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 distance of the boundary of the upper surface from the centre of the base of the frustum is $\sqrt{h^2+\frac{r^2}{9}}$. Show that the volume of the frustum is $\frac{13}{27}\pi r^{2}h$

Answer 1

We have
$OA=r,O{O}^{\prime }=h,O{B}^{\prime }\sqrt{h^2+\frac{r^2}{9}}$

In $△O{O}^{\prime }B$ we have

${O{B}^{\prime }}^2={O{O}^{\prime }}^2+{O^{\prime }B}^2$

$\Rightarrow h^2+\frac{r^2}{9}=h^2+{O^{\prime }{B}^{\prime }}^2$

$\Rightarrow O^{\prime }{B}^{\prime }=\frac{r}{3}$

Volume of the frustum $=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h$

volume of the frustum $=\frac{1}{3}\pi \{r^2+{\left(\frac{r}{3}\right)}^2+r\times \frac{r}{3}\}h$

$=\frac{1}{3}\pi \{r^2+\frac{r^2}{9}+\frac{r^2}{3}\}h$

$=\frac{13}{27}\pi r^{2}h$ $\left[Henceproved\right]$