Ex.12.2 Q3 Areas Related to Circles Solution - NCERT Maths Class 10

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Question

The length of the minute hand of a clock is \(14\, \rm{cm.}\) Find the area swept by the minute hand in \(5\) minutes.

Text Solution

 

What is known?

Length of the minute hand of the clock

What is unknown?

Area swept by the minute hand in \(5\) minutes.

Reasoning:

Since the minute hand completes one rotation in \(1\) hour or \(60\) minutes

Therefore,Area swept by the minute hand in \(60\) minutes \(=\)  Area of the Circle with radius equal to the length of the minute hand \(\begin{align}= \pi {{r^2}}.\end{align}\)

Using Unitary method

Area swept by minute hand in \(1\) minute: \(\begin{align}\frac{{\pi {r^2}}}{{60}}\end{align}\)

Area swept by minute hand in \(5\) minutes: \(\begin{align}\frac{{\pi {r^2}}}{{60}} \times 5 = \frac{{\pi {r^2}}}{{12}}\end{align}\)

Steps:

Length of the minute hand \(({r}) = 14\,\rm{cm}\)

We know that the minute hand completes one rotation in \(1 \)hour or \(60\) minutes

Therefore,Area swept by the minute hand in \(60\) minutes \(=\) Area of circlewith radius '\(r\)'

\[\begin{align} &= \pi {r^2}\\& = \frac{{22}}{7} \times 14 \times 14\\ &= 22 \times 28{\text{c}}{{\text{m}}^2}\\ &= 616\;{\text{cm}^2}\end{align}\]

Therefore,Area swept by the minute hand in \(5\) minutes \(=\begin{align}\frac{5}{{60}}\pi {r^2} \Rightarrow \frac{1}{{12}}\pi {r^2}\end{align}\)

\[\begin{align}&= \frac{1}{{12}} \times \frac{{22}}{7} \times 14 \times 14\rm c{m^2}\\&= \frac{{154}}{3}\rm c{m^2}
\end{align}\]