Question

Let the radius of the each circular park be r and the distance to be traversed by the sprinters a, b and c be

Answer 1

Answer:

Distance travelled by A = a = 3 × 2r = 6r.

ΔA1B1D is a 30°, 60°, 90° triangle.

So, B1D = √3r/2

⇒ B1B2 = 2r + 2 × √3r/2 = r (2 + √33)

⇒ Distance travelled by B

= b = 3 × r (2 + √3) = 3r (2 + √3)

ΔA1C1E is a 30°, 60°, 90° triangle.

So, C1E = √33r.

⇒ C1C2 = 2√3 r + 2 r = 2r (1 + √3)

⇒ Distance travelled by C = c = 3 × 2r (1 + √3) = 6r (1 + √3)

Now, b - a = 3√33r and c - b = 3√3r.