Kunci Jawaban Matematika Kelas 9 Halaman 280, 281, 282, 283, Menghitung Luas dan Volume

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

Soal 3. Nilai 1/r + 1/t adalah sebagai berikut.

Contoh Jawaban:

r = r
t = t
r < t
V = V
Lp = L
V = πr²t
Lp = 2πr(r + t)
V = L
πr²t = 2πr(r + t)
rt = 2(r + t)
1/2 = (r + t)/rt
r/rt + t/rt = 1/2
1/t + 1/r = 1/2
Maka, ada kemungkinan nilai V = L jika 1/r + 1/t = 1/2 dan r < t

Soal 4. a. Luas Permukaan Magnet

Jawaban: