Consider the race track as shown below,

The inner and outer radius of track is ‘r’ cm and ‘R’ cm respectively.

Let inner and outer circumference be ‘C₁’ and C₂’ respectively.

C₁ = 352 m and C₂ = 396 m.

We know,

Circumference of circle = 2πr

Where r = radius of the circle

C₁ = 2πr and C₂ = 2πR

⇒ 2πr = 352 and 2πR = 396

On rearranging,

⇒ r = 56 m and R = 63 m

So, the width of the race track = R – r,

⇒ Width of the race track = 63 – 56

⇒ Width of the race track = 7 m

Area of race track = area of outer circle – area of inner circle

⇒ Area of track = πR² – πr² (put values of r and R)

⇒ Area of track = π(63²) – π(56²)

⇒ Area of track = π(63² – 56²) (taking π common from R.H.S)

⇒ Area of track = π(3969 – 3136)

⇒ Area of track = π×833

= 22×119

= 2618 m²

The width of tack is 7 m and area of track is 2618 m².