Question

An athletic track 14 m wide consists of two straight sections 120 m long joining semi-circular ends whose inner radius is 35 m. Calculate the area of the track.

Answer 1

We have

$OB=35m$ and $AB=CD=14m,OA=(35+14)m=49m$

Now, Area of the track = Area of rectangle ABCD +Area of rectangle EFGH + 2{Area of the semi-circle with radius 49m} - 2{ Area of the semi-Circle with radius 35m}

$=(AD\times AB)+(EH\times EF)+2\left[\frac{1}{2}\pi \right(49{)}^2]-2[\frac{1}{2}\times \pi \left(35{)}^2\right]$

$=(14\times 120)+(14\times 120)+2\{\frac{1}{2}\times \frac{22}{7}\times (49{)}^2\}-2\{\frac{1}{2}\times \frac{22}{7}\times (35{)}^2\}$

$=\{3360+\frac{22}{7}(49+35\left)\right(49-35\left)\right\}=7056m^2$