A cone of radius 8 cm and height 12 cm


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 two parts.


Let ORN be the cone then given, radius of the base of the cone r1 = 8cm

and height of the cone, $(h) O M=12 \mathrm{~cm}$ Let $P$ be the mid-point of $O M$, then

$O P=P M=\frac{12}{2}=6 \mathrm{~cm}$

Now, $\triangle O P D \sim \triangle O M N$

$\therefore \quad \frac{O P}{O M}=\frac{P D}{M N}$

$\Rightarrow \quad \frac{6}{12}=\frac{P D}{8} \Rightarrow \frac{1}{2}=\frac{P D}{8}$

$\Rightarrow \quad P D=4 \mathrm{~cm}$

The plane along CD divides the cone into two parts, namely

(i) a smaller cone of radius 4 cm and height 6cm and (ii) frustum of a cone for which

Radius of the top of the frustum, r1 = 4 cm

Radius of the bottom,r2 = 8 cm

and             height of the frustum, h = 6 cm

$\therefore \quad$ Volume of smaller cone $=\left(\frac{1}{3} \pi \times 4 \times 4 \times 6\right)=32 \pi \mathrm{cm}^{3}$

and volume of the frustum of cone $=\frac{1}{3} \times \pi \times 6\left[(8)^{2}+(4)^{2}+8 \times 4\right]$

$=2 \pi(64+16+32)=224 \pi \mathrm{cm}^{3}$

∴  Required ratio = Volume of frustum : Volume of cone = 24 π: 32 π = 1:7



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