Question

A training field is formed by joining a rectangle and two semcircles, as shown below. The rectangle is 85 m long and 64 m long and 64 m wide. Find the area of the training field. Use the value 3.14 for pi, and do not round your answer. Be sure to include unit in your answer.

Answer 1

$\text{8655.36 m}^2$

Explanation

Step 1

to find the area of the shape , we can divide the area in more known shapes

for area 1 and 2,that is the area ofa circle with diameter 54 m

$\begin{gathered} Area\text{ of the circle=}\pi\cdot\frac{diameter^2}{4} \\ Area\text{ of the circle=3.14}\cdot\frac{(64m)^2}{4} \\ Areaofthecircle=3215.36m^2 \end{gathered}$

Step 2

area of the rectangle

$\begin{gathered} \text{Area of the rectangle=width}\cdot length \\ \text{Area of the rectangle=}\cdot85m\cdot64m \\ \text{Area of the rectangle=5440 m}^2 \end{gathered}$

Step 3

the total area is the sum of the rectangle area plus the circle area

$\begin{gathered} \text{total area=3215.36 m}^2+5440m^2 \\ \text{total area=8655.36 m}^2 \end{gathered}$

I hope this helps you