Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. Find the radius of the base of each smaller cone, if height of each is 108 cm.

Let the radius of the smaller cone be r cm.

Given,

Height of smaller cone (h) = 108 cm

Diameter of bigger cone = 40 cm

So, radius (R) = $\dfrac{40}{2}$ = 20 cm

Height of bigger cone (H) = 9 cm.

According to question,

Volume of big cone = 3 × Volume of each smaller cone.

$\Rightarrow \dfrac{1}{3}πR^2H = 3 \times \dfrac{1}{3}πr^2h \\[1em] \Rightarrow R^2H = 3 \times r^2h \\[1em] \Rightarrow (20)^2 \times 9 = 3 \times r^2 \times 108 \\[1em] \Rightarrow r^2 = \dfrac{20^2 \times 9}{3 \times 108} \\[1em] \Rightarrow r^2 = \dfrac{3600}{324} \\[1em] \Rightarrow r^2 = \dfrac{100}{9} \\[1em] \Rightarrow r = \sqrt{\dfrac{100}{9}} \\[1em] \Rightarrow r = \dfrac{10}{3} = 3\dfrac{1}{3}\text{ cm}.$

Hence, radius of base of each cone = $3\dfrac{1}{3}$ cm.