Ex.12.3 Q7 Areas Related to Circles Solution - NCERT Maths Class 10

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In the given figure, \(ABCD\) is a square of side \(\text{14 cm.}\) With Centers \(A, B, C\) and \(D,\) four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Text Solution

What is known?

\(ABCD\) is a square of side \(\text{= 14 cm.}\)

With centers\( A, B, C, D\) four circles are drawn such that each circle touches externally \(2\) of the remaining \(3\) circles.

What is unknown?

Area of the shaded region.


Since the circles are touching each other externally,visually it is clear that

Radius of each circle \(\begin{align} {r} = \frac{1}{2}\,\, \times \end{align} \)(side of square)

Also, \(ABCD\) being a square all angles are of measure \({90^ \circ },\,\,\, \)

Therefore,all sector are equal as they have same radii and angle.

\(\therefore\;\) Angle of each sector which is part of the square \(\left( {\theta } \right) = {90^ \circ }\)

\(\therefore\;\)\(\text{Area of each sector}\)\[\begin{align}&= \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\\&= \frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}\\&= \frac{{\pi {r^2}}}{4}\end{align}\]

From the figure it is clear that:

Area of shaded region \(=\)  Area of square \(– \)  Area of \(4\) sectors

\[\begin{align} &= {\left( {{\text{side}}} \right)^2} - 4 \times {\text{Area of each sector}}\\ &= {\left( {14} \right)^2} - 4 \times \frac{{\pi {r^2}}}{4}\\ &= {\left( {14} \right)^2} - \pi {r^2}\end{align}\]


Area of each of the \(4\) sectors is equal as each sector subtends an angle of \({90^ \circ }\) at the centre of a circle with radius \(\text{= 7 cm}\) 

\(\therefore\) \(\text{Area of each sector}\)\[\begin{align}&= \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\\ &= \frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {(7)^2}\\ &= \frac{1}{4} \times \frac{{22}}{7} \times 7 \times 7\\ &= \frac{{77}}{2}{\text{c}}{{\text{m}}^2}\end{align}\]

Area of shaded region \(=\)  Area of square \(- \,4 \;\times\) Area of each sector

\[\begin{align}&= {(14)^2} - 4 \times \frac{{77}}{2}\\ &= 196 - 154\\ &= 42\,\,{\text{c}}{{\text{m}}^2}\end{align}\]