The outer circumference of a circular race-track is 528 m.

Solution:

It is given that the outer circumference C of circular track is 528 m.

We know that the circumference of circle of radius r is

$C=2 \pi r$

Substituting the value of C,

$528=2 \times \frac{22}{7} \times r$

$528 \times 7=44 r$

$r=\frac{528 \times 7}{44}$

$r=84 \mathrm{~m}$

Thus, the radius of outer circle is.

Since circular race track is 14 m wide everywhere. Then

radius of inner circle $=$ radius of outer circle $-14$

$=84-14$

 

$=70 \mathrm{~m}$

We know that the area of circle of radius $\mathrm{r}$ is $A=\pi r^{2}$

So, Area of outer circle $=\frac{22}{7} \times 84 \times 84$

Area of inner circle $=\frac{22}{7} \times 70 \times 70$

Now,

Area of circular track = Area of outer circle-Area of inner circle

$=\frac{22}{7} \times 84 \times 84-\frac{22}{7} \times 70 \times 70$

$=\frac{22}{7}(84 \times 84-70 \times 70)$

$=\frac{22}{7} \times 2156$

Area of circular track $=6776 \mathrm{~m}^{2}$

It is given that,

The cost of levelling the track per square meter $=$ Rs $0.50$

So, The cost of levelling the track 6776 square meter $=$ Rs $0.50 \times 6776$

$=\operatorname{Rs} 3388 /-$