Diketahui sebuah kerucut memiliki luas alas 64π dan luas selimut 136π. berapakah tinggi kerucut tersebut ?
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64phi = phi . r²
r= √64 = 8cm
Ls = phi . r (r + s)
136phi = phi . ( r² + rs)
136 = 8² + 8s
136 = 64 + 8s
8s = 136 - 64 = 72cm
s = 72 : 8 = 9
t = √ s² - r²
t = √ 9² - 8²
t = √ 81 - 64 = √ 17 cm