SOLUTION: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.

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Question 73092: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
Found 2 solutions by checkley75, rmromero:
Answer by checkley75(3666)   (Show Source): You can put this solution on YOUR website!
L=2W+2
2L+2W=PERIMETER
2(2W+2)+2W=52
4W+4+2W=52
6W=52-4
6W=48
W=48/6
W=8 ANSWER FOR THE WIDTH. THUS:
L=2*8+2
L=16+2
L=18 ANSWER FOR THE LENGTH
PROOF
2*18+2*8=52
36+16=52
52=52
Answer by rmromero(383)   (Show Source): You can put this solution on YOUR website!


The length of a rectangle is 2 cm more than twice its width.  

If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.



What is asked in the problem?

 find the dimensions of the rectangle.



Given:

 The length of a rectangle is 2cm more than twice its width

 The perimeter of the rectangle is 52cm



Representation:

  Let x = the width of the rectangle

     2x+2 = the length of the rectangle



Equation

          2l + 2w = P

     2(2x+2) + 2x = 52        Multiply (Distributive Property)

      4x + 4 + 2x = 52        Combine like terms

           6x + 4 = 52

       6x + 4 - 4 = 52 - 4    subtract 4 both sides

               6x = 48        divide both sides by 6

                x = 8 cm -----> Width



     length = 2x + 2 

            = 2(8) + 2  

            = 16 + 2

            = 18 cm    ----- length



Checking



         2L + 2W = P

    2(18) + 2(8) = 52

         36 + 16 = 52

              52 = 52





 

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