A race track is in the shape of a ring whose inner and outer circumferences are 440 m and 506 m, respectively. What is the cost of levelling the track at ₹6/m² ? (Take π = $\frac{22}{7}$)

₹ 29,799

We know that,

Internal circumference of the track = 2πr

2πr = 440

= 2 × \(\frac{22}{7}\)× r = 440

= r = 70 m

External circumference of the track = 2πR

2πR = 506

= 2 × \(\frac{22}{7}\)× R = 506

= R = 506 × \(\frac{7}{44}\) = 80.5 m

Area of track = π (R² – r²) = \(\frac{22}{7}\)× (80.5² – 70²) = \(\frac{22}{7}\)× 150.5 × 10.5 = 4966.5 m²

Cost of 1 m² = Rs. 6

Cost of 4966.5 m² = 4966.5 × 6 = Rs. 29799