Question

a training field is formed by joining a rectangle and two semicircles, as shown below. the rectangle is 100 m long and 74 m wi
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 ir
100 m
74 m

Explanation:

Step1: Calculate area of rectangle

The rectangle has length \( l = 100 \, \text{m} \) and width \( w = 74 \, \text{m} \). The area of a rectangle is \( A_{\text{rectangle}} = l \times w \).
\( A_{\text{rectangle}} = 100 \times 74 = 7400 \, \text{m}^2 \)

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

The two semicircles have a diameter equal to the width of the rectangle, so the radius \( r = \frac{74}{2} = 37 \, \text{m} \). The area of a circle is \( A_{\text{circle}} = \pi r^2 \), with \( \pi = 3.14 \).
\( A_{\text{circle}} = 3.14 \times 37^2 = 3.14 \times 1369 = 4298.66 \, \text{m}^2 \)

Step3: Calculate total area of training field

Add the area of the rectangle and the area of the circle (from the two semicircles).
\( A_{\text{total}} = A_{\text{rectangle}} + A_{\text{circle}} = 7400 + 4298.66 = 11698.66 \, \text{m}^2 \)

Answer:

\( 11698.66 \, \text{square meters} \) (or \( 11698.66 \, \text{m}^2 \))