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  • Explain solution RD sharma class 12 chapter 20 Areas Of Bounded Region exercise Fill in the blanks question 14

Explain solution RD sharma class 12 chapter 20 Areas Of Bounded Region exercise Fill in the blanks question 14

Answers (1)

 

20\pisq.units

Hint:

Use this formula to integrate: \int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}

Given:

x^{2}+y^{2}=1

Explanation:

Area ABCD=4(Area OAB)                                           … (i)

Now,     \frac{x^{2}}{25}+\frac{y^{2}}{16}=1

\frac{y^{2}}{16}=1-\frac{x^{2}}{25}\\ \frac{y^{2}}{16}=\frac{25-x^{2}}{25}\\ y=\frac{4}{5}\sqrt{25-x^{2}}

Now, from (i) we have

Required area

=4\int_{5}^{0}\frac{4}{5}\sqrt{25-x^{2}}dx\\ =\frac{16}{5}\left [ \frac{x}{2}\sqrt{25-x^{2}}+\frac{25}{2}\sin^{-1}\frac{x}{5} \right ]_{0}^{5}\\ =\frac{16}{5}\left [ \frac{5}{2}\sqrt{25-5^{2}}+\frac{25}{2}\sin^{-1}1 \right ]-\frac{16}{5}\left [ \frac{0}{2}\sqrt{25-0}+\frac{25}{2}\sin^{-1}0 \right ]\\ =\frac{16}{5}\left [ \frac{25}{2}\times \frac{\pi}{2} \right ]\\ =20\pi \;\;sq.units

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