Question

The slant height of the frustum of a cone is $4$cm, and the perimeter of its circular bases are $18$cm and $6$cm respectively. Find the curved surface area and total surface area of the frustum.

Answer 1

Given, $2\pi r_1=18cm,$ $2\pi r_2=6cm,l=4cm$
$\therefore r_1=\frac{9}{\pi }cm$, $r_2=\frac{3}{\pi }$cm
Curved surface area of the frustum$=\pi (r_1+r_2)l$
$=\pi \left(\frac{9}{\pi }+\frac{3}{\pi }\right)4=\pi \times \frac{12}{\pi }\times 4$
$\therefore$ CSA of the frustum $=48c{m}^2$
Total surface area of the frustum$=\pi (r_1+r_2)l+\pi r_1^2+\pi r_2^2$

$=48+\pi \times \frac{9}{\pi }\times \frac{9}{\pi }+\pi \times \frac{3}{\pi }\times \frac{3}{\pi }=48+\frac{81}{\pi }+\frac{9}{\pi }$

$=48+\frac{81\times 7}{22}+\frac{9\times 7}{22}=48+25.8+2.9$
$\therefore$ TSA of the frustum$=76.7c{m}^2$.