CBSE 2026 · Region 5 · Set 3 · Q34 · 5 marks
Sketch the graph defined by $\displaystyle \left\{(\mathrm{x}, \mathrm{y}): \frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{25}=1\right\}$. Find the area of the region of minor segment cut off by the line $\displaystyle \mathrm{x}=\frac{5}{2}$, using integration.
Marking-scheme solution
Required area $\displaystyle =2 \int_{\frac{5}{2}}^{5} \sqrt{25-x^{2}}\, d x$
$\displaystyle =2\left[\dfrac{x}{2} \sqrt{25-x^{2}}+\dfrac{25}{2} \sin^{-1} \dfrac{x}{5}\right]_{\frac{5}{2}}^{5}$
$\displaystyle =25\left(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}\right)$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.