UPSC CDS 1 2023 Math Paper

Concept:The problem involves areas of triangles formed in a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle. The side ratios opposite these angles are $1 : \sqrt{3} : 2$. Area of $\triangle AEC$ is obtained by subtracting area of $\triangle ADE$ from area of $\triangle ADC$.Explanation:In $\triangle ADC$, $\angle D = 90^{\circ}$, $\angle A = 30^{\circ}$, $\angle C = 60^{\circ}$. Hence, the side opposite $30^{\circ}$ is $AD$, opposite $60^{\circ}$ is $DC$, and opposite $90^{\circ}$ is $AC$. Given $AD = r$, the ratios give $DC = r\sqrt{3}$ and $AC = 2r$. Area of $\triangle ADC = \frac{1}{2} \times AD \times DC = \frac{1}{2} \times r \times r\sqrt{3} = \frac{\sqrt{3}}{2} r^2$. Now, in $\triangle ADE$, $\angle A = 30^{\circ}$ and $\angle D = 90^{\circ}$. $\tan 30^{\circ} = \frac{DE}{AD} = \frac{DE}{r} \implies \frac{1}{\sqrt{3}} = \frac{DE}{r} \implies DE = \frac{r}{\sqrt{3}}$. Area of $\triangle ADE = \frac{1}{2} \times AD \times DE = \frac{1}{2} \times r \times \frac{r}{\sqrt{3}} = \frac{r^2}{2\sqrt{3}}$. Thus, area of $\text{Area}(\triangle ADC) - \text{Area}(\triangle ADE)$. $= \frac{\sqrt{3}}{2} r^2 - \frac{r^2}{2\sqrt{3}} = r^2 \left( \frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}} \right)$. $= r^2 \left( \frac{3 - 1}{2\sqrt{3}} \right) = r^2 \left( \frac{2}{2\sqrt{3}} \right) = \frac{r^2}{\sqrt{3}}$. Answer:$\frac{r^2}{\sqrt{3}}$ (Option A).