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

## 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]

Area of the quadrant

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

Area of the right triangle

\(\begin{align} = \frac{1}{2} \times \text{base} \times \text{height} \end{align}\)

**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]

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