Solving for the Width of a Rectangle Given Its Perimeter

Have you ever encountered a problem where you're given the perimeter of a rectangle and need to find its dimensions? Let's work through a classic example: if a rectangle's width is 3 less than its length and the perimeter is 26 feet, what is the width of the rectangle? We'll break down the solution step-by-step to help you understand how to solve similar problems.

Step-by-Step Solution

The first step is to set up an equation using the information provided. Let's assume the length of the rectangle is denoted as L and the width as W. According to the problem, the width (W) is 3 less than the length (L). So:

1. Define Variables

Let L length of the rectangle
tThen W L - 3

2. Use the Perimeter Formula

The perimeter (P) of a rectangle is given by the formula:

[text{Perimeter} 2 times (text{length} text{width})]

Given that the perimeter is 26 feet, the equation becomes:

[2L 2W 26]

Substitute W L - 3 into the perimeter formula:

[2L 2(L - 3) 26]

3. Simplify and Solve the Equation

Simplify the equation:

[2L 2L - 6 26] [4L - 6 26]

Add 6 to both sides:

[4L 32]

Divide both sides by 4:

[L 8]

Hence, the length of the rectangle is 8 feet.

The width of the rectangle is:

[W L - 3 8 - 3 5]

So, the width of the rectangle is 5 feet.

Alternative Method: System of Equations

Here's another approach using a system of equations:

Let W be the width and L be the length of the rectangle. We know two things:

1. The width is 3 less than the length (W L - 3)

2. The perimeter of the rectangle is 26 feet (2(L W) 26)

Substitute W L - 3 into the perimeter equation:

[2(L (L - 3)) 26]

Simplify the equation:

[2(2L - 3) 26] [4L - 6 26]

Add 6 to both sides:

[4L 32]

Divide by 4:

[L 8]

The length is 8 feet. The width can be found as:

[W L - 3 8 - 3 5]

Hence, the width is 5 feet.

Conclusion

Solving the width of a rectangle given its perimeter can be done systematically by setting up equations and simplifying them. For the given problem where the width is 3 less than the length and the perimeter is 26 feet:

Length (L) 8 feet Width (W) 5 feet

This method can be applied to similar problems involving the perimeter and dimensions of rectangles.