The length of the hour hand of a clock is $\displaystyle 10$ cm. Find the area of the minor sector swept by the hour hand of the clock between $\displaystyle 5$ a.m. to $\displaystyle 8$ a.m. Also, find the area of the major sector.

Central angle subtended by hour hand between $\displaystyle 5$ am to $\displaystyle 8$ am \(\displaystyle =\frac{360^{\circ}}{12} \times 3=90^{\circ}\) \[\begin{aligned} & \text { Area of minor segment }=\frac{90}{360} \times \frac{22}{7} \times(10)^{2} \\ & \qquad=\frac{550}{7} \text { or } 78.57 \mathrm{~cm}^{2} \text { approx. } \\ & \text { Area of circle }=\frac{22}{7} \times(10)^{2}=\frac{2200}{7} \mathrm{~cm}^{2} \\ & \text { Area of major segment }=\frac{2200}{7}-\frac{550}{7} \\ & \qquad=\frac{1650}{7} \text { or } 235.71 \mathrm{~cm}^{2} \text { approx. } \end{aligned} \]