Question

The height of a cone is 10 cm. The cone is divided into two parts usign a plane parallel to its base at the middle of its height. Find the ration of the volumes of two parts.

Answer 1

As the cone is divided into two equal parts by the axis so, AQ = $\frac{AP}{2}$
with the help of similarity theory,

$\frac{QD}{PC}=\frac{AQ}{AP}$

$So,\frac{QD}{PC}=\frac{1}{2}$

So, radius of QD = $\frac{PC}{2}=\frac{R}{2}$

now,the volume of frustum$=\frac{1}{3}\pi R^{2}H-\frac{1}{3}\pi \left(\frac{R}{2}{)}^2\right(\frac{H}{2})$
$=\frac{1}{3}\pi R^{2}H-\frac{1}{3}\pi \frac{R^{2}H}{8}$
$=\frac{1}{3}\pi R^{2}H(1-\frac{1}{8})$
$=\frac{1}{3}\pi R^{2}H\times \frac{7}{8}$
compare the two parts in cone
$\frac{Volumeofthesmallcone}{Volumeoffrustum}=\frac{\frac{1}{3}\pi \left(\frac{R}{2}{)}^2\right(\frac{H}{2})}{\frac{1}{3}\pi R^{2}H\times \frac{7}{8}}=\frac{1}{7}$