A solid is in the shape of a frustum of a cone,

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 height is 9 cm. Find the area of its whole surface and the volume.

Solution:

The height of the frustum of a cone is h=9cm. The radii of the upper and lower circles of the frustum of the cone are r1 =30cm and r2 =18cm respectively.

The slant height of the frustum of the cone is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(30-18)^{2}+9^{2}}$

$=\sqrt{225}$

$=15 \mathrm{~cm}$

The volume of the frustum of the cone is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(30^{2}+30 \times 18+18^{2}\right) \times 9$

$=\frac{1}{3} \times 1764 \times 9 \times \pi$

$=5292 \pi \mathrm{cm}^{3}$

The total surface area of the frustum of the cone is

$S_{1}=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{1}^{2}+\pi r_{2}^{2}$

$=\pi \times(30+18) \times 15+\pi \times 30^{2}+\pi \times 18^{2}$

$=\pi \times 48 \times 15+900 \pi+324 \pi$

$=1944 \pi \mathrm{cm}^{2}$