Luas permukaan sebuat kerucut adalah 282,86cm². Jika jari jari 5cm dam π = [tex] \frac{22}{7} [/tex], maka tentukan tinggi kerucut tersebut ! poin seimbang deng

Lp = 282,86 cm²
r = 5 cm

Lp = πr(r + s)
282,86 = [tex] \frac{22}7 (5) [/tex] (5 + s)
282,26 = [tex] \frac{110}{7} [/tex] (5 + s)
282,26 x [tex] \frac{7}{110} [/tex] = 5 + s
s ≈ 13 cm

Maka
t = [tex] \sqrt{13^2 - 5^2} [/tex]
t = [tex] \sqrt{144} [/tex]
t = 12 cm

Lp = Lalas + Lselimut
Lp = πrs + πr²
Lp = πr(s + r)
282,86 = (3,14)(5)(s + 5)
18 = s + 5
s = 13 cm

Tinggi kerucut
[tex] = \sqrt{s^2 - r^2} \\ = \sqrt{(13 cm)^2 - (5 cm)^2} \\ = \sqrt{169 cm^2 - 25 cm^2} \\ = \sqrt{144 cm^2} \\ = 12 cm [/tex]