# Ex.12.3 Q9 Areas Related to Circles Solution - NCERT Maths Class 10

Go back to  'Ex.12.3'

## Question

In Figure, $$AB$$ and $$CD$$ are two diameters of a circle (with center $$O$$) perpendicular to each other and $$OD$$  is the diameter of the smaller circle. If $$OA = \rm{7 cm,}$$ find the area of the shaded region. ## Text Solution

What is known?

$$AB$$ and $$CD$$ are two diameters of a circle with centre $$O,$$ perpendicular to each other that is $$\,\angle {BOC} = {90^ \circ }$$

$$OD$$ is diameter of smaller circle,and $$OA = \text{7 cm.}$$

What is unknown?

Reasoning:

Since $$AB$$ and $$CD$$ are diameter of the circle.

\begin{align}\therefore \quad OD&= OC = OA=OB=R=7\rm{}cm\end{align} \,\,\,(being radii of the circle)

$$\therefore$$  $$\rm{}AB=CD=2R=14\, cm$$

$$\therefore$$  Radius of shaded smaller circular region \begin{align}r = \frac{{OD}}{2} = \frac{7}{2}\rm{cm} \end{align}

Area of the shaded smaller circular region \begin{align} = \pi {\left( {\frac{7}{2}} \right)^2}\end{align}

Area of shaded segment of larger circular region $$=$$ Area of semicircle $$ACB$$ $$–$$  Area of $$\Delta {ABC}$$

\begin{align} &= {\frac{\pi }{2}{{(OA)}^2} - \frac{1}{2} \times {AB} \times {OC}\\\qquad \qquad\left( {\therefore \angle BOC = {{90}^\circ }} \right)}\\ &= {\frac{\pi }{2}{7^2} - \frac{1}{2} \times 14 \times 7\\\qquad \qquad({AB = OA + OB = 20A} = 14)}\end{align}

Area of shaded region $$=$$ Area of shaded smaller cricular region $$\rm{} \,+$$ Area of shaded sagment of larger circular region

Steps:

$$\text{OA = 7 cm}$$

$$AB$$ and $$CD$$ are diameter of the circle with center $$O$$

\begin{align}&\therefore OD = OC = OA = OB = R = 7cm({\text{being radii of the circle }})\\&\therefore AB = 2R = 14cm\end{align}

Radius of shaded circular region $$= \frac{{OD}}{2} = \frac{7}{2}{\text{cm}}$$

Area of shaded smaller circular region \begin{align}= \pi {r^2}\end{align}

\begin{align}&= \pi {\left( {\frac{7}{2}cm} \right)^2}\\&= \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2}c{m^2}\\&= \frac{{77}}{2}c{m^2}\\&= 38.5c{m^2}\end{align}

Area of shaded segment of larger circular region $$=$$  Area of semicircle $$ACB\; -$$ Area of  $$\Delta {ABC}$$

\begin{align}&= \frac{1}{2}\pi {\left( {OA} \right)^2} - \frac{1}{2} \times AB \times OC\left(\therefore {OC \bot AB} \right)\\&= \frac{1}{2}\pi {R^2} - \frac{1}{2} \times 2R \times R\\&= \frac{1}{2} \times \frac{{22}}{7} \times {\left( {7cm} \right)^2} - \frac{1}{2} \times 14cm \times 7cm\\&= 77c{m^2} - 49c{m^2}\\&= 28c{m^2} \end{align}

Area of shaded region = Area of the shaded smaller circular region + Area of the shaded segment of larger circular region.

\begin{align}&= 38.5c{m^2} + 28c{m^2}\\&= 66.5c{m^2}\end{align}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school