SOLUTION: the lenght of a rectangle exceeds its width by 2yd. if the perimneter of the recatangle is 88 ft find the dimensions

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Question 24733: the lenght of a rectangle exceeds its width by 2yd. if the perimneter of the recatangle is 88 ft find the dimensions

Found 2 solutions by Paul, rapaljer:
Answer by Paul(988) (Show Source):
You can put this solution on YOUR website!
Length = x+2
Width = x
Formula = 2(l+w) = P

2(x+x+2)=88
4x+4=88
4x=84
x = 21
21+2=23
Hence the lenght is 23m, and the width is 21m.
Paul.

Answer by rapaljer(4671) (Show Source):
You can put this solution on YOUR website!
Before beginning this problem, there is a bit of a trick to it. Notice that the length of the rectangle exceeds the width by 2 yards, but the perimeter is in FEET. It might be a good idea to change the 2 yards to 6 feet.

Let x = width
x+6 = length
2(W) + 2(L) = Perimeter
2(x) + 2(x+6) = 88
2x + 2x + 12 = 88
4x + 12 = 88
4x +12 - 12 = 88-12
4x = 76

Divide both sides by 4:

ft = width
feet = length

Check:
2(W) + 2(L)
2(19) + 2(25)
38 + 50 = 88 feet
It checks!

R^2 at SCC