If the areas of circular bases of a frustum of a cone are 16 ^cm2 and 36 ^cm2 respectively and the height of the frustum is 15 cm, what is the volume of the frustum?

$$\begin{gathered} Area of bas{e}_1=16 c{m}^2 \\ Area of bas{e}_2= 36 c{m}^2 \\ height of the frustum = 15 cm \end{gathered}$$

$$\begin{gathered} volume V of a frustum of right circular cone is \\ V = \frac{1}{3}\times \mathrm{\pi }\times \mathrm{h}\times ({{\mathrm{r}}_1}^2+{{\mathrm{r}}_2}^2+{\mathrm{r}}_1{\mathrm{r}}_2) \\ = \frac{\mathrm{h}}{3}\times ({{\mathrm{\pi r}}_1}^2+{{\mathrm{\pi r}}_2}^2+{\mathrm{\pi r}}_1{\mathrm{r}}_2) \\ =\frac{\mathrm{h}}{3}\times ({{\mathrm{\pi r}}_1}^2+{{\mathrm{\pi r}}_2}^2+\sqrt{{{\mathrm{\pi r}}_1}^2}\times \sqrt{{{\mathrm{\pi r}}_2}^2}) \\ =\frac{\mathrm{h}}{3}\times (\mathrm{area} \mathrm{of} {\mathrm{base}}_1+\mathrm{area} \mathrm{of} {\mathrm{base}}_2+\sqrt{\mathrm{area} \mathrm{of} {\mathrm{base}}_1}\times \sqrt{\mathrm{area} \mathrm{of} {\mathrm{base}}_2}) \\ = \frac{15 \mathrm{cm}}{3}\times (16 {\mathrm{cm}}^2+36 {\mathrm{cm}}^2+\sqrt{16 {\mathrm{cm}}^2}\times \sqrt{36 {\mathrm{cm}}^2}) \\ = 5 \mathrm{cm} \times 76 {\mathrm{cm}}^2 =380 {\mathrm{cm}}^3 \left(\mathrm{Ans}\right) \end{gathered}$$



 Alternate procedure  :-
One can also first find out the values of $r_{1 }and r_2$ from the values of areas of circular bases with the help of formula given below and then putting that values in the formulae of volume      
       $r_1=\sqrt{\frac{(Area of bas{e}_1)}{\mathrm{\pi }}} and r_2=\sqrt{\frac{(Area of bas{e}_2)}{\mathrm{\pi }}}$