Question

A cone of radius 10 cm divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volume of the two parts.

Answer 1

Let the height of the cone be H and the radius be R.
This cone is divided into two parts through the mid-point of its axis. Therefore $AQ=\frac{1}{2}AP$ since $QD\parallel PC$
Therefore triangle AQD is similar to the triangle APC.
By the condition of similarity.
$\frac{QD}{PC}=\frac{AQ}{AP}=\frac{AQ}{2.AQ}\frac{QD}{R}=\frac{1}{2}\Rightarrow QD=\frac{R}{2}$
volume of the cone ABC $=\frac{1}{3}\pi R^{2}H$
volume of the frustum = volume of the cone ABC - volume of the cone AED
$=\frac{1}{3}\pi R^{2}H-\frac{1}{3}\pi {\left(\frac{R}{2}\right)}^2\left(\frac{H}{2}\right)=\frac{1}{3}\pi R^{2}H-\frac{1}{3}\pi \frac{R^{2}H}{8}=\frac{1}{3}\pi R^{2}H(1-\frac{1}{8})=\frac{1}{3}\pi R^{2}H\times \frac{7}{8}$
volume of the cone AED $=\frac{1}{8}\times \frac{1}{3}\pi R^{2}H$
therefore
$\frac{volumeoftheparttakenout}{volumeoftheremovingpartofthecone}=\frac{\frac{1}{8}\times \frac{1}{3}\pi R^{2}H}{\frac{7}{8}\times \frac{1}{3}\pi R^{2}H}=\frac{1}{7}$