c. V = πr²t
t = V : (πr²)
t = 224π : (π x 4²)
t = 224π : (16π)
t = 14 m
d. Lp = 2πr(r + t)
(r + t)r = Lp : 2π
(r + 13)r = Lp : 2π
13r + r² = 528π : 2π
13r + r² = 264
r² + 13r - 264 = 0
(r + 24)(r - 11) =
r = -24 atau r = 11 maka r = 11
e. Lp = 2πr(r + t)
(r + t)r = Lp : 2π
(r + 15)r = 450π : 2π
15r + r² = 450π : 2π
15r + r² = 225
r² + 15r - 225 = 0
r1 r2 = (-b ± √(b² - 4ac))/2a
r1 r2 = (-15 ± √((-15)² - 4x1x(-225))/2x1
r1 r2 = (-15 ± √(225 + 4x225))/2
r1 r2 = (-15 ± √(225 x5))/2
r1 r2 = (-15 ± 15√5))/2
r1 r2 = -15/2 + (15√5)/2
r = -15/2 + (15√5)/2
f. V = πr²t
r² = V : (πt)
r = √(V : (πt))
r = √(294π : (π6))
r = √49 = 7 m
3. Berpikir Kritis. Terdapat suatu tabung dengan jari-jari r cm dan tinggi tabung t cm, dimana r < t. Misalkan tabung tersebut memiliki volume V cm³ dan luas permukaan L cm1/r + 1/t. Apakah mungkin V = L?
Jika ya, tentukan nilai 1/r + 1/t
r = r
t = t
r < t
V = V
Lp = L
V = πr²t
Lp = 2πr(r + t)