The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find :

(i) the ratio of their volumes.

(ii) the ratio of their lateral surface areas.

(i) Let the radius and height of the bigger cone be r and h respectively.

So, smaller cone's radius = $\dfrac{r}{2}$ and height = $\dfrac{h}{2}$.

We know volume of cone = $\dfrac{1}{3}πr^2h$.

Ratio of volume of two cones = $\dfrac{\text{Vol. of smaller cone}}{\text{Vol. of bigger cone}}$

$\dfrac{\text{Vol. of smaller cone}}{\text{Vol. of bigger cone}} = \dfrac{\dfrac{1}{3}π \times \Big(\dfrac{r}{2}^2\Big) \times \dfrac{h}{2}}{\dfrac{1}{3}π \times r^2 \times h} \\[1em] = \dfrac{\dfrac{1}{3}π \times \dfrac{r^2}{4} \times \dfrac{h}{2}}{\dfrac{1}{3}π \times r^2 \times h} \\[1em] = \dfrac{\dfrac{1}{3}π \times r^2 \times h}{8 \times \dfrac{1}{3}π \times r^2 \times h} \\[1em] = \dfrac{1}{8}.$

Hence, the ratio of the volumes of cone = 1 : 8.

(ii) Let the slant height of bigger cone be l .

l = $\sqrt{r^2 + h^2}.$

Let slant height of smaller cone be l₁.

$l_1 = \sqrt{\Big(\dfrac{r}{2}\Big)^2 + \Big(\dfrac{h}{2}^2\Big)} \\[1em] = \sqrt{\dfrac{r^2}{4} + \dfrac{h^2}{4}} \\[1em] = \sqrt{\dfrac{r^2 + h^2}{4}} \\[1em] = \dfrac{1}{2}\sqrt{r^2 + h^2} \\[1em] = \dfrac{1}{2}l.$

We know that lateral surface area of cone = π × radius × slant height.

Ratio of lateral surface area of two cones

$= \dfrac{\text{Curved surface area of smaller cone}}{\text{Curved surface area of bigger cone}} \\[1em] = \dfrac{π \times \dfrac{r}{2} \times \dfrac{l}{2}}{π \times r \times l} \\[1em] = \dfrac{πrl}{4πrl} \\[1em] = \dfrac{1}{4}.$

Hence, the ratio of the lateral surface area of cones = 1 : 4.