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a. Soal a
96π / 12 = 12π/t
96πt = 144π
t = 144/96 = 1,5cm
L = 2πr(r + t)
12π = 2πr(r + 1,5)
12 = 2r(r + 1,5)
12 = 2r² + 3r
2r² + 3r - 12 = 0
r₁,₂ = -b ± (√(b²-4ac))/(2a)
r₁,₂ = -3 ± ((√3²-4(2)(-12))/(2(2))
r₁,₂ = -3 ± ((√9+96)/(4)) = -3 ± (√105)/4
r₁ = -3 - (√105)/(4) = -3,3125 cm
r₂ = -3 + (√105)/(4) = 1,8125 cm
Karena ukuran dari bangun ruang selalu positif, maka r = 1,8125 cm
Baca Juga: Kunci Jawaban IPS Kelas 9 Halaman 192 Aktivitas Kelompok Teka Teki Silang: Perdagangan Bebas 10 Soal Mendatar
b. Soal b
v = ⅓ πr²t
12π = ⅓ π(4)²t
12 = ⅓(16t)
12(3) = 16t
36 = 16t
t = 36/16 =2,25 m
12π/2,25 = 324π/t
12πt = 729π
12t = 729
t = 729/12
t = 60,75 m
v = ⅓ πr²t
324π = ⅓ πr²(60,75)
324 = 20,25r²
r² = 324/20,25 = 16
r = √16 = 4 m
s² = t² + r²
s² = (60,75)² + (4)²
s² = 3690,5625 + 16
s² = 3706,5625
s² = √3706,5625 = 60,88 m
