A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.Find the area of the training field. Use the value 3,14 for pie , and do not round your answer. Be sure to include the correct unit in your answer.

Answer:[tex]11402.66 m^{2}[/tex]Step-by-step explanation:The width of rectangle is the diameter of the semi-circle partArea 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 obtainTotal 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 widthRectangular 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]