# Ex.13.7 Q4 Surface Areas and Volumes Solution - NCERT Maths Class 9

Go back to  'Ex.13.7'

## Question

If the volume of a right circular cone of height $$9\rm\, cm$$ is \begin{align}48\pi \,\,\rm\,c{m^3} \end{align}, find the diameter of its base.

## Text Solution

Reasoning:

Volume of the cone is \begin{align}\frac{1}{3}\end{align} time of the volume of a cylinder having radius $$r$$ and height $$h$$ \begin{align}\frac{1}{3}\pi {r^2}h \end{align}

What is known?

Volume and height of the cone.

What is unknown?

Base diameter.

Steps:

Height $$(h) = 9 \rm\,cm$$

\begin{align} \text{Volume of cone} &= \frac{1}{3}\pi {r^2}h \\48\pi &= \frac{1}{3}\pi {r^2}h \\48\pi &= \frac{1}{3}\pi \times {r^2} \times 9 \\{r^2} &= \frac{{48 \times 3}}{{3 \times 3}} \\&= 16\\r &= \sqrt {16} \\&= 4\,\,\rm\,cm \end{align}

Base diameter \begin{align} = 2\,r = 2 \times 4 = 8\,\,\rm\,cm \end{align}

The diameter of the box of the night circular cone is $$8\rm\, cm.$$