SOLUTION: The area of a rectangle is 135 square feet. The width is 6 feet less than the length. What are the dimensions of the rectangle? The area of a rectangle is 135 square feet. The

Algebra ->  Rectangles -> SOLUTION: The area of a rectangle is 135 square feet. The width is 6 feet less than the length. What are the dimensions of the rectangle? The area of a rectangle is 135 square feet. The       Log On


   

Question 337741: The area of a rectangle is 135 square feet. The width is 6 feet less than the length. What are the dimensions of the rectangle?
The area of a rectangle is 135 square feet. The width is 6 feet less than the length. What are the dimensions of the rectangle?

what we know:
W=L-6
p = 2L +2W
A= L*W
135 = A
135 = L *(L-6)
135 = L^2 -6L
0= L^2-6L -135
0=(L-15)(L+9)
L=15 OR -9 L CAN NOT BE NEGATIVE DONT MAKE SENSE.
PLUG IN ABOVE EQUATIONS: W=L-6 AND W=15-6=9
DIMENSION IS L=15 FEET AND W=9 FEET
Found 2 solutions by nerdybill, chiexpert:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is 135 square feet. The width is 6 feet less than the length. What are the dimensions of the rectangle?
.
Let x = length
then
x-6 = width
.
since
width * length = area
we have:
x(x-6) = 135
x^2 - 6x = 135
x^2 - 6x - 135 = 0
(x-15)(x+9) = 0
.
x = {-9, 15}
we can throw out the negative solution (doesn't make sense) leaving
x = 15 feet (length)
.
width:
x-6 = 15-6 = 9 feet
.
dimensions:
9 by 15 feet

Answer by chiexpert(48) About Me  (Show Source):
You can put this solution on YOUR website!

The area of a rectangle is 135 square feet. The width is 6 feet less than the length. What are the dimensions of the rectangle?

what we know:
W=L-6
p = 2L +2W
A= L*W
135 = A
135 = L *(L-6)
135 = L^2 -6L
0= L^2-6L -135
0=(L-15)(L+9)
L=15 OR -9 L CAN NOT BE NEGATIVE DONT MAKE SENSE.
PLUG IN ABOVE EQUATIONS: W=L-6 AND W=15-6=9
DIMENSION IS L=15 FEET AND W=9 FEET