Question

a training field is formed by joining a rectangle and two semicircles, as shown below. the rectangle is 94 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 area of a rectangle is given by \(A_{rectangle}=length\times width\). Here, \(length = 94\space m\) and \(width = 56\space m\). So, \(A_{rectangle}=94\times56 = 5264\space m^{2}\)

Step2: Calculate the area of the two semicircles (which together form a circle)

The two semicircles combine to form a circle. The diameter \(d = 56\space m\), so the radius \(r=\frac{d}{2}=\frac{56}{2}=28\space m\). The area of a circle is \(A_{circle}=\pi r^{2}\). Using \(\pi = 3.14\), we have \(A_{circle}=3.14\times28^{2}=3.14\times784 = 2461.76\space m^{2}\)

Step3: Calculate the total area of the training field

The total area \(A = A_{rectangle}+A_{circle}\). So, \(A=5264 + 2461.76=7725.76\space m^{2}\)

Answer:

\(7725.76\space m^{2}\)