UPSC CDS II 2025 Mathematics Paper

Concept:Use the area formula and Pythagoras theorem for an isosceles triangle to find possible base lengths. Then test each statement to see if it gives a unique answer.Explanation:Let base $BC = b$ and height $h$.Area: $\frac{1}{2} b h = 48$ ⇒ $b h = 96$.From the isosceles triangle, half‑base and height form a right triangle with the equal side $10$:$h^2 + \left(\frac{b}{2}\right)^2 = 10^2$ ⇒ $h^2 + \frac{b^2}{4} = 100$.Substitute $h = \frac{96}{b}$:$\left(\frac{96}{b}\right)^2 + \frac{b^2}{4} = 100$ ⇒ $\frac{9216}{b^2} + \frac{b^2}{4} = 100$.Multiply by $4b^2$: $36864 + b^4 = 400 b^2$ ⇒ $b^4 - 400 b^2 + 36864 = 0$.Let $x = b^2$, then $x^2 - 400x + 36864 = 0$.Solve: $x = \frac{400 \pm \sqrt{160000-147456}}{2} = \frac{400 \pm 112}{2}$ ⇒ $x = 256$ or $144$.So $b = 16$ or $b = 12$.Statement‑I: BC is an even integer – both 16 and 12 are even, so no unique base is found. Not sufficient alone.Statement‑II: height > half of base.For $b=16$, $h = 96/16 = 6$, half‑base $=8$ ⇒ $6>8$ is false.For $b=12$, $h = 96/12 = 8$, half‑base $=6$ ⇒ $8>6$ is true.Only $b=12$ satisfies the condition. Sufficient alone.Answer:The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone. Option A.