A cone of radius 8 , cm and height 12 , cm is | vedclass

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(C) Let the cone be $ORN$ with base radius $r_1 = 8 \, cm$ and height $OM = 12 \, cm$.Let $P$ be the mid-point of $OM$,then $OP = PM = \frac{12}{2} =

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$1: 7$

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(C) Let the cone be $ORN$ with base radius $r_1 = 8 \, cm$ and height $OM = 12 \, cm$.Let $P$ be the mid-point of $OM$,then $OP = PM = \frac{12}{2} = 6 \, cm$.Since the plane is parallel to the base,$	riangle OPD \sim 	riangle OMN$.By properties of similar triangles,$\frac{OP}{OM} = \frac{PD}{MN}$.$\frac{6}{12} = \frac{PD}{8} \Rightarrow \frac{1}{2} = \frac{PD}{8} \Rightarrow PD = 4 \, cm$.The plane divides the cone into a smaller cone (top) and a frustum (bottom).Volume of smaller cone $V_1 = \frac{1}{3} \pi (PD)^2 (OP) = \frac{1}{3} \pi (4)^2 (6) = 32 \pi \, cm^3$.Volume of the original cone $V = \frac{1}{3} \pi (MN)^2 (OM) = \frac{1}{3} \pi (8)^2 (12) = 256 \pi \, cm^3$.Volume of the frustum $V_2 = V - V_1 = 256 \pi - 32 \pi = 224 \pi \, cm^3$.The ratio of the volume of the smaller cone to the frustum is $V_1 : V_2 = 32 \pi : 224 \pi = 1 : 7$.

$A$ cone of radius $8 \, cm$ and height $12 \, cm$ is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of the two parts.

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