Question

A 14 m wide athletic track consists of two straight sections each 120 m long joined by semi-circular ends with inner radius is 35 m. Calculate the area of the track.

Answer 1

Given: The radius of the inner semi-circle, $r=35m$

Width of the track $=14m$

$\therefore$ Radius of the outer semi-circle, $R=35+14=49m$

The area of the track is the sum of the areas of the semicircular tracks and the areas of the rectangular tracks.

Area of the rectangular tracks ABCD and EFGH $=2\times (l\times b)$

$=2\times 14\times 120=3360m^2$

Area of the semicircular tracks $=2\times$ (Area of the outer semicircle - Area of the inner semicircle)

$=2\times (\frac{1}{2}\pi R^2-\frac{1}{2}\pi r^2)$

$=2\times \frac{1}{2}\times \pi (R^2-r^2)$

$=\frac{22}{7}\times ({49}^2-{35}^2)..........(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

$=\frac{22}{7}(49+35)(49-35)$

$=\frac{22}{7}\times 84\times 14=3696m^2$

$\therefore$ Area of the track $=3360+3696=7056m^2$.