Question

The area of a rectangle is 33ft, and the length of the rectangle is 5 ft less than twice the width. Find the dimensions of the rectangle

Answer 1

The dimensions are 164 ft (33x5=165)

Answer 2

Hi there!

`\large\boxed{\text{ L = 6 ft, w = 5.5 ft}}`

Use the following equation for the area of a rectangle:

A = l × w

Since the length is 5 ft less than twice the width, we can write an expression for the length in terms of the width:

l = 2w - 5

Plug the given area and expression into the equation:

33 = (2w - 5) × w

Distribute:

33 = 2w² - 5w

Move everything over to one side:

0 = 2w² - 5w - 33

Factor the equation:

0 = (2w - 11)(w + 3)

Find each w value by setting each expression equal to 0:

0 = 2w - 11

11 = 2w

w = 5.5

0 = w + 3

-3 = w

Since dimensions must be positive, the width is 5.5 ft.

Plug this value into the initial equation to solve for length:

33 = 5.5l

33/5.5 = l

l = 6 ft.

The dimensions are length = 6ft, and width = 5.5ft.