Solving for the Width of a Rectangle with Given Area

Understanding and solving algebraic problems like finding the dimensions of a rectangle can be an interesting and challenging task. Let's explore a specific example involving a rectangle where the length is 3 feet more than its width and the area is 54 square feet. How do we find the width of the rectangle?

Setting Up the Problem Using Algebra

We begin by defining the width of the rectangle as w feet. The problem states that the length l is 3 feet longer than the width, so we can express it as:

l w 3

The area A of a rectangle is given by the formula:

A l × w

Given that the area is 54 square feet, we can set up the equation:

54 (w 3) × w

Formulating and Solving the Equation

Expanding the right-hand side of the equation, we get:

54 w^2 3w

To simplify the equation, we rearrange it as:

w^2 - 3w - 54 0

This is a quadratic equation, which can be solved using the quadratic formula:

W frac{-b pm sqrt{b^2 - 4ac}}{2a}

Here, the coefficients are: a 1, b -3, and c -54. Plugging these values into the quadratic formula, we have:

b^2 - 4ac (-3)^2 - 4(1)(-54) 9 216 225

W frac{-(-3) pm sqrt{225}}{2(1)} frac{3 pm 15}{2}

This gives us two possible solutions for W:

W frac{18}{2} 9 and W frac{-12}{2} -6

Since the width cannot be negative, we take the positive solution:

W 9 feet

Verifying the Solution

To verify this solution, let's consider the length of the rectangle. Since the length is 3 feet more than the width, we have:

l 9 3 12 feet

The area is then:

A w × l 9 × 12 108

Oops, we need to check our calculation. The correct width should be 6 feet, as:

W 6

Then the length is:

l 6 3 9 feet

The correct area is:

A 6 × 9 54

Therefore, the width of the rectangle is 6 feet and the length is 9 feet.