Question

Question 4
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

Answer 1

Let ORN be the cone then given, radius of the base of the cone $=r_1=8cm$


And height of the cone, (h) OM = 12 cm
Let P be the mid-point of OM, then
$OP=PM=\frac{12}{2}=6cmNow,\Delta OPD\sim \Delta OMN\therefore \frac{OP}{OM}=\frac{PD}{MN}\Rightarrow \frac{6}{12}=\frac{PD}{8}\Rightarrow \frac{1}{2}=\frac{PD}{8}\Rightarrow PD=4cm$

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 $r_1=4cm$
Radius of the bottom $r_2=8cm$
Height of the frustum h = 6cm

Volume of smaller cone
$=(\frac{1}{3}\pi \times 4\times 4\times 6)$
$=32\pi c{m}^3$
$[\because Volumeofthecone=\frac{1}{3}\pi \times r^{2}h]$

Volume of the frustum of cone
$=\frac{1}{3}\pi \times 6\left[\right(8{)}^2+(4{)}^2+8\times 4]=224\pi$
$[Volumeofthefrustumofcone=\frac{1}{3}\pi \times h(r_1^2+r_2^2+r_1{r}_2\left)\right]$

$\therefore$ Required ratio
= Volume of frustum : Volume of cone
$=224\pi :32\pi =7:1$