06 Kunci Jawaban dan Pembahasan MAT IX
06 Kunci Jawaban dan Pembahasan MAT IX
06 Kunci Jawaban dan Pembahasan MAT IX
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s=<br />
=<br />
2 2<br />
t r<br />
+<br />
2 2<br />
48 14<br />
+<br />
= 2.304 + 196<br />
= 2.500<br />
= 50 cm<br />
Lselimut = πrs<br />
= 22<br />
× 14 × 50<br />
7<br />
= 44 × 50<br />
= 2.200 cm2 Jadi, luas tumpeng yang akan dihias makanan<br />
2.200 cm2 .<br />
9. <strong>Jawaban</strong>: d<br />
V = 120 cm3 t = 10 cm<br />
V= 1<br />
3 πr2t ⇔ 120 = 1<br />
3 πr2 × 10<br />
⇔ r 2 = 36<br />
π<br />
⇔ r= 36<br />
π<br />
10. <strong>Jawaban</strong>: c<br />
V kerucut = 1<br />
3 πr2 t<br />
= 6<br />
π cm<br />
⇔ 314 = 1<br />
3 × 3,14 × 52 × t<br />
⇔ t = 12 cm<br />
Panjang garis pelukis:<br />
s2 = r2 + t2 = 52 + 122 = 169<br />
⇔ s = 169 = 13 cm<br />
11. <strong>Jawaban</strong>: a<br />
L = 4πr2 = 4 × 22<br />
7 × 212 = 5.544 cm2 12. <strong>Jawaban</strong>: b<br />
d = 10 cm → r = 5 cm<br />
L= 3πr 2<br />
= 3 × 3,14 × 5 2<br />
= 235,5 cm 2<br />
13. <strong>Jawaban</strong>: c<br />
s=<br />
2 2<br />
12 5<br />
+<br />
= 144 + 25<br />
= 169<br />
= 13 cm<br />
20 <strong>Kunci</strong> <strong>Jawaban</strong> <strong>dan</strong> <strong>Pembahasan</strong> PR Matematika Kelas <strong>IX</strong><br />
14. <strong>Jawaban</strong>: a<br />
L selimut = πrs<br />
= 3,14 × 5 × 13<br />
= 204,1 cm 2<br />
L setengah bola = 1<br />
2 × 4πr2 = 2πr 2<br />
= 2 × 3,14 × 6 2 = 226,08 cm 2<br />
2 2<br />
s = 8 + 6 = 100 = 10 cm<br />
L selimut kerucut = πrs = 3,14 × 6 × 10 = 188,4 cm 2<br />
L bandul = 226,08 + 188,4 = 414,48 cm 2<br />
15. <strong>Jawaban</strong>: a<br />
d = 6 cm ⇔ r = 3 cm<br />
ttabung = 10 cm<br />
tkerucut = 4 cm<br />
Panjang garis pelukis kerucut:<br />
s=<br />
2 2<br />
r t<br />
+<br />
= 2 2<br />
3 4<br />
+<br />
= 25<br />
= 5 cm<br />
Luas seluruh permukaan benda<br />
= ( 1<br />
2 × 4πr2 ) + 2πrt + πrs<br />
= ( 1<br />
2 × 4 × 3,14 × 32 ) + (2 × 3,14 × 3 × 10)<br />
+ (3,14 × 3 × 5)<br />
= 56,52 + 188,4 + 47,1 = 292,02 cm2 B. Uraian<br />
1. Misalkan r1 = jari-jari bola putih,<br />
r2 = jari-jari bola hitam<br />
r1 = 2r2 Lbola putih = 4πr 2<br />
1<br />
Lbola hitam = 4πr 2<br />
2<br />
Lbola putih : Lbola hitam = 4πr 2<br />
1 : 4πr2 2<br />
= 4π(2r2 ) 2 : 4πr 2<br />
2<br />
= 4r 2<br />
2 : r2<br />
2<br />
= 4 : 1<br />
Jadi, luas permukaan bola putih dibanding luas<br />
permukaan bola hitam 4 : 1. Atau, luas permukaan<br />
bola hitam : luas permukaan bola putih = 1 : 4.<br />
2. Volume air = volume setengah tabung<br />
= 1<br />
2 × πr2 × t<br />
= 1<br />
2<br />
× 22<br />
7 × 62 × 42<br />
= 2.376 cm 3