- Set up the perimeter equation: \(2w + 2(4 + w) = 60\).
- Simplify the equation to solve for the width: \(4w + 8 = 60\), which gives \(w = 13\).
- Factoring the linear equation as \(4(w - 13) = 0\) helps verify the result.
- Calculate the length: \(13 + 4 = 17\).
- The length of the rectangle is \(\boxed{17}\) yards.
Explanation
- Understanding the Problem
The problem defines the width of a rectangle as \(w\) (in yards) and the length as \(4 + w\). The perimeter of a rectangle is given by the formula \(P = 2 \times \text{width} + 2 \times \text{length}\). We are given that the perimeter is \(60\) yards, which leads to the equation \(2w + 2(4 + w) = 60\). Our goal is to solve for \(w\) and then determine the length of the rectangle. Although some methods involve quadratic forms, we will follow the given equation directly.
- Solving the Equation
Starting with the perimeter equation:
\[2w + 2(4 + w) = 60\]
Distribute the 2 into the parenthesis:
\[2w + 8 + 2w = 60\]
Combine the terms containing \(w\):
\[4w + 8 = 60\]
Subtract 8 from both sides of the equation:
\[4w = 52\]
Divide by 4:
\[w = \frac{52}{4} = 13\]
Thus, the width of the rectangle is \(13\) yards. Although the instructions mentioned the zero product property, we can express the linear equation in a factored form as \(4(w - 13) = 0\), which allows us to identify the root \(w = 13\).
- Determining the Length
Now that we have the width \(w = 13\), we can find the length of the rectangle. The problem states that the length is \(4 + w\). Substituting the value of \(w\):
\[\text{Length} = 4 + 13 = 17\]
The length of the rectangle is \(17\) yards.
Examples
Knowing how to solve for dimensions based on perimeter is essential in construction and landscaping. For example, if you are building a fence around a rectangular garden where the length must be 4 feet longer than the width and you have 60 feet of fencing, you can use these algebraic steps to determine exactly how long to cut each side of your fence to ensure the layout is perfect and the materials are used efficiently.