Question

A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 86m long and 64m 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. 64m 86m

Answer 1

To solve this problem, we have to find the area of the semi circles and the area of the rectangle and add them up together.

The area of the field is equal to 8719.36m^2

Area of the Field

Let us start by finding the area of the rectangle.

Data;

• width = 86m
• length = 64m

Area of a rectangle is given as

$a = length * width \\a = 86 * 64\\a = 5504m^2$

The area of the rectangle is 5504m^2

Area of the semicircle

The formula of a semi circle is given as

$A = \frac{1}{2} \pi r^2$

But since we have two semi circles here, we can simply multiply it by two and the formula becomes

$A = 2 * \frac{1}{2} \pi r^2\\A = \pi r^2$

But in the question, we have the diameter of the semi circle as the length of the rectangle.

$radius = \frac{diameter}{2} \\radius = \frac{64}{2} \\radius = 32m$

The radius of the semicircle is 32m

Area of the semicircle is

$A = \pi r^2\\A = 3.14 * 32^2\\A = 3215.36m^2$

The area of the field is

$area of field = 3215.36 + 5504\\area = 8719.36m^2$

The area of the field is equal to 8719.36m^2

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