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

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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 \(\Delta {ABC}\)

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

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 \(O\) 

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

\(\begin{align} \therefore AB = 2R = 14cm\end{align}\)

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  \(\Delta {ABC}\)

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

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

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