Question

The inside perimeter of a running track (shown in the following figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track.

Answer 1

It is given that, and

We know that the circumference C of semicircle of radius be r is

$C=\mathrm{\pi }r$

The inside perimeter of running track is the sum of twice the length of straight portion and circumferences of semicircles. So,
inside perimeter of running track = 400 m
$2l+2\mathrm{\pi }r\mathit{=}400m$
$\Rightarrow 2\times 90+2\times \frac{22}{7}\times r=400m$
$\Rightarrow r=\frac{220\times 7}{2\times 22}=35m$
Thus, radius of inner semicircle is 35 m.

Now,
radius of outer semi circle r' = 35 + 14 = 49 m

Area of running track = $2\times \mathrm{Area}\mathrm{of}\mathrm{rectangle}+2\times \mathrm{Area}\mathrm{of}\mathrm{outer}\mathrm{semi}\mathrm{circle}-2\times \mathrm{Area}\mathrm{of}\mathrm{inner}\mathrm{semicircle}$
$=2\times 90\times 14+2\times \frac{\mathrm{\pi }(49{)}^2}{2}-2\times \frac{\mathrm{\pi }(35{)}^2}{2}$
$=2520+\mathrm{\pi }\times \left(49+35\right)\left(49-35\right)$
$=2520+\frac{22}{7}\times 84\times 14$
$=2520+3696=6216m^2$
Hence, the area of running track = 6216 m²

Now, length L of outer running track is
L = 2 × l + 2$\mathrm{\pi }$r'
$=2\times 90+2\mathrm{\pi }\times 49$
$=180+2\times \frac{22}{7}\times 49$
$=180+308=488m$

Hence, the length L of outer running track is 488 m.