06 Kunci Jawaban dan Pembahasan MAT IX

5. Jawaban: b
d = 12 cm ⇒ r = 6 cm
t = 8 cm
Panjang garis pelukis:
s=
2 2
t + r =
2 2
8 6
+
= 100 = 10 cm
Luas permukaan kerucut
= luas selimut + luas alas
= πr(s + r)
= 3,14 × 6(10 + 6)
= 301,44 cm 2
6. Jawaban: a
r 1 : r 2 = 8 : 6
L 1 : L 2 = 4πr 1 2 : 4πr2 2
= r 1 2 : r2 2 = 8 2 : 6 2
= 64 : 36 = 16 : 9
L1 = 240 cm2 L2 = 9
× 240 = 135 cm2
16
7. Jawaban: c
Jari-jari = r = 70 cm = 7 dm
Volume = V = 2.310 liter = 2.310 dm3 V = πr2t ⇔ 2.310 = 22
7 × 72 × t
⇔ 2.310 = 154t
⇔ t = 15 dm
Lselimut = 2πrt = 2 × 22
× 7 × 15 = 660 dm2
7
Lalas = πr2 = 22
7 × 72 = 154 dm2 Lplat besi = Lselimut + Lalas = 660 + 154 = 814 dm2 8. Jawaban: c
Luas bola = 4πr2 ⇔ 1.256 = 4 × 3,14 × r2 ⇔ 1.256 = 12,56r2 ⇔ r2 = 100
⇔ r = 10 cm
Volume bola = 4
3 πr3
= 4
× 3,14 × 103
3
= 4.186,67 cm3 9. Jawaban: b
Luas sebuah parasut = 1
× 4πr2
2
= 1
× 4 × 3,14 × 22
2
= 25,12 m2 Luas plastik minimal = 15 × 25,12
= 376,8 m2 24 Kunci Jawaban dan Pembahasan PR Matematika Kelas IX
10. Jawaban: b
Volume bangun ruang
= volume tabung + volume kerucut
= πr2ttabung + 1
3 πr2tkerucut = (3,14 × 62 × 16) + ( 1
3 × 3,14 × 62 × 5)
= 1.997,04 cm3 11. Jawaban: a
V = 1
3 πr2t ⇔ 1.232 = 1 22
×
3 7 × 72 × t
⇔ t = 24 cm
Panjang garis pelukis:
s = 2 2
t + r = 2 2
7 + 24 = 625 = 25 cm
Luas selimut kerucut = πrs
12. Jawaban: c
r A : r B = 2 : 4
V A : V B = 2 3 : 4 3 = 1 : 8
= 22
7
× 7 × 25 = 550 cm2
13. Jawaban: b
Luas tabung tanpa tutup = luas alas + luas selimut
⇔ 3.454 = πr2 + 2πrt
⇔ 3.454 = (3,14 × 102 ) + (2 × 3,14 × 10 × t)
⇔ 3.454 = 314 + 62,8t
⇔ t= 3.140
= 50 cm
62,8
Volume tabung = πr2t = 3,14 × 102 × 50
= 15.700 cm3 14. Jawaban: a
Diketahui t = 9 cm, r1 = 11 cm, dan r2 = 15 cm
∆V = 1
3 πt(r2 2 – r 2
1 )
= 1
3 × 3,14 × 9(152 – 112 ) = 979,68 cm3 15. Jawaban: c
ttabung = 2r → r = 1
2 t
rbola = rtabung = 1
2 t
Vtabung = πr2t Vbola = 4
3 πr3
= 4
3 πr2 × 1
2 t
= 2
3 πr2t = 2
3 V tabung