The minute hand of a clock is

Question:

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 $7.05 \mathrm{AM}$.

Solution:

We know that the area A of a sector of circle at an angle θ of radius r is given by

$A=\frac{\theta}{360^{\circ}} \pi r^{2}$

We have,

Angle described by the minute hand in one minute $=6 \%$

So, angle described by the minute hand in five minute $=6^{\circ} \times 5=30^{\circ}$

Thus,

Area swept by the minute hand in 5 minute

$=$ Area of a sector of angle $30^{\circ}$ in the circle of radius $\sqrt{2 I} \mathrm{~cm}$

$=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times \sqrt{21} \times \sqrt{21} \mathrm{~cm}^{2}$

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

 

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