SOLUTION: The Perimeter of a certain playing field is 222ft the field is a rectangle, and the length is 49 ft longer then the width. Find the dimensions. Personal note: If I don

Algebra ->  Rectangles -> SOLUTION: The Perimeter of a certain playing field is 222ft the field is a rectangle, and the length is 49 ft longer then the width. Find the dimensions. Personal note: If I don      Log On


   

Question 1014678: The Perimeter of a certain playing field is 222ft the field is a rectangle, and the length is 49 ft longer then the width. Find the dimensions.




Personal note: If I done my math correct, I think you are suppose to add 49+49 which gets 98. subtract 98 from 222 ft and then divide it by 49. which equals to 2.530612245, now if you times that with 49 you should get 122.5.
I been working on this problem for the past couple of days, and maybe I'm either overthinking it or I'm just doing it wrong.
Found 2 solutions by addingup, macston:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
The Perimeter of a certain playing field is 222ft the field is a rectangle, and the length is 49 ft longer then the width. Find the dimensions
2l+2w = 222 Your problem says that l= w+49, so substitute:
2(w+49)+2w = 222
2w+98+2w= 222 Add w on left and subtract 98 on both sides:
4w= 124 Divide both sides by 4:
w= 31 and the length:
31+49= 80

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
W=width; L=length=W+49
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P=2(L+W)
222 ft=2(L+W) Divide each side by 2
111 ft=L+W . Substitute for L
111 ft=(W+49ft)+W
111ft=2W+49ft Subtract 49 ft from each side.
62ft=2W
31ft=W
ANSWER 1: The width is 31 feet.
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L=W+49ft=31ft+49ft=80ft
ANSWER 2: The length is 80 feet.
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CHECK:
P=2(L+W)
222ft=2(80ft+31ft)
222ft=2(111ft)
222ft=222ft