SOLUTION: The length of a rectangle is 1ft. more than twice the width, and the area of the rectangle is 66 ft^2. Find the dimensions of the rectangle. length: width:

Algebra ->  Surface-area -> SOLUTION: The length of a rectangle is 1ft. more than twice the width, and the area of the rectangle is 66 ft^2. Find the dimensions of the rectangle. length: width:      Log On


   

Question 859053: The length of a rectangle is 1ft. more than twice the width, and the area of the rectangle is 66 ft^2. Find the dimensions of the rectangle.
length:
width:
Answer by Awesom3guy(31) About Me  (Show Source):
You can put this solution on YOUR website!
Let's call the sides of the rectangle L and W (length and width).

"The length of a rectangle is 1ft. more than twice the width."
So we write L = 2W + 1.

We know that the area of any triangle is L*W.
L*W = 66
We substitute the value for L from the first equation in the place of L here. Instead of L we write (2W + 1) and we have:
(2W + 1)*W = 66.

Now we have a quadratic equation. Expanding:
2W^2 + W = 66
2W^2 + W - 66 = 0

Now we factor:
(W + 6)(2W - 11) = 0
From this we get that W + 6 = 0, which would give us W = -6. This is impossible.
Or 2W - 11 = 0, which gives W = 5.5.
The width is 5.5 feet.

L = 2W + 1
L = 2*5.5 + 1 = 12
The length is 12 feet.

Check this, 5.5*12 = 66.