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ATQ, Given: Inner circumference of racetrack (in the ring form) = 220m Outer circumference of racetrack =352m Let the radius of inner track be 'r' 2πr=220m[ Circumference of the circle =2πr] 2 × 22/7 × r = 220m r = (220m × 7)/(2 × 22) = 35m Let the radius of inner track be 'R' 2πr = 352m [ Circumference of the circle =2πr] 2 × (22/7) × R = 220m R = (352m × 7)/(2 × 22) = 56m ∴Width of track = radius of outer track − radius of inner trac = 56 - 35 = 21m Area of the track = Area of outer track − Area of the inner track πR2 -πr2 [∵ Area of the circle = π(radius)2} π(R2 - r2) π(R− r)(R+r)[∵(a2−b2)=(a−b)(a+b)] = (22/7) × (56m - 35m)(56m+35m) = (22/7) × 21m × 91m = (42042/7) = 6006 m²
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