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  • Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 37 maths textbook solution

Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 37 maths textbook solution

Answers (1)

Answer:

 (b)

Hint:

 \int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\left ( \frac{x}{a} \right )+c

Given:

 In first quadrant, x- axis (y = 0), y = x, x2 + y2 = 32

Explanation:

Intersection point,

\begin{aligned} &y = x\\ &x^{2}+x^{2}=32\\ &2x^{2}=32\\ &x^{2}=16\\ &x=\pm 4 \end{aligned}

x = -4 is not possible as we have to find in first quadrant and in first quadrant x > 0

\begin{aligned} &x=4\\ &y^{2}+x^{2}=32\\ &x^{2}=32\\ &x=4\sqrt{2} \end{aligned}

Now,     y = x

Therefore, y = 4

Area of ABCD = Area of ABD + Area of BCD

\begin{aligned} &=\int_{0}^{4}xdx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}dx\\ &=\left [ \frac{x^{2}}{2} \right ]_{0}^{4}+\left [ \frac{x}{2}\sqrt{32-x^{2}}+\frac{1}{2}(32)sin^{-1}\frac{x}{4\sqrt{2}} \right ]_{4}^{4\sqrt{2}}\\ &=8+\frac{4\sqrt{2}}{2}\sqrt{32(4\sqrt{2})^{2}}+\frac{1}{2}(32)sin^{-1}\frac{4\sqrt{2}}{4\sqrt{2}}-\frac{4}{2}\sqrt{32-(4)^{2}}-16sin-1\frac{4}{4\sqrt{2}}\\ &=8+0+16sin^{-1}1-2\sqrt{32-16}-16sin^{-1}\frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} &=8+16\left ( \frac{\pi }{2} \right )-2(4)-16 \left ( \frac{\pi }{4} \right )\\ &=8+8\pi -8-4\pi \\ &=4\pi \text { sq.units } \end{aligned}

Posted by

Gurleen Kaur

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