To find:
The area of the training field.
Solution:
The training field is made of two semicircles and a rectangle.
The length and width of the rectangle is 96 m and 64 m. So, the area of the rectangle is:
[tex]\begin{gathered} A=l\times w \\ =96\times64 \\ =6144\text{ m}^2 \end{gathered}[/tex]
The diameter of the semicircle is 64 m. SO, the radius of the semicircle is 32 m.
The area of two semicircles is:
[tex]\begin{gathered} A=2\times\frac{1}{2}\pi r^2 \\ =3.14\times(32)^2 \\ =3.14\times1024 \\ =3215.36 \end{gathered}[/tex]
So, the area of the training field is:
[tex]\begin{gathered} A=6144+3215.36 \\ =9359.36 \end{gathered}[/tex]
Thus, the area of the training field is 9359.36 m^2.
