The minute hand of a clock is $ \sqrt{21} \mathrm{~cm} $ long. Find the area described by the minute hand on the face of the clock between $ 7.00 \mathrm{AM} $ and $ 7.05 \mathrm{AM} $.

Given:

The minute hand of a clock is \( \sqrt{21} \mathrm{~cm} \) long.

To do:

We have to find the area described by the minute hand on the face of the clock between \( 7.00 \mathrm{AM} \) and \( 7.05 \mathrm{AM} \).

Solution:

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

Length of the minute hand of a clock $r =\sqrt{21}\ cm$.
Time period from 7 AM to 7.05 AM $=5$ minutes

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

$= 30^{\circ}$

Area described by the minute hand on the face of the clock between \( 7.00 \mathrm{AM} \) and \( 7.05 \mathrm{AM} = \) Area of the sector formed at the centre.

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

$=\frac{22}{7} \times(\sqrt{21})^{2} \times \frac{30^{\circ}}{360^{\circ}}$

$=\frac{22}{7} \times 21 \times \frac{1}{12}$

$=\frac{11}{2}$

$=5.5 \mathrm{~cm}^{2}$

The area described by the minute hand on the face of the clock between \( 7.00 \mathrm{AM} \) and \( 7.05 \mathrm{AM} \) is $5.5 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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