The minute hand of a clock is $ 10 \mathrm{~cm} $ long. Find the area of the face of the clock described by the minute hand between $ 8 \mathrm{AM} $ and $ 8.25 \mathrm{AM} $

Given:

The minute hand of a clock is \( 10 \mathrm{~cm} \) long.

To do:

We have to find the area of the face of the clock described by the minute hand between \( 8 \mathrm{AM} \) and \( 8.25 \mathrm{AM} \).

Solution:

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

Length of the minute hand of a clock $r =10\ cm$.
Time period from 8 AM to 8.25 AM $=25$ minutes

Central angle $\theta=\frac{25}{60} \times 360^{\circ}$

$= 150^{\circ}$

Area of the face of the clock described by the minute hand between \( 8 \mathrm{AM} \) and \( 8.25 \mathrm{AM} = \) Area of the sector formed at the centre.

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

$=\frac{22}{7} \times(10)^{2} \times \frac{150^{\circ}}{360^{\circ}}$

$=\frac{22}{7} \times 100 \times \frac{5}{12}$

$=\frac{11000}{84}$

$=130.95 \mathrm{~cm}^{2}$

The area of the face of the clock described by the minute hand between \( 8 \mathrm{AM} \) and \( 8.25 \mathrm{AM} \) is $130.95 \mathrm{~cm}^{2}$. 

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Updated on: 10-Oct-2022

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