rectangle perim. 588, length 5X greater than width; find dim

M

macninni

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Feb 19, 2009
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I just cannot get started on this problem:

The perimeter of the rectangle is 588m. The length is five times greater than the width. What is the width? What is the length?

I think I need one more number to do this question......how can you do 588 = 5x-x?
 
Re: pre-algebra question

The perimeter of a rectangle is 2L + 2W, where L is the length and W is the width.
 
Re: pre-algebra question

Hello, "macninni!

I assume you need a walk-through . . . with baby-steps . . .

The perimeter of the rectangle is 588m.
The length is five times greater than the width.
What is the width? What is the length?
Click to expand...

A rectangle has a length (L) and a width (W).
Code: 
             L
      * - - - - - - *
      |             |
    W |             |W
      |             |
      * - - - - - - *
             L

We are told: the length is 5 times the width.
Hence: .\(\displaystyle L \,=\,5W\)
Code: 
             5W
      * - - - - - - *
      |             |
    W |             |W
      |             |
      * - - - - - - *
             5W

The perimeter (the total distance around) is: .\(\displaystyle 5W + W + 5W + W \:=\:12W\)

And we are told that the perimeter is 588 m.

So we have the equation: .\(\displaystyle 12W \,=\,588\)


Now solve for \(\displaystyle W\) . . . then for \(\displaystyle L.\)

 
Thank you Soroban, I got it......W = 49, and L= 245.....I am a 65-year-old CT resident, who has not done an algebra problem since I graduated from Wayland, MA High School in 1961.....my daughter is just beginning college, with the ultimate goal of becoming a pharmacist, and she had this problem that neither of us could get started on....although I am an R.N., it has been years since I have had to calculate by hand.......no practice.....thanks again!!!!
 
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