CBSE 2025 · Region 7 · Set 1 · Q33 · 5 marks
A woman discovered a scratch along a straight line on a circular table top of radius $\displaystyle 8$ cm . She divided the table top into $\displaystyle 4$ equal quadrants and discovered the scratch passing through the origin inclined at an angle $\displaystyle \frac{\pi}{4}$ anticlockwise along the positive direction of x -axis. Find the area of the region enclosed by the $\displaystyle x$-axis, the scratch and the circular table top in the first quadrant, using integration.
Marking-scheme solution
Equation of the circular tabletop: $\displaystyle x^2 + y^2 = 64$Equation of line (scratch): $\displaystyle x = y$The line and circle intersect at $\displaystyle x = 4\sqrt{2}$Area of the shaded region = \int_{0}^{4\sqrt{2}} x\,dx + \int_{4\sqrt{2}}^{8} \sqrt{64 - x^2}\,dx = \left.\frac{x^2}{2}\right]_{0}^{4\sqrt{2}} + \left[\frac{x}{2}\sqrt{64 - x^2} + 32\sin^{-1}\frac{x}{8}\right]_{4\sqrt{2}}^{8} = \frac{32}{2} + 32\sin^{-1}1 - 2\sqrt{2}\cdot 4\sqrt{2} - 32\sin^{-1}\frac{1}{\sqrt{2}} = 16 + 16\pi - 16 - 8\pi = 8\pi\ \text{cm}^2$$Application of IntegralsArea under Simple CurvesApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.