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 a^2 + c^2 = b^2. We are given AB + AC = 3, which means c + b = 3 b = 3 - c. Substituting b into the Pythagorean theorem: a^2 + c^2 = (3 - c)^2 a^2 + c^2 = 9 - 6c + c^2 a^2 = 9 - 6c a = 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: S = 1 4 c^2(9 - 6c) = 9 4 c^2 - 3 2 c^3 Differentiating S with respect to c and equating to zero: dS dc = 9 2 c - 9 2 c^2 = 0 9 2 c(1 - c) = 0 Since c > 0, we get c = 1. For c = 1, d^2S dc^2 = 9 2 - 9c = - 9 2 Substituting c = 1 into the expression for area: = 1 2 (1) 9 - 6(1) = 3 2 square units. Answer: 3 2 square unit