Question:
A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25 cm and 20 cm respectively. Find its height and area of tin used in its construction.
Solution:
Since volume of frustum
$=\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right)$
$=15250 \mathrm{~cm}^{3}$
$h \times \frac{\pi}{3} \times\left(\left(\frac{25}{2}\right)^{2}+(10)^{2}+\frac{25}{2} \times 10\right)$
$=15250$
$h \times \frac{\pi}{3}(156.25+22.5)$
$=15250$
$h=\frac{3 \times 7 \times 15250}{22 \times 381.25}$
$=\frac{320250}{8387.5}$
$=38.18 \mathrm{~cm}$
Area of the required
$=\pi(12.5+10) \sqrt{(12.5-10)+38.18}$
$=3017 \mathrm{~cm}^{2}$