SOLUTION: The length of a rectangle is 5 in more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.
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Question 88492: The length of a rectangle is 5 in more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Let x = the width
:
Write what it says using x = width:
"The length of a rectangle is 5 in more than twice its width."
Length = 2x + 5
:
"If the perimeter of the rectangle is 52 in.",
We know the perimeter is: 2L + 2W = 52:
:
substitute (2x+5) for length and x for width:
:
2(2x+5) + 2x = 52
:
We can simplify, divide equation (each term) by 2:
(2x+5) + x = 26
2x + x = 26 - 5
3x = 21
x = 21/3
x = 7 is the width of the rectangle
:
Find the length so we can check our solution:
2(7) + 5 = 19 is the length
:
2(19) + 2(7) = 52; proves our solution
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