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Chapter 1: Problem 145

The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find its length and width.

Short Answer

Expert verified
The width of the rectangular sign is 6 feet and the length is 9 feet.

Step by step solution

01

Define the variables

Let's identify what we know and define variables to represent unknowns. We know the area of the rectangle is 54 square feet and that the length is 3 feet longer than the width. Let's call the width \(w\) in feet and the length \(l\) in feet. Given the relationship of the length being 3 feet longer, \(l = w + 3\).
02

Set up the equation

Now let's use the formula for the area of a rectangle, which is \(Area = Width x Length\) or \(A = w \cdot l\). Substituting the known values, we get \(54 = w \cdot (w + 3)\).
03

Solve the equation

To solve the equation, we need to distribute \(w\) into \(w + 3\): \(54 = w^2 + 3w\). This is a quadratic equation. Rearrange to have 0 on one side: \(0 = w^2 + 3w - 54\).
04

Factor the quadratic equation

We can factor the quadratic equation: \(0 = (w - 6)(w + 9)\).
05

Solve for \(w\)

Setting each factor equal to zero gives the potential solutions \(w = 6\) and \(w = -9\). Since width can't be negative, we dismiss \(w = -9\) and so \(w = 6\) feet.
06

Solve for \(l\)

Substitute \(w\) back into the linear equation from step 1: \(l = w + 3 = 6 + 3 = 9\) feet. Therefore, the length is 9 feet.

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