06 Kunci Jawaban dan Pembahasan MAT IX

24. Jawaban: d
Vb : Vk = 4
3 πrb 3 : 1
3 πrk 2t = 4r 3
b : rk
2t = 4 × 33 : 32 × 3
= 4 : 1
25. Jawaban: d
Vtabung = πr 2
t
tt
⇔ 6.280 = 3,14 × r 2
t × 20
⇔ 6.280 = 62,8 × r 2
t
⇔ r 2
6.280
t =
62,8
⇔ r 2
t = 100
⇔ rt = 10 cm
Alas kerucut berimpit dengan alas tabung sehingga
rk = rt = 10 cm
Vk = 1
3 πrk 2tk ⇔ 2.512 = 1
3 × 3,14 × 102 × t k
⇔ 7.536 = 314 × t k
⇔ tk = 7.536
314
26. Jawaban: b
t k =
2 2
25 20
−
= 625 − 400
= 24 cm
= 225
= 15 cm
Tinggi tabung = tt = 2 × tk = 30 cm
Vt = πr2 × tt = 3,14 × (20) 2 × 30
= 3,14 × 400 × 30
= 37.680 cm3 Vk = 1
3 πr2tk = 1
3 × 3,14 × 202 × 15
= 5 × 3,14 × 400
= 6.280 cm3 Volume tabung di luar kotak
= Vt – 2 × Vk = 37.680 – 2 × 6.280
= 37.680 – 12.560 = 25.120 cm3 27. Jawaban: a
Diketahui jari-jari setengah bola = jari-jari tabung
= 30 cm
Tinggi tabung = 50 cm.
26 Kunci Jawaban dan Pembahasan PR Matematika Kelas IX
Volume benda
= volume tabung – volume setengah bola
= πr 2 t b – 1
2
× 4
3 πr3
= 3,14 × 302 × 50 – 2
× 3,14 × 303
3
= 3,14 × 302 × (50 – 2
× 30)
3
= 2.826 × 30
= 84.780 cm3 28. Jawaban: d
Volume tabung = Vt yaitu:
Vt = πr 2
1
t1
= 3,14 × 202 × 24,75
= 3,14 × 400 × 24,75
= 31.086 cm3 Volume kerucut = Vk , sehingga tinggi kerucut:
Vk = 1
3 πr2 2t2 ⇔ 31.086 = 1
3 × 3,14 × 302 × t 2
⇔ 31.086 = 1
3 × 3,14 × 900 × t 2
⇔ 31.086 = 314 × 3 × t 2
⇔ t2 = 31.086
= 33 cm
942
29. Jawaban: a
Misalkan luas permukaan belahan kerucut = L
Panjang garis pelukis yaitu:
s =
2 2
r + t =
2 2
9 + 12 = 15 cm
L = luas setengah lingkaran alas + luas setengah
selimut + luas segitiga sama kaki
= 1
2 × πr2 + 1
2
× πrs + 1
2
× 2r × t
= 1
r × (πr + πs + 2t)
2
= 1
× 9 × (3,14 × 9 + 3,14 × 15 + 24)
2
= 4,5 × 99,36
= 447,12 cm2 30. Jawaban: c
Perhatikan gambar potongan bola berikut.
L= 1
1
× luas permukaan bola + 3 × luas
8 4 lingkaran
= 1
8 × (4πr2 ) + 3 × 1
× πr2
4
= 1
2 πr2 + 3
4 πr2
= 5
4 πr2
= 5
4 × 3,14 × 102 = 392,5 cm 2