The area of a rectangle is 33 ft2 , and the length of the rectangle is 5 ft less than double the width. Find the dimensions of the rectangle.

Question

Thank you for answering this

Answer 1

So you should start by remembering that the area of a rectangle is equal to the length X width

We know that the area is 33, so

33 = LW

Now we need to translate the second sentence into an equation. The length (L) is (=) 5ft less than (-5) double (x2) the width (W). Rewritten as an equation we get: L=2W-5.

Substitute your equation for length into your equation for area: 33 = (2W-5)W

Distribute: 33 = 2W^2 - 5W

From here you can set your problem equal to zero: 0 = 2W^2 - 5W - 33

And factor (using whatever method you best like): 0 = (2W - 11)(W + 3)

To solve for w, you set each set of parentheses equal to zero, and solve

2W - 11 = 0 W + 3 = 0

W = 11/2 W = -3

Since you can't have a negative length, our width is 11/2 ft

Plug back into our length formula to get it: L = 2 (11/2) -5 = 11-5 = 6ft

Check your answer by multiplying the length and width together: (11/2)x3 = 33ft

Answer 2

area=33^ft

l=2w-5

1089/(2w-5)=w

(2w-5)w=1089

2w^2-5w=1089

w=24.618ft

l=44.236'

Answer 3

Let W = width, then L=2W - 5

Area = LW = (2W-5)W = 33 => 2W² - 5W -33 = 0

This factors pretty easily and gives an extraneous root and the correct answer. Go for it!