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  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 25.

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 25.

Answers (1)

Answer:

\frac{8 \pi}{3}\; \text{sq. units}

Hint: :

\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -\frac{1}{a}+c \text {. } .

Given:

The given equations are x^2+y^2=16  and y=\sqrt{3}x .

Solution:

x^2+y^2=16   Represents a circle with centreo(0,0) and cutting the x axis at A(4,0)

 y=\sqrt{3}x  Represents straight passing through o(0,0) .

Point of intersection  is obtained by solving the two equations.

x^2+y^2=16 and y=\sqrt{3}x 

x^{2}+(\sqrt{3} x)^{2} =16

4x^2=16

x=\pm 2

y=\pm 2\sqrt{3}

B(2,2\sqrt{3}) and B'(-2,-2\sqrt{3}) are points of intersection of circle and straight line

Shaded area (OBQAO) =area(OBPO)+area(PBQAP)

 

\begin{aligned} &=\int_{0}^{2} \sqrt{3} x d x+\int_{2}^{4} \sqrt{16-x^{2}} d x \\ &=\sqrt{3}\left[\frac{x^{2}}{2}\right]_{0}^{2}+\left[\frac{1}{2} x \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1}\left(\frac{x}{4}\right)_{2}^{4}\right] \\ &=2 \sqrt{3}+8 \frac{\pi}{2}-2 \sqrt{3}-8 \frac{\pi}{6} \\ &=2 \sqrt{3}+4 \pi-2 \sqrt{3}-\frac{4 \pi}{3} \\ &=\frac{8 \pi}{3} \text { sq.units } \end{aligned}

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