# Ex.12.3 Q16 Areas Related to Circles Solution - NCERT Maths Class 10

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## Question

Calculate the area of the designed region in Figure common between the two quadrants of circles of radius $$\text{8 cm}$$ each. ## Text Solution

What is  known?

Designed region is the common area between the $$2$$ quadrants of circles of radius $$\text{8 cm.}$$

What is unknown?

Area of the designed region.

Reasoning:

In a circle with radius r and angle at the center with degree measure $$\theta$$;

(i) Area of the sector \begin{align} = \frac{{\rm{\theta }}}{{{{360}^{\rm{\circ}}}}} \times \pi {r^2}\end{align}

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

From the figure,it is observed that

Area of the designed region = $$2\times$$Area of the segment of the quadrant of radius $$\rm8\,cm$$ = $$2\times$$[Area of the quadrant - Area of the right triangle]

\begin{align} = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2} = \frac{1}{4}\pi {r^2}\end{align}

Area of the right triangle\begin{align} = \frac{1}{2} \times base \times height\end{align}

Steps:

From the figure, it is observed that

Area of the designed region $$=2\times$$Area of the segment of the quadrant of radius $$\rm{}8\,cm$$

$$=2\times$$[Area of the quadrant - Area of the right triangle]

\begin{align}&= 2 \times \left[ {\frac{1}{4}\pi {r^2} - \frac{1}{2} \times base \times height} \right]\\&= 2 \times \left[ {\frac{1}{4} \times \frac{{22}}{7} \times 8cm \times 8cm - \frac{1}{2} \times 8cm \times 8cm} \right]\\&= 2 \times \left[ {\frac{{352}}{7}c{m^2} - 32c{m^2}} \right]\\&= 2 \times \left[ {\frac{{352 - 224}}{7}c{m^2}} \right]\\&= 2 \times \frac{{128}}{7}c{m^2}\\&= \frac{{256}}{7}c{m^2}\end{align}

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