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

## 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 \(\begin{align} = 2\,r = 2 \times 4 = 8\,\,\rm\,cm \end{align}\)

**Answer:**

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