Answer : (b) T = S
∠PQA = 90º (Angle in a semi-circle)
∴ (MP) . (MA) = MQ2. (Mean Proportional) ...(i)
Let MP = MB = 1(Given MP = BM)
Since ΔABC is an equilateral Δ,
AM = \(\frac{\sqrt{3}}{2}\) × BC = \( \sqrt{3}\) .
∴ From (i) 1 . \( \sqrt{3}\) = MQ2 → MQ = \(3^{\frac{1}{4}}\)
Area of square = T = (MQ)2 = \( \sqrt{3}\)
Area of ΔABC = S = \(\frac{\sqrt{3}}{4}\) (BC)2
= \(\frac{\sqrt{3}}{4}\) × 4 = \( \sqrt{3}\)
∴ T = S.