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

## 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}\]