The area (in square units) of minor segment of the circle x2 + y2 = 2…

The correct answer is Option (2) → \(\displaystyle\frac{25}{12}\left(4\pi-3\sqrt{3}\right)\)

The minor segment lies between x = 5/2 and x = 5. The circle is symmetric about the x-axis, so the area is twice the part above the x-axis, where \(y = \sqrt{25-x^2}\):

    \( \displaystyle A = 2\int_{5/2}^{5}\sqrt{25-x^2}\,dx \)

Using the standard integral \(\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\) with a = 5:

    \( \displaystyle A = 2\left[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}\sin^{-1}\frac{x}{5}\right]_{5/2}^{5} \)

At x = 5:

    \( \displaystyle 0 + \frac{25}{2}\times\frac{\pi}{2} = \frac{25\pi}{4} \)

At x = 5/2, where \(\sqrt{25-\frac{25}{4}} = \frac{5\sqrt{3}}{2}\):

    \( \displaystyle \frac{5}{4}\times\frac{5\sqrt{3}}{2} + \frac{25}{2}\times\frac{\pi}{6} = \frac{25\sqrt{3}}{8} + \frac{25\pi}{12} \)

So

    \( \displaystyle A = 2\left(\frac{25\pi}{4} - \frac{25\pi}{12} - \frac{25\sqrt{3}}{8}\right) \)

    \( \displaystyle = 2\left(\frac{25\pi}{6} - \frac{25\sqrt{3}}{8}\right) \)

    \( \displaystyle = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} \)

    \( \displaystyle = \frac{25}{12}\left(4\pi-3\sqrt{3}\right) \)

Shortcut: A segment with central angle \(\theta\) has area \(\frac{r^2}{2}(\theta - \sin\theta)\). The chord x = 5/2 is at distance 5/2 from the centre, so \(\cos\frac{\theta}{2} = \frac{5/2}{5} = \frac{1}{2}\) and \(\theta = \frac{2\pi}{3}\):

    \( \displaystyle A = \frac{25}{2}\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} \)

NCERT Exemplar:
Application of Integrals