If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^{3}$,

Solution:

Height (h) of cone = 9 cm

Let the radius of the cone be r.

Volume of cone $=48 \pi \mathrm{cm}^{3}$

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

$\Rightarrow r^{2}=16 \mathrm{~cm}^{2}$

$\Rightarrow r=4 \mathrm{~cm}$

Diameter of base $=2 r=8 \mathrm{~cm}$