A race track is in the form of a rig whose inner circumference is 352 m and outer circumference is 396 m.

Solution:

Let r m and R m be the radii of the inner and outer tracks.
Now,

Circumference of the outer track $=2 \pi R$

$\Rightarrow 396=2 \times \frac{22}{7} \times R$

$\Rightarrow R=\frac{396 \times 7}{44}$

$\Rightarrow R=63$

Circumference of the inner track $=2 \pi r$

$\Rightarrow 352=2 \times \frac{22}{7} \times r$

$\Rightarrow r=\frac{352 \times 7}{44}$

$\Rightarrow r=56$

Width of the track = Radius of the outer track - Radius of the inner track

$=63-56$

$=7 \mathrm{~m}$

Area of the outer circle $=\pi R^{2}$

$=\frac{22}{7} \times 63 \times 63$

$=12474 \mathrm{~m}^{2}$

Area of the inner circle $=\pi R^{2}$

$=\frac{22}{7} \times 56 \times 56$

$=9856 \mathrm{~m}^{2}$

Area of the track $=12474-9856$

$=2618 \mathrm{~m}^{2}$