bisa bantuin kawand.... no 10 B kerucut didalam bola

Kerucut DaLam boLa

Penampang bola ketika dibelah tepat ditengah ---> lingkaran

Penampang kerucut ---> segitiga (∆ABC)

AB = BC = 16 cm
sudut ABC = x

r bola = AO = OC = 10 cm
sudut AOC = 2x

AC = diameter kerucut

AC² = AC²
AB² + BC² - 2. AB . BC cos x = AO² + OC² - 2 AO OC cos 2x
16² + 16² - 2. 16.16 cos x = 10² + 10² - 2.10.10 cos 2x
200 cos 2x - 512 cos x + 312 = 0
25 cos 2x - 64 cos x + 39 = 0
25(2 cos² x - 1) - 64 cos x + 39 = 0
25 cos² x - 32 cos x + 7 = 0
(25 cos x - 7)(cos x - 1) = 0

yg memenuhi :
cos x = 7/25
sin x = 24/25

r = (AB × BC × AC) / (4 × L.ABC)
r = (AB × BC × AC) / (4 × 1/2 × AB × BC × sin x)
r = AC / (2 sin x)
AC = 10 × 2 × 24/25
AC = 19,2 cm

jari2 kerucut
= 1/2 AC
= 9,6 cm ✔