The length of the minute hand of a clock is $14 \,cm$. Find the area of the region swept by it between $10.10\, AM$ to $10.30 \,AM.$
The length of the minute hand of a clock is $14 \,cm$. Find the area of the region swept by it between $10.10\, AM$ to $10.30 \,AM.$
During $60$ minutes $(1$ hour $),$ the minute hand completes one rotation, i.e., it subtends an angle of measure $360 .$
$\therefore$ The measure of the angle for the minor sector swept during $20$ minutes duration from $10.10$ $AM$ to $10.30$ $AM$
$\theta=\frac{360}{60} \times 20=120$
Radius of the minor sector $r$
$=$ Length of the minute hand $=14 \,cm$
Area of a minor sector $=\frac{\pi r^{2} \theta}{360}$
$=\frac{22}{7} \times \frac{14 \times 14 \times 120}{360}$
$=\frac{616}{3}$
$=205 \frac{1}{3} cm ^{2}$
Thus, the area of the region swept by the minute hand during $10.10$ $AM$ to $10.30$ $AM$ is $205 \frac{1}{3} cm ^{2}$
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