# The area of a rectangle is 45 square feet. The length is 4 feet longer than the width. What...

## Question:

The area of a rectangle is 45 square feet. The length is 4 feet longer than the width.

What equation could be used to find the length of the rectangle in feet?

## Rectangle: Identifying Sides

A rectangle has two pairs of equal sides: two widths and two lengths. Assume that the length of a rectangle is x and the width of that rectangle is y. The area of such a rectangle can be calculated by {eq}A = xy {/eq}.

Assume that the length of the given rectangle is x and the width is y. The problem states that the length is 4 feet longer than the width:

{eq}x - y = 4...(1) {/eq}

Additionally, we know that the area is:

{eq}A = xy = 45...(2) {/eq}

Let's express the width from equation (1), so that we obtain an equation in terms of length:

{eq}y = x - 4...(3) {/eq}

Substituting equation (3) for y in equation (2), we obtain:

{eq}x(x - 4) = 45...(4) {/eq}

Distributing the terms:

{eq}x^2 - 4x = 45\\ x^2 - 4x - 45 = 0 {/eq}

Solving this quadratic equation with a requirement of:

{eq}x > 0 {/eq}

Will yield the solution for the length of the rectangle in feet.

Therefore, the equation is {eq}\boxed{x^2 - 4x - 45 = 0} {/eq} 