A reservoir is in the shape of a frustum of a right circular cone. It is 8m across at the top and 4m across the bottom. It is 6m deep. Find the area of its curved surface.
Solution
Curved surface area of a frustrum = $\pi (R+r)$
Here, $R= \dfrac{8}{2} = 4$ cm, $r= \dfrac{4}{2} = 2$ cm
And $L= \sqrt{h^2+(R−r)^2}$
$L= \sqrt{6^2 +2^2} = \sqrt{40}$
$\therefore$ Curved surface area,
$= \dfrac{22}{7} \times (2+4) \times \sqrt{40} = 119.26 =118.4$ m$^2$ (approx.)