71
The length of the hour hand of a clock is $10$ cm. Find the area of the minor sector swept by the hour hand of the clock between $5$ a.m. to $8$ a.m. Also, find the area of the major sector.
Show SolutionHide Solution↓
Central angle subtended by hour hand between $5$ am to $8$ am $= \frac{360^\circ}{12} \times 3 = 90^\circ$
Area of minor segment $= \frac{90}{360} \times \frac{22}{7} \times (10)^2$
$= \frac{550}{7}$ or $78.57$ cm$^2$ approx.
Area of circle $= \frac{22}{7} \times (10)^2 = \frac{2200}{7}$ cm$^2$
Area of major segment $= \frac{2200}{7} - \frac{550}{7} = \frac{1650}{7}$ or $235.71$ cm$^2$ approx.
Area of minor segment $= \frac{90}{360} \times \frac{22}{7} \times (10)^2$
$= \frac{550}{7}$ or $78.57$ cm$^2$ approx.
Area of circle $= \frac{22}{7} \times (10)^2 = \frac{2200}{7}$ cm$^2$
Area of major segment $= \frac{2200}{7} - \frac{550}{7} = \frac{1650}{7}$ or $235.71$ cm$^2$ approx.