Answer:
[tex]11402.66 m^{2}[/tex]
Step-by-step explanation:
The width of rectangle is the diameter of the semi-circle part
Area of one semicircle is given by [tex]\frac {0.5\pi d^{2}}{4}[/tex]
Total area of semi circle will be [tex]2\times\frac {0.5\pi d^{2}}{4}[/tex]
Substituting 74 m for d and [tex]\pi[/tex] as 3.14 we obtain
Total area semi-circle=[tex]2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}[/tex]
Area of rectangle is given by the product of length and width
Rectangular area=[tex]96 m*74 m=7104 m^{2}[/tex]
Total area of rectangular and semi-circles will be
[tex]4298.66 m^{2}+7104 m^{2}=11402.66 m^{2}[/tex]
Therefore, area of training field is [tex]11402.66 m^{2}[/tex]
