Question

The height of a cone is $10cm$. The cone is divided into two parts by drawing a plane through the midpoint of the axis of the cone, parallel to the base. Compare the volume of the two parts.

• A. $1:7$
• B. $2:9$
• C. $3:11$
• D. $3:5$

Answer 1

The correct option is A $1:7$

Let the height of the cone be $H$ and the radius be $R$.
This cone is divided into two parts through the midpoint of its axis. therefore $AQ=\frac{1}{2}AP$
Since $QD\parallel PC$, therefore $△AQD\sim △APC$ (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 $AED$
$=\frac{1}{3}\pi r^{2}h-\frac{1}{3}\pi {\left(\frac{r}{2}\right)}^2\frac{h}{2}$

$=\frac{1}{3}\pi r^{2}h(1-\frac{1}{8})$

$=\frac{1}{3}\pi r^{2}h\times \frac{7}{8}$
Volume of the cone AED $=\frac{1}{3}\pi r^{2}h\times \frac{1}{8}$
$\frac{\text{Volume of part taken out}}{\text{Volume of remaining part of the cone}}=\frac{1/3\pi r^{r}h\times 1/8}{1/3\pi r^{r}h\times 7/8}=\frac{1}{7}$