# 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.

Video Solution
Areas Related To Circles
Ex 12.2 | Question 3

## 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}

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