What is the area of the circular Track? Statement-I: The area of equilateral triangle that can be inscribed in circular Track is 48√3 m2 Statement-II: The area of square of maximum size that can be inscribed in a circular Track is 196 m2.

Statement I:Area of equilateral triangle = 48√3 m2. (√3/4) × (side) 2 = 48√3 Side = √192 = 8√3 m Statement I: Area of equilateral triangle = 48√3 m2. (√3/4) × (side) 2 = 48√3 Side = √192 = 8√3 m Therefore, AD = √[(8√3)² – (4√3)²] = √144 = 12 m So, radius of circular field = OA = (2/3) × 12 = 8 m Therefore, area of the circular garden = πT × 8 × 8 = 64π m² Therefore, the question can be solved using statement-l alone. Statement-II: Are of square = 196 m² So, side of square =√196 = 14 m So, diagonal of square = √2 × side of square = 14√2 m Also, diagonal of square = diameter of circular garden So, radius of circular garden = 14√2/2 = 7√2 m Therefore, area of circular garden = π (7√2)² = 308 m² Therefore, questions can be solved using statement-ll alone. Hence, option c.