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# The minute hand of a clock is $12\;cm$ long. Then the area of the face of the clock described by the minute hand in $35$ minutes is:A. $210\;c{m^2}$ B. $264\;c{m^2}$ C. $144\;c{m^2}$ D. $200\;c{m^2}$

Last updated date: 17th Jun 2024
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Hint: Find the angle that the minute hand completes after $35$ minutes. Then find the area of the sector which the minute hand swept while completing $35$ minutes by taking the angle made by the minute hand.

The minute hand of a clock is $12\;cm$ long.
As we know that the minute hand of a clock completes a complete angle i.e. $360^\circ$ of angle in one hour i.e. $60\;\min$ . So, the minute hand completes
$\dfrac{{360}}{{60}} = 6^\circ$ angle after each minute.
So, the angle made by the minute hand after completing $35$ minutes is equal to
$35 \times 6 = 210^\circ$ .
Now the given minute hand of the clock is behaving as the radius and the area swept is the area of the sector with angle $210^\circ$ and radius $12\;cm$.
The area of the sector with radius $r$ and angle $\theta$ is equal to
$\dfrac{\theta }{{360^\circ }} \times \pi {r^2}$ .
Substitute the angle and radius in the formula for area of the sector:
$\Rightarrow A = \dfrac{\theta }{{360^\circ }} \times \pi {r^2} \\ = \dfrac{{210^\circ }}{{360^\circ }} \times \dfrac{{22}}{7} \times 12\;cm \times 12\;cm \\ = \dfrac{7}{{12}} \times \dfrac{{22}}{7} \times 12\;cm \times 12\;cm \\ = 264\;c{m^2} \;$
So, the area of the sector is equal to $264\;c{m^2}$ . So, the area of the face of the clock described by minute hand in $35$ minutes is equal to $264\;c{m^2}$ .
So, the correct answer is “Option B”.

Note: The minute hand in the clock always acts as radius of the circle as it extends from the centre to the circle. The area of the sector with radius $r$ and angle $\theta$ is equal to $\dfrac{\theta }{{360^\circ }} \times \pi {r^2}$ .