Need help with a perimeter of retangle problem

italianprincess6807

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Jun 17, 2009
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I need help with this question:

A retacngular dog pen is enclosed with 82 feet of fencing. If the width is 3 feet longer than length, what are the dimensions of the dog pen?

I know the formula is P=2L+2W
I put so far 82=2(3+L)+2W after that I don't even think that is right. Anybody that understands this can you please help me to understand how to do this.

I appreciate it.
Thanks
 
perimeter P = 2x width[W] plus length [L]
P=2[w+L]

given:
P = 82 ft.
w=L+3 ft

substitute for P
82=2[w+l]
substitute for W
82=2[L+3+L]
simplify
82=2[2L+3]
clear bracket
82=4L+6
subtract 6 from each side
76=4L
divide both sides by 4
L=19 ft answer
then W= 22 ft answer

CHECK
2[W+L]=?82
2[19+22]=?82
2[41]=?82 yes

Arthur
 
Usually the length is greater than the width. Therefore, I feel most people would assign the variables as l=w+3 rather than the other way around. Small thing. Not all that important.
 
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