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Question 212869: The length of a rectangular room is 2 feet longer than twice the width. If the room's perimeter is 136 feet, what are the room's dimensions?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! Start by letting x= the width of the room
2x+2 = length of the room.
The formula is 2W+WL = P
2(x) + 2(2x+2) = 136
2x+ 4x + 4 = 136
6x + 4= 136
6x=132
x= 132/6=22 Width
2x+2= Length
2(22) +2= 46= Length
Check:
2W+2L=P
2(22) + 2(46)
44+92=136
It checks!!
If you need an informal explanation of Word Problems, in particular Perimeter Problems, together with examples, exercises and answers, please see my own website. Do a "Bing" search for my last name "Rapalje" and look for "Rapalje Homepage" near the top of the search list. Click on the link "Basic, Intermediate and College Algebra: One Step at a Time" near the top of my Homepage. Select "Basic Algebra", and look in "Chapter 1" for the topic "Word Problems". Also, check out my "MATH IN LIVING COLOR" pages on this topic to see solutions to a lot of these problems, IN COLOR of course! I have a LOT of explanation and examples of PERIMETER problems!
R^2
Dr. Robert J. Rapalje, Retired
Seminole Community College
Altamonte Springs Campus
Florida
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