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

## 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?**

Area of the shaded region.

**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.

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