SOLUTION: Solve and show work. 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.

Algebra ->  Rectangles -> SOLUTION: Solve and show work. 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.      Log On


   

Question 72630: Solve and show work.
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.
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
Solve and show work.
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.

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

"Write down twice the width, 2w, and then 
add 2 to make it 2 cm more, getting
2w + 2".  So LENGTH = 2w+2

Draw a rectangle and place w
on the widths and 2w+2 on the lengths: 

        ___________________
       |       2w+2        |
      w|                   |w
       |                   |
       |___________________| 
               2w+2  

Perimeter means add all the sides all the way around 
the rectangle:

Perimeter = Left side + Top side + Right side + Bottom side

   52     =      w    +    2w+2  +      w     +    2w+2

       52 = w + 2w + 2 + w + 2w + 2

Solve that and get w = 8 

Now find the length by substituting in

LENGTH = 2w+2 
LENGTH = 2(8) + 2 = 16+2 = 18

So the length = 18 cm and the width is 8 cm.

Edwin