# If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^{3}$, find the diameter of its base.

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Given:

The volume of a right circular cone of height $9\ cm$ is $48 \pi\ cm^3$.

To do:

We have to find the diameter of its base.

Solution:

Height of the cone $h = 9\ cm$

Volume of the right circular cone $=48 \pi cm^{3}$

This implies,

$\frac{1}{3} \pi r^{2} h =48 \pi$

$\frac{1}{3} \pi r^{2} \times 9 = 48 \pi$

$r^{2} = \frac{48}{3}$

$r^2= 16$

$r=\sqrt{16}$

$r= 4\ cm$

This implies,

Diameter of the base $=2r$

$=2\times4$

$=8\ cm$

Hence, the diameter of the right circular cone is $8\ cm$.

Updated on 10-Oct-2022 13:46:38