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The area (in square units) of minor segment of the circle x2+y2=25x^2 + y^2 = 25 cut off by the line x=52x = \frac{5}{2} is

Solution

✅ Correct Option: 2

Minor segment area =2∫5/2525−x2 dx=2[x225−x2+252sin⁡−1(x/5)]5/25= 2\int_{5/2}^{5} \sqrt{25 - x^2}\, dx = 2\left[\frac{x}{2}\sqrt{25-x^2} + \frac{25}{2} \sin^{-1}(x/5)\right]_{5/2}^{5}. Evaluating: at x=5x=5, value is 25π2\frac{25\pi}{2}; at x=5/2x = 5/2, value is 5754+25π6\frac{5\sqrt{75}}{4} + \frac{25\pi}{6}. Difference =25π2−25π6−2534=25π3−2534=2512(4π−33)= \frac{25\pi}{2} - \frac{25\pi}{6} - \frac{25\sqrt{3}}{4} = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} = \frac{25}{12}(4\pi - 3\sqrt{3}).

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