Question

The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area. $[Use\pi =\mathrm{3.14.}]$

Answer 1

$V=\frac{1}{3}\pi \times h\times (R^2+r^2+R\times r)$
where:
R = larger radius
r = smaller radius
h = height
V = volume


V =$\frac{1/3}{\pi }\times 8[{18}^2+{12}^2+18(12\left)\right]V=5730.265c{m}^3$


Total surface area = $\pi \left[\right(R+r)s+(R^2+r^2)$
where:
S = slant height

R - r = 18 - 12 = 6 cm

Pythagorean theorem:
$S^2=(R-r{)}^2+h^2$
$S=\sqrt{6^2+8^2}$
S = 10 cm
Total surface area = $\pi \times \left[\right(18+12\left)\right(10)+{18}^2+{12}^2]$
Total surface area = 2412.743 $c{m}^2$