UPSC CDS I 2024 Math Paper

Concept:The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.Explanation:From the given figure, $DC$ is the diameter of a semicircle.So $DO = OC = EO = FO$, all equal to the radius of the semicircle.It is given that $\angle OEF = \angle AOQ = 45^{\circ}$.In $\triangle EOF$, using Pythagoras theorem:$EF = \sqrt{EO^2 + FO^2} = EO\sqrt{2}$.Since the rectangle length is $10$ cm, $EO = 5$ cm, therefore $EF = 5\sqrt{2}$ cm.$AB = 10$ cm (given).The height of trapezium $AEFB$ is half of $EF$, i.e., $\frac{5}{\sqrt{2}}$ cm.Now, area = $\frac{1}{2} \times (AB + EF) \times \text{height}$= $\frac{1}{2} \times (10 + 5\sqrt{2}) \times \frac{5}{\sqrt{2}}$= $\frac{5}{\sqrt{2}} \times \frac{10 + 5\sqrt{2}}{2}$= $\frac{5 \times (10\sqrt{2} + 10)}{\sqrt{2} \times 2\sqrt{2}}$= $\frac{50\sqrt{2} + 50}{4}$= $\frac{50(\sqrt{2}+1)}{4}$On simplifying, the numerical value is approximately $30$.Answer:$30$ square cm (Option A).