Question

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.

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=1/2 AP

since QD || PC, therefore, triangle AQD is similar to the triangle APC.

by the condition of similarity

$\frac{QD}{PC}=\frac{AQ}{AP}=\frac{AQ}{2AQ}\frac{QD}{R}=\frac{1}{2}QD=\frac{R}{2}$

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

$=\frac{1}{3}\pi R^{2}H-\frac{1}{3}\pi \left(\frac{R}{2}{)}^2\right(\frac{H}{2})=\frac{1}{3}\pi R^{2}H-\frac{1}{3}\pi \frac{R^{2}H}{8}=\frac{1}{3}\pi R^{2}H(1-\frac{1}{8})=\frac{1}{3}\pi R^{2}H\times \frac{7}{8}$

volume of the cone AED=$\frac{1}{8}\times \frac{1}{3}\pi R^{2}H$

therefore$\frac{\text{Volume of part taken out}}{\text{Volume of remaining part of the cone}}=\frac{\frac{1}{8}\times \frac{1}{3}\pi R^{2}H}{\frac{7}{8}\times \frac{1}{3}\pi R^{2}H}=\frac{1}{7}$

Answer $1:7$