The minute hand of a clock is 10 cm long.

We know that the area A of a sector of circle at an angle θ of radius r is given by

$A=\frac{\theta}{360^{\circ}} \pi r^{2}$

We have,

Angle described by the minute hand in one minute $=6^{\circ}$

So, Angle described by the minute hand in 25 minute $=6^{\circ} \times 25=150^{\circ}$

$\therefore$ Required area

$=\frac{150^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(10)^{2}$

$=130.95 \mathrm{~cm}^{2}$