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  • Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 34 Maths.

Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 34 Maths.

Answers (1)

Answer:

\left (\frac{\sqrt{5}\pi}{4}-\frac{1}{2} \right ) sq. Units.

Hint:

Given: \left\{(x, y):|x-1| \leq y \leq \sqrt{\left.\left(5-x^{2}\right)\right\}}\right.

Solution: Equation of the curve is y=\sqrt{(5-x)^2}      or  y^2+x^2=5 ,which is a circle with

centre at (0, 0) and? radius? \frac{5}{2}.

         

Equation of the line is y=\left | x-1 \right |

Consider, y =x -1  and y=\sqrt{5-x^2}

Eliminating y ,we get

\begin{aligned} &x-1=\sqrt{5-x^{2}} \\ &x^{2}+1-2 x=5-x^{2} \\ &2 x^{2}-2 x-4=0 \\ &x^{2}-x-2=0 \\ &(x-2)(x+1)=0 \\ &x=2,-1 \end{aligned}

The required area is

\begin{aligned} &A=\int_{-1}^{2} \sqrt{5-x^{2}} d x-\int_{-1}^{1}(-x+1) d x-\int_{1}^{2}(x-1) d x \\ &=\left[\frac{x}{2} \sqrt{5-x^{2}}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{x}{\sqrt{5}}\right)\right]_{-1}^{2}+\left[\frac{x^{2}}{2}-x\right]_{-1}^{1}+\left[\frac{-x^{2}}{2}+x\right]_{1}^{2} \\ &=\left[\frac{2}{2} \sqrt{5-4}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{2} \sqrt{5-1}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]+\left[\frac{1}{2}-1-\frac{1}{2}-1\right]+\left[\frac{-4}{2}+2+\frac{1}{2}-1\right] \end{aligned}

\begin{aligned} &=\left[1+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)+\frac{2}{2}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]-2-\frac{1}{2} \\ &=2+\left[\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]-2-\frac{1}{2} \\ &=\frac{\sqrt{5}}{2} \times \frac{\pi}{2}-\frac{1}{2} \\ &=\frac{\sqrt{5} \pi}{4}-\frac{1}{2} \end{aligned}

=\left (\frac{\sqrt{5}\pi}{4}-\frac{1}{2} \right ) sq.units

Therefore, the area of the region is \left (\frac{\sqrt{5}\pi}{4}-\frac{1}{2} \right ) sq.units.

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