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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 27

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 27

Answers (1)

best_answer

Answer:

\frac{a^{2}}{12}(4 \pi-3 \sqrt{3}) { sq } \cdot { unit }

96\; s q \cdot { unit }

Hint:

Use definite integrals.

Given:

Find area of circle  x^{2}+y^{2}=a^{2} cut by the line x=\frac{a}{2}

Solution:

By solving equation \frac{a}{2}, \frac{\sqrt{3} a}{2}and \frac{a}{2}, \frac{-\sqrt{3} a}{2}

Hence form diagram, we get

Required area =2 \times \text { Area of } \mathrm{AOB}

                         \begin{aligned} &=2 \int_{a / 2}^{a} \sqrt{a^{2}-x^{2}} d x \\\\ &=2\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{a / 2}^{a} \end{aligned}

                        \begin{aligned} &=2\left[\frac{a^{2}}{2}\left(\frac{\pi}{2}\right)-\frac{a}{4}\left(\frac{a \sqrt{3}}{2}\right)-\frac{a^{2}}{2}\left(\frac{\pi}{6}\right)\right] \\\\ &=\frac{a^{2}}{12}[6 \pi-3 \sqrt{3}-2 \pi] \\\\ &=\frac{a^{2}}{12}[4 \pi-3 \sqrt{3}] { sq\;. unit } \end{aligned}

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