The length of minute hand of a clock is $ 5 \mathrm{~cm} $. Find the area swept by the minute hand during the time period 6:05 am and $ 6: 40 $ am.

Given:

The length of minute hand of a clock is \( 5 \mathrm{~cm} \).

To do:

We have to find the area swept by the minute hand during the time period 6:05 am and \( 6: 40 \) am.

Solution:

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

Length of the minute hand of a clock $r =5\ cm$.
Time period from 6:05 AM to 6:40 AM $=35$ minutes.

This implies,

Central angle $\theta=\frac{35}{60} \times 360^{\circ}$

$= 210^{\circ}$

Area swept by the minute hand during the time period 6:05 am and 6: 40 am $=$ Area of the sector formed at the centre.

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

$=\frac{22}{7} \times(5)^{2} \times \frac{210^{\circ}}{360^{\circ}}$

$=\frac{22}{7} \times 25 \times \frac{7}{12}$

$=\frac{275}{6}$

$=45.83 \mathrm{~cm}^{2}$

The area swept by the minute hand during the time period 6:05 am and \( 6: 40 \) am is $45.83 \mathrm{~cm}^{2}$.  

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Updated on: 10-Oct-2022

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