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Question
A cone of base radius 4 cm is divided into two parts by drawing a plane through the midpoint of its height and parallel to its. Compare the volume of the two parts.
Sum
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Solution
ΔABC ∼ ΔAPQ
∴ `h/(2h) = r_1/4`
⇒ r1 = 2 cm

Volume of smaller cone = `1/3 πr_1^2h`
= `1/3 π(2)^2h`
= `1/3 xx 4πh`
Volume of frustum = `1/3 πh(r_1^2 + r_2^2 + r_1r_2)`
= `1/3 π xx h(2^2 + 4^2 + 2 xx 4)`
= `1/3 πh xx 28`
∴ Required ratio = `(1/3 xx 4πh)/(1/3 xx 28 πh)`
= `4/28`
= `1/7`
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