The length of the minute hand of a clock is $ 14 \mathrm{~cm} $. Find the area swept by the minute hand in 5 minutes.

Given:

The length of the minute hand of a clock is \( 14 \mathrm{~cm} \).

To do:

We have to find the area swept by the minute hand in 5 minutes.

Solution:

Let the angle formed at the centre be $\theta$.

Length of the minute hand of a clock $r =14\ cm$.
Time period $=5$ minutes.

This implies,

Central angle $\theta=\frac{5}{60} \times 360^{\circ}$

$= 30^{\circ}$

Area swept by the minute in 5 minutes $=$ Area of the sector formed at the centre.

Area of the sector $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$

$=\frac{22}{7} \times(14)^{2} \times \frac{30^{\circ}}{360^{\circ}}$

$=\frac{22}{7} \times 196 \times \frac{1}{12}$

$=\frac{11\times28}{6}$

$=51.33 \mathrm{~cm}^{2}$

The area swept by the minute hand in 5 minutes is $51.33 \mathrm{~cm}^{2}$.   

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Updated on: 10-Oct-2022

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