A bucket is in the form of a frustum of a cone and holds 15.25 litres of water.

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

$=\pi(12.5+10) \sqrt{(12.5-10)+38.18}$

$=3017 \mathrm{~cm}^{2}$