The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm

Solution:

We have,

Height, $h=8 \mathrm{~cm}$,

Base radii, $R=18 \mathrm{~cm}$ and $r=12 \mathrm{~cm}$

Also, the slant height, $l=\sqrt{(R-r)^{2}+h^{2}}$

$=\sqrt{(18-12)^{2}+8^{2}}$

$=\sqrt{6^{2}+8^{2}}$

$=\sqrt{36+64}$

$=\sqrt{100}$

$=10 \mathrm{~cm}$

Now,

Total surface area of the solid frustum $=\pi(R+r) l+\pi R^{2}+\pi r^{2}$

$=3.14 \times(18+12) \times 10+3.14 \times 18^{2}+3.14 \times 12^{2}$

$=3.14 \times 30 \times 10+3.14 \times 324+3.14 \times 144$

$=3.14 \times(300+324+144)$

$=3.14 \times 768$

$=2411.52 \mathrm{~cm}^{2}$

So, the total surface area of the solid frustum is 2411.52 cm2.