06 Kunci Jawaban dan Pembahasan MAT IX

3. d = 12 cm ⇒ r = 6 cm
t tabung = 16 cm
t kerucut = 8 cm
V pasak = V tabung + V kerucut
= πr 2 t tabung + 1
3 πr2 t kerucut
= 3,14 × 62 × 16 + 1
3 × 3,14 × 62 × 8
= 1.808,64 + 301,44
= 2.110,08 cm3 4. d : t = 4 : 7 ⇒ d = 4
7 t
Luas selimut tabung = 2πrt = πdt
⇔ 448π = π × 4
t × t
5.
⇔
4
7 t2 = 448π
π
⇔ t2 = 448 × 7
= 784
4
⇔ t= 784 = 28
d = 4
× 28 = 16
7
r = 1 1
d = × 16 = 8
2 2
Luas permukaan tabung
= 2πr(r + t)
= 2π × 8(8 + 28) cm2 = 576π cm2 t = 80 cm
7
d = 56 cm
Luas permukaan tugu yang terkena cat hijau
= 2πrt + πr 2
= (2 × 3,14 × 28 × (80 – 12)) + (3,14 × 28 2 )
= 11.957,12 + 2.461,76
= 14.418,88 cm 2
A. Pilihan Ganda
1. Jawaban: d
V1 : V2 = πr 2
1 t: πr2 2t = r 2
1 : r2
2
= 102 : 152 = 100 : 225
= 4 : 9
12 cm
bagian yang dicat
2. Jawaban: a
V1 : V2 = r 2
1 : r2
2 1
= (
2 r2 )2 : r 2
2
= 1
4 r2 2 : r 2 1
2 = : 1 = 1 : 4
4
Jadi, perbandingan volume kerucut pertama dan
kerucut kedua 1 : 4.
3. Jawaban: a
Volume air yang tumpah = volume 3 kelereng
3Vk = 12 4
7
⇔ 3 × ( 4
3 πr3 )= 12 4
7
⇔ 4 × 22
7 × r3 = 88
7
⇔ r3 = 1
⇔ r= 1
4. Jawaban: b
r1 = 14 cm, t1 = 6 cm
r2 = 1
2 r1 = 7 cm
Perubahan volume = 1
3 πt(r1 2 – r 2
2 )
= 1 22
×
3 7 × 6(142 – 72 )
= 924 cm3 5. Jawaban: c
r1 = r2 = 5 cm
t1 = 6 cm
t2 = 4 cm
Perbandingan volume:
V1 : V2 = πr 2
1
t1 : πr 2
2
t2
= t1 : t2 = 6 : 4 = 3 : 2
6. Jawaban: a
Selisih volume = 244,92 cm3 r1 = 8 cm
t = 6 cm
Selisih volume = 1
3 πt(r1 2 – r 2
2 )
⇔ 244,92 = 1
3 × 3,14 × 6(82 – r 2
2 )
⇔ 244,92 = 6,28(64 – r 2
2 )
⇔ 39 = 64 – r 2
2
⇔ r 2
2 = 25
⇔ r2 = 5 cm
7. Jawaban: c
d1 = 8 dm ⇒ r1 = 4 dm
V1 : V2 = 8 : 1
⇔ 4
3 πr1 3 : 4
3 πr2 3 = 8 : 1
⇔ r 3
1 : r2
3 = 8 : 1
⇔ 43 : r 3
2 = 8 : 1 ⇔ r2
3 1
=
8 × 43 ⇔ r2 = 2
Jadi, jari-jari bola setelah diubah adalah 2 dm.
Kunci Jawaban dan Pembahasan PR Matematika Kelas IX 21