Answer:
6343.5 m^2
Explanation:
If you cut off the semicircular ends of this track and fit them together, you get a perfect circular area of radius 25 m.
The area of the circle is A = (pi)(25 m)^2, or A = 625pi m^2.
The rectangular middle portion of the track area is 2(25 m), or 50 m wide and 87.6 m long. Thus the area of this portion is (50 m)(87.6 m), or 4380 m^2.
Adding together the areas of the rectangle and circle, we get the total area:
A = 4380 m^2 + 625pi m^2, which evaluates to 6343.5 m^2