Question

A solid is in the shape of a frustum of a cone. The diameters of the two circular ends are 60 cm and 36 cm and the leight is 9 cm. Find the area of its whole surface and the volume.

Answer 1

Soln:

given height of the frustum cone = 9 cm

Lower end radius $r_1=\frac{60}{2}=30cm$

Upper end radius$r_2=\frac{36}{2}=18cm$

Let slant height of the frustum cone be L

$L=\sqrt{(r_1-r_2{)}^2+h^2}$

$L=\sqrt{(18-30{)}^2+9^2}=\sqrt{144+81}=15cm$

Volume of the frustum cone$=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h$

$=\frac{1}{3}\pi ({30}^2+{18}^2+30\times 18)\times 9=5262\pi c{m}^3$

Total surface area of frustum cone$=\pi (r_1+r_2)L+\pi r_1^2+\pi r_2^2$

$=\pi (30+18)\times 15+\pi \times {30}^2+\pi \times {18}^2=\pi (720+900+324)=1944\pi c{m}^2$

∴∴ total surface area = 1944 $\pi c{m}^2$