SOLUTION: The perimeter of a rectangle is 208 inches. The length exceeds the width by 54 inches. Find the length and the width

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Question 320175: The perimeter of a rectangle is 208 inches. The length exceeds the width by 54 inches. Find the length and the width
Found 2 solutions by checkley77, Sunny Day:
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
L=W-54
P=2L+2W
208=2(W-54)+2W
208=2W-108+2W
208=4W-108
4W=208+108
4W=316
W=316/4
W=79 ANS. FOR THE WIDTH.
L=79-54=25 ANS. FOR THE LENGTH.
PROOF:
208=2*25+2*79
208=50+158
208=208

Answer by Sunny Day(15) (Show Source):
You can put this solution on YOUR website!
if the length and width are represented by l and w respectively,
perimeter = 2(l+w) = 208 ------- (1)
length is greater than width by 54 inches.
So, l = 54 + w ------- (2)
Substituting for l in (1), we get
2(54 + w + w) = 208
2(54 + 2w) = 208
Removing the brackets,
2X54 + 2X2w = 208
108 + 4w = 208
4w = 208 - 108
or 4w = 100
or w = 100/4 = 25
So l = 54 + w = 54 + 25 = 79
Hence length = 79 inches and width = 25 inches.
[On cross checking, you can see that the length is 54 inches greater than the width and that the perimeter is 208 inches]