A frustum of a cone is 9 cm thick and the diameters
Question:

A frustum of a cone is 9 cm thick and the diameters of its circular ends are 28 cm and 4 cm. Find the volume and lateral surface area of the frustum.
(Take π = 22/7).

Solution:

Volume

$=\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right)$

$=9 \times \frac{\pi}{3}\left(\left(\frac{28}{2}\right)^{2}+\left(\frac{4}{2}\right)^{2}+\frac{28}{2} \times \frac{4}{2}\right)$

$=3 \pi\left((14)^{2}+4+14 \times 2\right)$

$=684 \pi \mathrm{cm}^{3}$

S.A. $=\pi(14+2) \sqrt{(14-2)^{2}}+9=240 \pi \mathrm{cm}^{2}$

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