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

Given :
Height (h) of cone = 9 cm
Also, Volume of cone = \(48 {\pi} {cm}^3\)

Let the radius of the cone be r.
But, Volume of cone = \(\frac{1}{3} {\pi} {r}^2 h\)
\(\Rightarrow \) \(48 {\pi} {cm}^3\) = \(\frac{1}{3} × {\pi} × {r}^2 × 9\) \(cm\)
\(\Rightarrow \)\({r}^2 = \frac{48 {\pi} × 3}{9 × {\pi}} {cm}^2\)
\(\Rightarrow \) \({r}^2 = {4}^2 {cm}^2\)
\(\Rightarrow \) r = 4cm
\(\therefore \) Diameter of base = 2r = 8 cm

Therefore, the diameter of the base of cone is 8 cm.