Question

The drawing below (not to scale) shows a running track. The ends are semicircles. The distance between the two inner parallel straight lines is 60 meters, and they are each 100 meters long. The track is 10 meters wide.

Find the area of the actual track.(Shaded region)

Answer 1

100×10×2=2000
60+10+10=80
80/2=40
40²π=1600π
60/2=30
30²π=900π
1600π-900π=700π
Area of shaded region=700π+2000

Answer 2

Refer to the attached image.

We have to find the area of the shaded region.

Area of shaded region = 2(Area of rectangle ABCD) +2(Area of semicircle with radius AO - Area of semicircle with radius BO)

Area of rectangle with length 'l' and width 'w' is given by
Now, area of rectangle ABCD =
`AC * CD`

=
`100 * 10`

= 1000 sq. meters (equation 1)

Area of semicircle with radius 'r' is given by
`(\Pi r^(2))/(2)`

Therefore, Area of semicircle with radius AO - Area of semicircle with radius BO

( AO=AB+BO = 10+30 = 40 m and BO= 30 m)

=
`((\Pi (40)^(2))/(2)-(\Pi (30)^(2))/(2))`

=
`(\Pi )/(2)(40^(2)-30^(2))`

`=(\Pi )/(2)(700)` (equation 2)

Therefore,

Area of shaded region

= 2(1000) + 2(
`(\Pi )/(2)(700)`) (from equations 1 and 2)

= (2000 + 700
`\Pi`) square meters.