SOLUTION: Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 56 feet.
Algebra ->
Rectangles
-> SOLUTION: Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 56 feet.
Log On
Question 804699: Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 56 feet. Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! If you spent a little time on the problem instead of wasting your time writing this.
[The length is______________________feet
The width is_________________________feet]
L=1+2W
56=2W+2L
56=2W+2*(1+2W)
56=2W+2+4W
54=6W
9=W
L=1+2*9
L=19
check
56=2*9+2*19
56=18+38
56=56
ok