Question

a training field is formed by joining a rectangle and two semicircles, as shown below. the rectangle is 87 m long and 58 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: Find area of rectangle

The area of a rectangle is given by \( A_{rectangle} = length \times width \). Here, length \( = 87 \, \text{m} \) and width \( = 58 \, \text{m} \). So, \( A_{rectangle} = 87 \times 58 \). Calculating that: \( 87 \times 58 = 5046 \, \text{m}^2 \).

Step2: Find area of the two semicircles (which make a full circle)

The two semicircles have a diameter equal to the width of the rectangle, \( d = 58 \, \text{m} \), so the radius \( r = \frac{d}{2} = \frac{58}{2} = 29 \, \text{m} \). The area of a circle is \( A_{circle} = \pi r^2 \). Using \( \pi = 3.14 \), we get \( A_{circle} = 3.14 \times (29)^2 \). Calculating \( 29^2 = 841 \), then \( 3.14 \times 841 = 2640.74 \, \text{m}^2 \).

Step3: Find total area of training field

The total area is the sum of the area of the rectangle and the area of the circle (from the two semicircles). So, \( A_{total} = A_{rectangle} + A_{circle} = 5046 + 2640.74 \). Adding these together: \( 5046 + 2640.74 = 7686.74 \, \text{m}^2 \).

Answer:

\( 7686.74 \, \text{square meters} \) (or \( 7686.74 \, \text{m}^2 \))