UPSC NDA I 2018 Mathematics Solved Paper

Concept:The area of a trapezium is $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$. To maximize the area, use differentiation and set the derivative to zero.Explanation:Let the two equal slant sides be $6$ cm each. Let $x$ be the horizontal distance from the foot of the height to the end of the bottom base. Then the top parallel side is $6$ cm, the bottom parallel side is $(6+2x)$ cm. The height is $\sqrt{6^2 - x^2} = \sqrt{36-x^2}$ cm.Area $A = \frac{1}{2} \left(6 + (6+2x)\right) \sqrt{36-x^2} = (6+x)\sqrt{36-x^2}$.Differentiate $A$ with respect to $x$:$\frac{dA}{dx} = \sqrt{36-x^2} + (6+x) \cdot \frac{-2x}{2\sqrt{36-x^2}} = \sqrt{36-x^2} - \frac{(6+x)x}{\sqrt{36-x^2}}$.Set $\frac{dA}{dx} = 0$:$\sqrt{36-x^2} = \frac{(6+x)x}{\sqrt{36-x^2}}$ ⟹ $36-x^2 = (6+x)x$.Simplify: $36 - x^2 = 6x + x^2$ ⟹ $36 = 6x + 2x^2$ ⟹ $x^2 + 3x - 18 = 0$.Factor: $(x-3)(x+6)=0$ ⟹ $x=3$ (since $x>0$).From the geometry, $\cos \alpha = \frac{x}{6} = \frac{3}{6} = \frac{1}{2}$.Therefore, $\alpha = \cos^{-1}\left(\frac{1}{2}\right) = 60^{\circ} = \frac{\pi}{3}$.Answer:$\frac{\pi}{3}$