NDA Mathematics Question (2025) — Solution

Step-by-step solution

Let AB = c, BC = a, and AC = b. Given that the triangle is right-angled at B, we have b^2 = a^2 + c^2. We are also given AB + AC = 3 c + b = 3 b = 3 - c. Substituting b in the Pythagoras theorem: a^2 = (3 - c)^2 - c^2 = 9 - 6c + c^2 - c^2 = 9 - 6c The area of the triangle is = 1 2 ac = 1 2 c 9 - 6c . To maximize the area, we can maximize its square: S = ^2 = 1 4 c^2(9 - 6c) = 9 4 c^2 - 3 2 c^3 Differentiating S with respect to c and equating to zero for maximum area: dS dc = 9 2 c - 9 2 c^2 = 0 c(1 - c) = 0 c = 1 (since c > 0) For c = 1, b = 3 - 1 = 2. In the right-angled ABC, we have: A = AB AC = c b = 1 2 Therefore, A = 3 . Answer: 3