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.

The slant height of the frustum is $\sqrt{h^2+\frac{4}{9}{r}^2}$ . Show that the volume of the frustum is $\frac{13}{27}\pi r^{2}h$

Answer 1

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

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

=$\frac{\left(13\pi r^{2}h\right)}{27}$.