NDA 1 2025 Maths, question 74

Topic: Differential calculus

Passage · Q73 to Q74

Let $ABC$ be a triangle right-angled at $B$ and $AB + AC = 3$ units.

Q74

What is the maximum area of the triangle?

(a) $\sqrt{3}/2$ square unitCorrect answer
(b) $\sqrt{3}$ square units
(c) $\sqrt{6}/2$ square units
(d) $\sqrt{6}$ square units

Solution.

Answer: (a) $\sqrt{3}/2$ square unit

With $AB = u$ , the area squared is $K^{2} = u^{2}(9 - 6u)/4$ .
It is maximised at $u = 1$ , so $AB = 1$ , $AC = 2$ and $BC = \sqrt{3}$ .
Therefore $K_{\max} = \frac{1}{2}AB \cdot BC = \sqrt{3}/2$ square unit.