06 Kunci Jawaban dan Pembahasan MAT IX

A. Pilihan Ganda
1. Jawaban: b
L = 2πr(r + t)
⇔ 2.992 = 2 × 22
× r(r + 20)
7
⇔ 20.944 = 44r(r + 20)
⇔ 476 = r(r + 20)
⇔ r2 + 20r – 476 = 0
2
− b± b −4ac
r1· 2 =
2a
2
− 20 ±
=
20 −4⋅1 ⋅( −476)
2⋅1 = 20 400 + 1.904
− ±
2
= 20 2.304
− ±
2
= 20 48
− ±
2
− 20 + 48
r1 = = 14
2
−20 −48
r2 = = –34 (tidak memenuhi)
2
Jadi, jari-jari tabung 14 cm.
2. Jawaban: a
Diketahui: jari-jari tabung besar = r1 = 70 cm
t1 = 100 cm
r2 = 35 cm
t2 = 50 cm
V1 = πr 2
1
t1
= 22
7 × 702 × 100
= 22 × 700 × 100 = 1.540.000 cm 3
V 2 = πr 2 2 t2 = 22
7 × 352 × 50 = 192.500 cm 2
Banyak tabung kecil =
3. Jawaban: c
L= πr 2 + 2πrt
= 22
7 × 282 + 2 × 22
7
= 2.464 + 17.600
= 20.064 cm2 V
V
2
1
= 8 buah.
× 28 × 100
4. Jawaban: a
L= 2πr(r + t)
⇔ 79.200 = 2 × 22
7 × 70 × (70 + t)
⇔ 79.200 = 30.800 + 440t
⇔ 440t = 48.400
⇔ t = 110 cm
Vtabung = πr2t = 22
7 × 702 × 110
= 1.694.000 cm3 = 1.694 dm3 = 1.694 liter
5. Jawaban: b
Tabung I: d1 = 20 cm ⇔ r1 = 10 cm
t1 = 15 cm
Tabung II:d2 = 30 cm ⇔ r2 = 15 cm
t2 = 25 cm
Vtabung I = πr 2
1
t1 = 3,14 × 102 × 15 = 4.710 cm3 Vtabung I dimasukkan ke tabung II, artinya diketahui
V = 4.710 cm3 , r2 = 15 cm, dan ditanyakan tair .
V = πr 2
2
tair
⇔ 4.710 = 3,14 × 152 × tair 4.710
⇔ tair = 2
3,14 × 15
⇔ tair = 4.710
≈ 6,67 cm
706,5
Jadi, tinggi air pada tabung II 6,67 cm.
6. Jawaban: c
d = 28 cm ⇔ r = 14 cm
t = 50 cm
Vbotol = 220 ml
Vtabung = πr2t = 22
7 × 142 × 50
= 30.800 cm3 = 30.800 ml
Banyak botol = 30.800
= 140 buah
220
7. Jawaban: d
d = 10 cm ⇔ r = 5 cm
t = 12 cm
s= 2 2
t r
+
=
2 2
12 5
+
= 144 + 25
= 169
= 13 cm
Luas selimut kerucut = πrs
= 3,14 × 5 × 13
= 204,1 cm2 10 cm
8. Jawaban: b
r = d
= 14 cm
2
t = 48 cm
12 cm
Kunci Jawaban dan Pembahasan PR Matematika Kelas IX 19