Question

Area involving rectangles and circlesEspañolVA training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 82 m long and 64 m wide.Find the area of the training field. Use the value 3.14 for , and do not round your answer. Be sure to include the correct unit in your answer.

Answer 1

Answer:

8463.36 m^2

Explanation:

Given the length(l) of the rectangle as 82m and the width(w) as 64m, we can go ahead and determine the area of the rectangle as follows;

$\begin{gathered} \text{Area of a rectangle }=l\times b \\ =82\times64 \\ =5248m^2 \end{gathered}$

The two semi-circles are equivalent to one circle with a diameter of 64m, so the radius of the circle will be;

$\text{Radius(r)}=\frac{Diameter(d)}{2}=\frac{64}{2}=32m$

Let's go ahead and determine the area of the circle given pi as 3.14;

$\begin{gathered} Area\text{ of a circle=}\pi\times r^2 \\ =3.14\times(32)^2 \\ =3.14\times1024 \\ =3215.36m^2 \end{gathered}$

Therefore, the area of the training field can be determined by adding the area of the rectangle and that of the two semicircles;

$\begin{gathered} \text{Area of the training field}=\text{Area of rectangle + Area of two semicircles} \\ =5248+3215.36 \\ =8463.36m^2 \end{gathered}$