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} $

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.

Complete step-by-step answer:
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} $ .