The radii of the circular bases of a frustum of a right circular cone are 12 cm

Solution:

The height of the frustum cone is h = 12 cm. The radii of the bottom and top circles are r1 = 12cm and r2 = 3cm respectively.

The slant height of the frustum cone is

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

$=\sqrt{(12-3)^{2}+12^{2}}$

$=\sqrt{225}$

 

$=15 \mathrm{~cm}$

The total surface area of the frustum cone is

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

$=\pi \times(12+3) \times 15+\pi \times 12^{2}+\pi \times 3^{2}$

 

$=\pi \times 225 \times 26+144 \pi+9 \pi$

$=378 \pi \mathrm{cm}^{2}$

Hence Total surface area $=378 \pi \mathrm{cm}^{2}$

The volume of the frustum cone is

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

$=\frac{1}{3} \pi\left(12^{2}+12 \times 3+3^{2}\right) \times 12$

$=\frac{1}{3} \times \pi \times 189 \times 12$

$=756 \pi \mathrm{cm}^{3}$

Hence Volume of frustum $=756 \pi \mathrm{cm}^{3}$