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

Answers (1)

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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