Question

The slant height of a frustum of a cone is $4\mathrm{cm}$ and the perimeters (circumference) of its circular ends are $18\mathrm{cm}\mathrm{and}6\mathrm{cm}.$ Find the surface area of the frustum.

Answer 1

Let the radius of smaller circle end $=\mathrm{r}$

Circumference of smaller circle$=2\mathrm{\pi r}$

$$\begin{gathered} 2\mathrm{\pi r}=6{\mathrm{cm}}^2 \\ \mathrm{r}=6/2 \\ \mathrm{r}=3\mathrm{cm} \end{gathered}$$

Let the radius of larger circle end$=\mathrm{R}$

Circumference of larger circle$=2\mathrm{\pi R}$

$$\begin{gathered} 18{\mathrm{cm}}^2=2\mathrm{\pi R} \\ \mathrm{R}=\frac{18}{2} \\ \mathrm{R}=9\mathrm{cm} \end{gathered}$$

Now, as curve surface area of frustum$=\mathrm{\pi }(\mathrm{r}+\mathrm{R})\mathrm{l}$

$$\begin{gathered} =\mathrm{\pi }\times (9+3)\times 4 \\ =12\times 4 \\ =48{\mathrm{cm}}^2 \end{gathered}$$

Hence, the curved surface area of the frustum $=48{\mathrm{cm}}^2$