Ex.12.3 Q15 Areas Related to Circles Solution - NCERT Maths Class 10

Go back to  'Ex.12.3'

Question

In Figure, \({ABC}\) is a quadrant of a circle of radius \(14 \,\rm{cm} \) and a semicircle is drawn with \({BC}\) as diameter. Find the area of the shaded region.

Text Solution

 

What is known?

\(ABC\) is a quadrant of a circle of radius \(\text{14 cm}\) and a semicircle is drawn with \( {BC}\) as diameter.

What is unknown?

Area of the shaded region.

Reasoning:

 To visualize the shaded region better mark point \(D\) on arc \({BC}\) of quadrant \({ABC}\) and \(E\) on semicircle drawn with \({BC}\) as diameter.

From the figure it is clear that

Area of the shaded region \( = \) Area of semicircle \( {BEC}\, - \) Area of segment \({BDC}\)  

  •  To find area of semicircle \({BEC}, \) we need to find radius or diameter\((BC)\) of the semicircle.

\(\Delta BAC\) is a right angled \(\Delta \), right angled at \({A}\) (\({ABC}\) being a quadrant)

Using Pythagoras theorem, we can find the hypotenuse (\({BC}\))

  • To find area of segment \(BDC\) 

Area of segment \({BDC}\;=\)  Area of quadrant \({ABC} \,- \) Area of \(\Delta \, {ABC}\) 

Area of quadrant \( {ABC}\) can be found by using the formula

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

\({\text{Area of  }}\Delta ABC\)\[\begin{align}&= \frac{1}{2} \times {\rm{ base }} \times {\rm{ height }}\\&= \frac{1}{2} \times AC \times AB\left( {\angle A = {{90}^\circ }} \right)\end{align}\]

Steps:

\(\Delta {ABC}\) is a right angled \(\Delta \), right angled at \( {A}\)

\[\begin{align} {BC^2} &= {AB^2} = {AC^2}\\ &= {14\rm{cm}^2} + {14\rm{cm}^2}\\ BC &= \sqrt {2 + {{14}^2}} \\ &= 14\sqrt 2 {\text{ cm}}\end{align}\]

\(\therefore\;\)Radius of semicircle \(BEC\),\(\begin{align}r = \frac{BC}{2} = \frac{14\sqrt 2 }{2}\rm cm = 7\sqrt 2 cm\end{align}\)

Area of the shaded region= Area of semicircle \(BEC\) - Area of segment \(BDC \)        

Area of shaded region \(= \)  Area of semicircle \({BEC} \,–\) (Area of quadrant  \({ABDC}\;-\) Area \(\Delta {ABC}\))

\[\begin{align}&= \frac{{\pi {r^2}}}{2} -  \begin{pmatrix} \frac{{90}}{{360}} \times \pi {{\left( {14} \right)}^2} \\ - \frac{1}{2} \times AC \times AB \end{pmatrix} \\&= \frac{{\pi {{\left( {7\sqrt 2 } \right)}^2}}}{2} - \begin{pmatrix} \frac{{{{\left( {14} \right)}^2}\pi }}{4} - \frac{1}{2} \\ \times 14 \times 14 \end{pmatrix}\\&= \begin{Bmatrix}  \frac{{22 \times 7 \times 7 \times 2}}{{7 \times 2}} - \\ \begin{pmatrix}  \frac{{22 \times 14 \times 14}}{{7 \times 4}} \\ - 7 \times 14  \end{pmatrix}  \end{Bmatrix} \\&= 154 - \left( {154 - 98} \right)\\&= 98 \rm{cm^2}\end{align}\]