Ex.12.2 Q2 Areas Related to Circles Solution - NCERT Maths Class 10

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Find the area of a quadrant of a circle whose circumference is \(22\, \rm{cm.}\)

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Areas Related To Circles
Ex 12.2 | Question 2

Text Solution

What is known?

A circle whose circumference is \(22\, \rm{cm.}\)

What is unknown?

Area of quadrant of a circle.


Find the radius of the circle \((r)\) from the circumference \((c),\) \({\bf{C}} = 2\pi r\)

Therefore, \(\begin{align} r = \frac{c}{{2\pi }}\end{align}\)

Since quadrant means one of the four equal parts. Using unitary method, since four quadrants corresponds to Area of a circle

Therefore, Area of a quadrant 

\[\begin{align}&= \frac{1}{4} \times {\text{Area of a circle}}\\&= \frac{1}{4} \times \pi {r^2}\end{align}\]


Circumference \(\begin{align}\left( c \right) = 22{\text{ cm}}\end{align}\)

Therefore,radius \(\begin{align}\rm{r} = \frac{C}{{2\pi }}\end{align}\) 

\begin{align}&= \frac{{22}}{{2 \times \frac{{22}}{7}}}\\&= \frac{{22 \times 7}}{{2 \times 22}}\\
&= \frac{7}{2}\,\,\rm cm\end{align}

Therefore,Area of a quadrant \(\begin{align}= \frac{1}{4} \times \pi {r^2}\end{align}\)

\[\begin{align}&= \frac{1}{4} \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2}\rm c{m^2}\\ &= \frac{{77}}{8}\rm c{m^2}\end{align}\]