Ex.12.3 Q5 Areas Related to Circles Solution - NCERT Maths Class 10

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Question

From each corner of a square of side \(\text{4 cm}\) a quadrant of a circle of radius \(\text{1 cm}\) is cut and also a circle of diameter \(\text{2 cm}\) is cut as shown in the given figure. Find the area of the remaining portion of the square.

 

Text Solution

What is the known?

From each corner of a square of side \(\text{= 4 cm}\) a quadrant of a circle of radius \(\text{1 cm}\) is cut and also a circle of diameter \(\text{2 cm.}\)

What is the unknown?

Area of remaining portion of the square.

Reasoning:

(i) Since diameter of circle which is cut out \(\text{= 2 cm}\) 

\(\therefore\;\) Radius of this circle \((r) =\text{ 1 cm}\) 

(ii) Logically since all quadrants cut out are of same radius.

Area of portions cut out of square \(=\)  Area of the circle \(+\) 4\(\times\)(Area of each quadrant)

\[\begin{align}&= \pi {r^2} + 4\left( {\frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}} \right)\\ &= { \pi {r^2} + 4 \times \frac{{\pi {r^2}}}{4}}\\ &= {\pi {r^2} + \pi {r^2}}\\ &= 2\pi {r^2}\end{align}\]

(iii) From the figure it is clear that

Area of remaining portion of the square \(=\;\)Area of square \(-\) Area of portion cut out of square

\[\begin{align}= {\left( \rm{side} \right)^2} - 2\pi {r^2}\end{align}\]

Steps:

Diameter of circle \(\text{= 2 cm}\)

Radius of circle \(\begin{align} (r) = \frac{{2\,{\text{cm}}}}{2} = \,1\,{\text{cm}} \end{align}\)

Radius of all quadrants cut out (r)=1cm\(\rm{}(r)=1\,cm\) 

Area of the portions cut out of the square \(=\)  Area of a full circle  \(+\, 4 \;\times\)  (Area of each quadrant)

\[\begin{align}&={\pi {r^2} + 4\left( {\frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}} \right)}\\ &={ \pi {{(1)}^2} + 4 \times \frac{{\pi {{(1)}^2}}}{4}}\\ &={\pi + \pi }\\ &={2\pi }\end{align}\]

Area of remaining portion of the square \(= \) Area of square \(- \) Area of portions cut out

\[\begin{align}&= \rm{}{\left( {side} \right)^2} - 2\pi {r^2}\\&=\rm{} {\left( {4cm} \right)^2} - \frac{{44}}{7}c{m^2}\\&= \rm{}16c{m^2} - \frac{{44}}{7}c{m^2}\\&= \rm{}\frac{{112 - 44}}{7}c{m^2}\\&=\rm{} \frac{{68}}{7}c{m^2}
\end{align}\]

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