Problem 44 - College Algebra

Question

A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions?

Step-by-Step Solution

Answer

The dimensions of the soccer field are a width of 50 yards and a length of 100 yards.

Step 1: Set up the Equation

Denote the width of the field as \( x \) and the length as \( 2x \), since it is twice as long as it is wide. The perimeter of a rectangle is given by \( P = 2l + 2w \), where \( l \) is length, \( w \) is width, and \( P \) is the perimeter. Therefore, the equation from the problem statement is \( 300 = 2(2x) + 2(x) \).

Step 2: Simplify the Equation

Simplify the equation to make it easier to solve for the variable. \( 300 = 4x + 2x \). Merging like terms will give: \( 300 = 6x \).

Step 3: Solve for \( x \)

Now that we have a simple equation, we can solve for \( x \). Divide both sides by 6 to get \( x \). We get \( x = 50 \). Therefore, the width ( \( x \) ) of the soccer field is 50 yards.

Step 4: Solve for \( 2x \)

Next we need to find the length of the soccer field. Using the relation \( 2x = l \) , substitute the value of \( x \) that was calculated in the previous step. We get, \( 2x = 2(50) = 100 \). Therefore, the length of the soccer field is 100 yards.