003d40000002 0000240000 002c4000000 a00000218 0000024746563 024746563680000 4000002040000001472 c AC 46c756d69000 00014646d6 000000146 6800000430000 AO = OC = D 88767565640 58595a0000021800000 04300000000c725 2 0146758595a0000022c 003f80000001 e640000025 000034c0000 758595a00 0000c72545243 452430000043c00 000000146258595a000 8 46d656173000 4000000706 00867669657 00022c000 0000043c00000 = = 4 cm0024000000014646d6e 00080c675452430 2A 0040c0000002 46d6464000 7000003d400 000146258 80c6754524300 000043c0000080c 6400000254000000706 474656368000 002c400000 0000246c756 595a00000 00043c0000080 46d6464000002c40000 B 625452430000043 BO = DO = 0088767565640000034 004300000000 0887675656 d69000003f8 240000000 c625452430000 2 c0000080c746578 D c000000867669657700 c72545243000 40000034c0 000000146d6 14646d6e6 043c0000080c7 7400000000436f7 6 = = 3 cm0003d4000000246c756 0043c0000080 0000086766 56173000004 400000254 4657874000000 2 d69000003f800000014 B 079726967687420 c67545243000 9657700000 0c000000247 000000706 00436f70797269 6d6561730000040c000 286329203139393 CD= 0002474656368000004 0043c0000080 3d40000002 46563680000 46d646400 6768742028632 8204865776c6574 300000000c725452430 c62545243000 46c756d690 04300000000 0002c4000 9203139393820 742d5061636b617 = 000043c0000080c6754 0043c0000080 00003f8000 c7254524300 000887675 4865776c65747 26420436f6d7061 52430000043c0000080 c74657874000 000146d656 00043c00000 656400000 42d5061636b61 c625452430000043c00 6e7900006465736 00080c7465787400000 00000436f707 1730000040 80c67545243 34c000000 726420436f6d70 300000000000000 000436f707972696768 972696768742 c000000247 0000043c000 867669657 616e790000646 127352474220494 7420286329203139393 028632920313 4656368000 0080c625452 7000003d4 5736300000000 54336313936362d 8204865776c6574742d 939382048657 0043000000 430000043c0 000000246 0000001273524 5061636b61726420436 322e31000000000 f6d70616e7900006465 76c6574742d5 00c7254524 000080c7465 c756d6900 7422049454336 000000000000012 7363000000000000001 061636b61726 30000043c0 78740000000 0003f8000 313936362d322 735247422049454 2735247422049454336 420436f6d706 000080c675 0436f707972 000146d65 e310000000000 336313936362d32 313936362d322e31000 16e790000646 4524300000 69676874202 617300000 0000000000001 0000000000000000000 2e3100000000000 1273524742204945433 573630000000 43c0000080 86329203139 40c000000 2735247422049 000000000000000 6313936362d322e3100 000000012735 c625452430 39382048657 247465636 4543363139363 000000000000000 0000000000000000000 247422049454 000043c000 76c6574742d 800000430 62d322e310000 0000000000000C 0000000000000000000 2 2 5061636b617 336313936362 0080c74657 0000000c7 0000000000000 00000000000000 C O + DO

ffd8ffe000104 ffd8ffe000104a ffd8ffe000 a464946000102 4649460001020 104a464946 0100c800c8000 100c800c80000 0001020100 ffd8ffe000104a4 0ffe20c5849434 ffe20c58494343 c800c80000 64946000102010 35f50524f4649 5f50524f46494c ffe20c58494 0c800c80000ffe2 4c45000101000 4500010100000 3435f50524 0c584943435f505 00c484c696e6f c484c696e6f02 f46494c450 Luas 24f46494c450001 prisma 021000006d6e7 1000006d6e747 0010100000 0100000c484c69 ABCD.EFGH = 2524742205859 2 × 4725247422058 c484c696e6 6e6f021000006d6 LABCD 595a2007ce000 +14 × LABFE 5a2007ce00020 f021000006 = 2 ×e7472524742205 ( × AC × BD) + 4 × 2 2000900060031 0090006003100 d6e7472524 8595a2007ce000 (BC ×0000616373704 1CG) 00616373704d5 7422058595 20009000600310 = 2 ×d534654000000 4 × 5 × 4 a2007ce000 2× 8 × 6 + 3465400000000 000616373704d5 = 48 + 80 = 128 4945432073524 m2 0049454320735 2000900060 34654000000004 2474200000000 7420000000000 0310000616 94543207352474 Jadi, luas kertas yang diperlukan 128 cm2. 0000000000000 0000000000000 373704d534 20000000000000 =6540000000 36. 0000000f6d600 00000f6d60001 00000000000000 22 7 0100000000d32 00000000d32d4 0f6d60001000000 * 0494543207 d485020200000 8502020000000 Jawaban:00d32d48502020 a 21 23524742000 2 0000000000000 0000000000000 * 0000000000 Lalas = π ×00000000000000 r 0000000000000 0000000000000 21 20000000000 00000000000000 =0000000000000 11 × 3 × 0000000000000 00000f6d60 00000000000000 2 2= 346,5 cm0000000000000 =2 0000000000000 0010000000 00000000000000 ×0000000000000 π × r × t 0000000000000 0d32d48502 00000000000000 Lselimut0000000000000 0000000000000 0200000000 00000000000000 tabung0000000000001 0000000000116 0000000000 00000000001163 22 21 = 2 × 3707274000001 × × 15 21 0000000000 1637072740000 72 70727400000150 ffd8ffe000104a46494600 2 0150000000336 00000033646573 = 22 ×5000000033646 3 01020100c800c80000ffe20 × 15 = 0000000000 4657363000001 5736300000184 0000000000 63000001840000 990 cm c584943435f50524f46494 Lselimut840000006c777 =π× r 0000006c77747 0000000000 kerucut 006c7774707400 c4500010100000c484c696 074000001f000 0000000000 × s 2247074000001f0 0001f0000000146 21 =× ×000014626b707 15e6f021000006d6e7472524 00000014626b7 0000000000 72 26b70740000020 7422058595a2007ce00020 2 0740000020400 4000002040000 40000001472585 = 11 × 3 009000600310000616373 × 15 = 0000000000 0000147258595 00147258595a0 0000000116 95a00000218000 495 cm 704d53465400000000494 Lbangu a000002180000 = 346,5 + 0000218000000 990 + 495 3707274000 000146758595a0 2543207352474200000000 = 1.831,5 cm n 00146758595a0 146758595a000 0015000000 000022c0000001 00000000000000000000f6 000022c000000 0022c00000014 0336465736 46258595a00000 37. Jawaban: c Jumlah data = 2+8+8 d6000100000000d32d485 6258595a00000 3000001840 24000000014646 + 5 + 4 +146258595a000 3 = 30 Median terletak di antara 020200000000000000000 0024000000014 2400000001464 000006c777 d6e64000002540 data ke-15 dan data ke 000000000000000000000 6d6e640000025 4707400000 0000070646d646 000000000000000000000 16. 646d6e6400000 2540000007064 400000070646d 1f00000001 4000002c400000 000000000000000000000 x15+ x 16 6d6464000002c 6464000002c40 4626b70740 Median = 08876756564000 000000000000000116370 2 4000000887675 0000088767565 0000204000 0034c000000867 727400000150000000336 77+ 65640000034c0 640000034c000 0001472585 = 6696577000003d 465736300000184000000 ST= 2 = 7 0000086766965 Jadi, median data di atas 0008676696577 95a0000021 4000000246c756 6c77747074000001f00000 adalah 7. 77000003d4000 000003d400000 8000000146 = d69000003f80000 0014626b7074000002040 000246c756d69 0246c756d6900 758595a000 00146d65617300 00000147258595a000002 000003f800000 0003f80000001 0022c00000 38. Jawaban: b Misal: n = 00040c00000024 18000000146758595a000 1 0146d65617300 46d6561730000 0146258595 74656368000004 banyak anak mula-mula 0022c000000146258595a = 20 00040c0000002 040c000000247 a000002400 300000000c7254 0000024000000014646d6 4746563680000 4656368000004 0000014646 52430000043c00 e64000002540000007064 x1 = rata-rata nilai mula-mula 04300000000c7 300000000c725 d6e6400000 00080c67545243 6d6464000002c40000008 n = banyak anak yang =2 6,5 2545243000004 452430000043c 2540000007 0000043c000008 8767565640000034c0000 x2 = rata-rata nilai anak-anak yang ditambahkan 3c0000080c675 0000080c67545 0646d64640 0c625452430000 33. Jawaban: d T ABCD berbentuk 008676696577000003d40 = 7 Rumus rata-rata ditambahkan 452430000043c 2430000043c00 00002c4000 043c0000080c74 persegi 00000246c756d69000003f gabungan: 0000080c62545 00080c6254524 0008876756 65787400000000 8000000146d6561730000 2430000043c00 30000043c0000 5640000034 nx + nx 436f70797269676 22 040c00000024746563680 xga =1 11 00080c7465787 080c746578740 c000000867 + 87420286329203 nn 20× 6,5 + n × 7 AE = 10 cm 200004300000000c725452 b = AC 12 2 ∆AET siku-siku 0000000436f70 400000000436f 6696577000 13939382048657 430000043c0000080c6754 ⇒ 6,67079726967687 = di E, 7972696768742 003d400000 20 + n 76c6574742d506 2 2 52430000043c0000080c62 4202863292031 0286329203139 1636b617264204 6,6(20 + n2) = 20 × 5452430000043c0000080c 6,5 + n20246c756d6 ×7 3939382048657 3938204865776 9000003f80 36f6d70616e7900 7465787400000000436f70 132 +76c6574742d50 6,6n2 = 130 + 7n2 c6574742d5061 00000146d6 00646573630000 797269676874202863292 A 61636b6172642 636b617264204 5617300000 00000000001273 B 031393938204865776c65 0436f6d70616e 36f6d70616e79 40c0000002 52474220494543 74742d5061636b6172642 7900006465736 0000646573630 4746563680 36313936362d32 0436f6d70616e790000646 3000000000000 0000000000000 0000430000 2e310000000000 573630000000000000012 0012735247422 1273524742204 0000c72545 00000000000012 735247422049454336313 0494543363139 9454336313936 2430000043 73524742204945 936362d322e3100000000 36362d322e310 362d322e31000 c0000080c6 4336313936362d 000000000000001273524 0000000000000 0000000000000 7545243000 322e3100000000 742204945433631393636 0000000012735 0000001273524 0043c00000 00000000000000 2d322e310000000000000 2474220494543 7422049454336 80c6254524 00000000000000 000000000000000000000 36313936362d3 313936362d322 30000043c0 00000000000000 00000000000000000000A 2 2 22e3100000000 e310000000000 000080c746 0000TU −SU C = BD = 20 cm

ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c696e6f021000006d6e74725247422058595a2007ce000200090006003 10000616373704d5346540000000049454320735247420000000000000000000000000000f6d6000100000000d32d4850202000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000001163707274000001500000003364657363000001840000006c77747074000001f000000014626b70740 0000204000000147258595a00000218000000146758595a0000022c000000146258595a0000024000000014646d6e640000025400000070646d6464000002c4000000887675 65640000034c0000008676696577000003d4000000246c756d69000003f8000000146d6561730000040c0000002474656368000004300000000c725452430000043c0000080c 675452430000043c0000080c625452430000043c0000080c7465787400000000436f70797269676874202863292031393938204865776c6574742d5061636b61726420436f6d 70616e790000646573630000000000000012735247422049454336313936362d322e31000000000000000000000012735247422049454336313936362d322e3100000000000 0000000000000000000000000000000000000000000

0

6,6n2 – 7n2 = 130 – 132 ⇔ – 0,4n2 = = –2 n2 =

−2

=5

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