Exams › JEE Advanced › Maths
Correct answer: 6*sqrt(3)*r
For a triangle with incircle radius r: Area = r * s, where s = semi-perimeter = P/2. So Area = r * P/2. For an isosceles triangle with base 2a and equal sides b: Area = a * sqrt(b² - a²), P = 2a + 2b, s = a + b. Constraint: Area = r * s gives a*sqrt(b²-a²) = r*(a+b). Let t = b/a. Then sqrt(t²-1) = r*(1+t)/a, so a = r*(1+t)/sqrt(t²-1). Perimeter P = 2a*(1+t) = 2r*(1+t)²/sqrt(t²-1). Minimize over t > 1. Let f(t) = (1+t)²/sqrt(t²-1). Set df/dt = 0: 2(1+t)*sqrt(t²-1) - (1+t)² * t/sqrt(t²-1) = 0. Dividing by (1+t)/sqrt(t²-1): 2(t²-1) - (1+t)*t = 0. 2t² - 2 - t - t² = 0. t² - t - 2 = 0. (t-2)(t+1) = 0. So t = 2 (since t > 1). At t = 2 (b = 2a): a = r*(1+2)/sqrt(4-1) = 3r/sqrt(3) = r*sqrt(3). P = 2a + 2b = 2*r*sqrt(3) + 4*r*sqrt(3) = 6*r*sqrt(3) = 6*sqrt(3)*r.