Question:
If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm respectively. What is the curved surface area of the frustum?
Solution:
The parameter of upper base
$=2 \pi r_{1}$
$2 \pi r_{1}=12$
$r_{1}=\frac{6}{\pi} \mathrm{cm}$
The parameter of lower base
$=2 \pi r_{2}$
$2 \pi r_{2}=24$
$r_{2}=\frac{12}{\pi} \mathrm{cm}$
The surface area of frustum
$=\pi\left(\frac{6}{\pi}+\frac{12}{\pi}\right) \times 6$
$=\pi \times \frac{18}{\pi} \times 6$
$=108 \mathrm{~cm}^{2}$