Question

The radius of the base of a right circular cone is r. It is cut by a plane parallel to the base at a height h from the base. What is the volume of the frustum obtained if it's slant height is $\sqrt{h^2+\frac{4}{9}{r}^2}$.

• A. $\frac{27}{13}\pi r^{2}h$
• B. $\frac{1}{3}\pi r^{2}h$
• C. $\frac{13}{27}\pi r^{2}h$
• D. $\frac{13}{26}\pi r^{2}h$

Answer 1

The correct option is C $\frac{13}{27}\pi r^{2}h$

Given r is the radius of the base.

The frustrum is having height "h"

Let the radius of the upper circle of frustrum be r'

Slant height of the frustum = $\sqrt{\left(h^2\right)+(r-r^{\prime }{)}^2}=\sqrt{(h^2+\frac{4r^2}{9})}$

$(r-r^{\prime }{)}^2=\frac{4r^2}{9}$

$(r-r^{\prime })=\frac{2r}{3}$

$r^{\prime }=\frac{r}{3}$

Volume of frustum of cone
$=\frac{\pi }{3}.h(r^2+r^{\prime 2}+r.r^{\prime })=\frac{\pi }{3}.h(r^2+(\frac{r}{3}{)}^2+r.\frac{r}{3})$

$=\frac{\pi }{3}h[r^2+\frac{r^2}{9}+\frac{r^2}{3}]$

$=\frac{13\pi r^{2}h}{27}$