Question

a training field is formed by joining a rectangle and two semicircles, as shown below. the rectangle is 96 m long and 64 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 area of a rectangle is given by \(A_{rectangle}=length\times width\). Here, \(length = 96\space m\) and \(width=64\space m\). So, \(A_{rectangle}=96\times64 = 6144\space m^{2}\)

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

The diameter of the circle \(d = 64\space m\), so the radius \(r=\frac{d}{2}=\frac{64}{2}=32\space m\). The area of a circle is \(A_{circle}=\pi r^{2}\). Using \(\pi = 3.14\) and \(r = 32\space m\), we have \(A_{circle}=3.14\times32^{2}=3.14\times1024 = 3215.36\space m^{2}\)

Step3: Calculate the total area of the training field

The total area \(A=A_{rectangle}+A_{circle}\). Substitute the values of \(A_{rectangle}\) and \(A_{circle}\): \(A=6144 + 3215.36=9359.36\space m^{2}\)

Answer:

\(9359.36\space m^{2}\)