06 Kunci Jawaban dan Pembahasan MAT IX

8. Jawaban: d
Selisih volume = 2.967,3 cm3 r2 = 8 cm
t = 9 cm
Selisih volume = πt(r 2
1 – r2
2 )
⇔ 2.967,3 = 3,14 × 9(r 2
1 – 82 )
⇔ 2.967,3 = 28,26(r 2
1 – 64)
⇔ 105 = r 2
1 – 64
⇔ r 2
1 = 169
⇔ r1 = 13 cm
9. Jawaban: c
d1 = 12 cm ⇒ r1 = 6 cm
d2 = 10 cm ⇒ r2 = 5 cm
t = 21 cm
Perubahan volume = π t(r 2
1 – r2
2 )
= 22
7 × 21(62 – 52 ) = 726 cm3 10. Jawaban: b
d = 6 dm = 60 cm ⇒ r1 = 30 cm
r2 = 30 – 3 = 27 cm
Volume bola yang tersisa = 4
3 π(r1 3 – r 3
2 )
11. Jawaban: d
V1 = 1.061,32 cm3 V2 = 759,88 cm3 t = 6 cm
V1 = 1
3 π r1 2 t
⇔ 1.061,32 = 1
3 × 3,14 × r1 2 × 6
⇔ 1.061,32 = 6,28r 2
1
1.061,32
⇔ r 2
1 =
6,28
⇔ r 2
1 = 169
⇔ r1 = 13
V2 = 1
3 πr2 2t ⇔ 759,88 = 1
3 × 3,14 × 6 × r2 2
⇔ 759,88 = 6,28r 2
2
= 4
3 π(303 – 27 3 )
= 9.756π cm 3
⇔ r 2
2 = 121
⇔ r1 = 11
Perbandingan jari-jarinya = r1 : r2 = 13 : 11.
12. Jawaban: d
Perubahan volume = 209,664π cm3 r1 = 6 cm
22 Kunci Jawaban dan Pembahasan PR Matematika Kelas IX
Perubahan volume = 4
3 π(r2 3 – r 3
1 )
⇔ 209,664π = 4
3 π(r2 3 – 63 )
⇔ 157,248 = r 3
2 – 216
⇔ r 3
2 = 373,248
⇔ r2 = 7,2
Tebal lapisan = r2 – r1 = 7,2 – 6 = 1,2 cm
13. Jawaban: c
V1 = πr 2
1 t
= 3,14 × 92 × 25
= 6.358,5
V2 = πr 2
2
t2
= 3,14 × 52 × ( 1
× 25)
4
= 490,625
Besar perubahan volume
= V1 – V2 = 6.358,5 – 490,625
= 5.867,9 cm3 14. Jawaban: d
V1 : V2 = 7.598,8 : 6.079,04
⇔πr2t1 : πr2t2 = 5 : 4
⇔ t1 : t2 = 5 : 4
15. Jawaban: d
V2 = 8
27 V1 ⇔ 4
3 πr 2 3 = 8
27
⇔ r 2 3 = 8
27 r 1 3
⇔ 27r 2 3 = 8r1 3
⇔ (3r 2 ) 3 = (2r 1 ) 3
× 4
3 πr 1 3
⇔ 3r2 = 2r1 ⇔
r2
r 1
2
=
3
Sehingga, r2 : r1 = 2 : 3.
B. Uraian
a. V 1 = πr 1 2 t1 = 3,14 × 2 2 × 3 = 37,68 cm 3
V 2 = πr 2 2 t2 = 3,14 × 2 2 × 6 = 75,36 cm 3
V 3 = πr 3 2 t3 = 3,14 × 4 2 × 3 = 150,72 cm 3
b. V 1 : V 2 : V 3 = r 1 2 t1 : r 2 2 t2 : r 3 2 t3
= (22 × 3) : (22 × 6) : (42 × 3)
= 1 : 2 : 4