Answer:
a
Step-by-step explanation:
We are told that each end of the track is a semicircle, and the inner part of the track is parallel. The width of the track is 10 m, and the length of the parallel lines is 100 m. Therefore, we can determine the area of the long middle portion of the track, as it is simply two rectangles.
The area of one inner part of the track is:
A = l x w
A = 100 m x 10 m
A = 1000 m²
Since we have two of these rectangles, we just multiply this value by two to get the entire area of the track for the inner section.
A = 2000 m²
Now we can tackle the semicircles. If we were to combine the two semicircles, we would get one large circle (shaded) with a smaller (unshaded) circle inside. We can determine the area of both circles and simply subtract the area corresponding to the inner circle.
We are told that the diameter of the inner circle is 60 m (radius of 30m), and since we are also told the track is 10 m wide, so the diameter of the outer circle is d = 60 m + 10 m + 10 m = 80 m, with a radius of 40 m.
The area of a circle is defined as A = πr²
Large circle:
A = π (40)² = 1600π
Inner circle:
A = π (30)² = 900π
The area of the shaded circle is simply the difference in these two areas.
Shaded circle:
A = 1600π - 900π = 700π = 2200 m²
To determine the area of the entire track, we now just combine the area of the shaded circle, with the shaded inner track area we already determined.
Total area of shaded track:
A = 2000 m² + 2200 m²
A = 4200 m²
I think the answer would be 26
since it is 25.6 and 6 is greater than 5 you round the 25 to 26
