The inside perimeter of a running track (shown in Fig. 20.24) is 400 m.

Solution:

It is given that the inside perimeter of the running track is $400 \mathrm{~m}$. It means the length of the inner track is $400 \mathrm{~m}$.

Let $\mathrm{r}$ be the radius of the inner semicircles.

Observe: Perimeter of the inner track=Length of two straight portions of $90 \mathrm{~m}+$ Length of two semicircles

$\therefore 400=(2 \times 90)+(2 \times$ Perimiter of a semicircle $)$

$400=180+\left(2 \times \frac{22}{7} \times \mathrm{r}\right)$

$400-180=\left(\frac{44}{7} \times \mathrm{r}\right)$

$\frac{44}{7} \times \mathrm{r}=220$

$\mathrm{r}=\frac{220 \times 7}{44}=35 \mathrm{~m}$

$\therefore$ Width of the inner track $=2 \mathrm{r}=2 \times 35=70 \mathrm{~m}$

Since the track is $14 \mathrm{~m}$ wide at all places, so the width of the outer track: $70+(2 \times 14)=98 \mathrm{~m}$

$\therefore$ Radius of the outer track semicircles $=\frac{98}{2}=49 \mathrm{~m}$

Area of the outer track $=($ Area of the rectangular portion with sides $90 \mathrm{~m}$ and $98 \mathrm{~m})+$ $(2 \times$ Area of two semicircles with radius $49 \mathrm{~m})$

$=(98 \times 90)+\left(2 \times \frac{1}{2} \times \frac{22}{7} \times 49^{2}\right)$

$=(8820)+(7546)$

$=16366 \mathrm{~m}^{2}$

And, area of the inner track $=($ Area of the rectangular portion with sides $90 \mathrm{~m}$ and $70 \mathrm{~m})+$ ( $2 \times$ Area of the semicircle with radius $35 \mathrm{~m}$ )

$=(70 \times 90)+\left(2 \times \frac{1}{2} \times \frac{22}{7} \times 35^{2}\right)$

$=(6300)+(3850)$

$=10150 \mathrm{~m}^{2}$

$\therefore$ Area of the running track $=$ Area of the outer track - Area of the inner track

$=16366-10150$

$=6216 \mathrm{~m}^{2}$

And, length of the outer track $=(2 \times$ length of the straight portion $)$ $+(2 \times$ perimeter of the semicircles with radius $49 \mathrm{~m})$

$=(2 \times 90)+\left(2 \times \frac{22}{7} \times 49\right)$

$=180+308$

$=488 \mathrm{~m}$