Question 1204535: The perimeter of a rectangular field is 296 meters. The length is 52 meters longer than the width. What are the dimensions of the field? Found 2 solutions by Theo, Alan3354:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! p = perimeter
l = length
w = width
l = w + 52
p = 2l + 2w
replace l with w + 52 to get:
p = 2(w + 52) + 2w = 2w + 104 + 2w = 4w + 104
replace p with 296 to get:
296 = 4w + 104
subtract 104 from both sides of the equation to get:
192 = 4w
solve for w to get:
w = 192/4 = 48
l = w + 52 = 48 + 52 = 100
2l + 2w = 200 + 96 = 296
values of l and w look good.
solution is that the dimensions of the field ares 100 by 48.
