Question

a training field is formed by joining a rectangle and two semicircles, as shown below. the rectangle is 83 m long and 56 m wide. find the area of the training field. use the value 3.14 for π, and do not round your answer. be sure to include the correct unit in your answer.

Explanation:

Step1: Calculate the area of the rectangle

The formula for the area of a rectangle is \(A_{rectangle}=length\times width\). Given \(length = 83m\) and \(width=56m\), so \(A_{rectangle}=83\times56 = 4648m^{2}\).

Step2: Calculate the area of the two - semicircles (which is equivalent to the area of one circle)

The diameter of the circle \(d = 56m\), so the radius \(r=\frac{d}{2}=\frac{56}{2}=28m\). The formula for the area of a circle is \(A_{circle}=\pi r^{2}\). Using \(\pi = 3.14\) and \(r = 28m\), we have \(A_{circle}=3.14\times28^{2}=3.14\times784 = 2461.76m^{2}\).

Step3: Calculate the total area of the training field

The total area \(A=A_{rectangle}+A_{circle}\). Substitute the values of \(A_{rectangle}\) and \(A_{circle}\) into the formula: \(A = 4648+2461.76=7109.76m^{2}\).

Answer:

\(7109.76m^{2}\)