Gambar 1 => no 1 samapi 5 gambar 2 => no 12,13,14,15 buat cara penyelesaiannya ! yang cuman asal2 doang akn saya laporkn ! kalo bertanya di comentar bukan di jwbnnya .
irsa214
1. tinggi drum = 1 m = 10 dm jari2 drum = 25 cm = 2.5 dm phi = 3.14 Vdrum tabung = phi * r * r * t = 3.14 * 2.5 * 2.5 * 10 = 196.25 dibagikan kepada 90 orang = 196.25 : 90 = 2,2 liter
tinggi drum = 1 m = 10 dm
jari2 drum = 25 cm = 2.5 dm
phi = 3.14
Vdrum tabung = phi * r * r * t = 3.14 * 2.5 * 2.5 * 10 = 196.25
dibagikan kepada 90 orang = 196.25 : 90 = 2,2 liter
2.
V tabung tersisa = Vt - Vk
(phi * r * r * t) - (1/3 * phi * r * r * t)
(22/7 * 3.5 * 3.5 * 14) - (1/3 * 22/7 * 3.5 * 3.5 * 6)
(11 * 3.5 * 14) - (22 * 3.5 * 3.5)
539 - 269.5 = 269.5
3.
V. ruang yang tersisa = V.tabung - (5 * V.bola)
= (phi * r² * t) - (5 * 4/3 * phi * r³)
= (3.14 * 1 * 1 * 10) - ( 5 * 4/3 * 3.14 * 1 * 1 * 1)
= 31.4 - 6.67
= 24.73
yang no 4 sama 5 maaf aku gk bisa
gambar 2
12.
luas selimut kerucut = phi * r * s
= 3.14 * 5 * 13
= 204.1 (a)
cara mencari s = √t² + r²
= √12² + 5²
= √144 +25
=√169
= 13
13.
luas kertas = 27500 cm²
tinggi topi = 24 cm
diameter = 14
jari2 = 7
s = √t² + r²
= √24² + 7²
= √576 + 49
= √625
= 25
luas selimut kerucut = phi * r * s
= 22/7 * 7 * 25
= 22 * 25
= 550
27500 : 550 = 50 buah (b)
15.
luas selimut kerucut = phi * r * s
= 22/7 * 7 * 25
= 22 * 25
= 550
yang no 14 maaf gk bisa bantu