need help w/ 'The length of a rectangular piece of cardboard

frenzy_252003

New member
Joined
Nov 9, 2006
Messages
2
The length of a rectangular piece of cardboard is 2 cm greater than its width. If the length and the width were each decreased by 1 cm, the area of the cardboard would be decreased by 27 square cm. What are the dimensions of the original piece of cardboard?

thanks a lot.. :D
 
Re: need help with word problem..

Hello, frenzy_252003!

Make sketches and baby-talk your way through it . . .


The length of a rectangular piece of cardboard is 2 cm greater than its width.
If the length and the width were each decreased by 1 cm,
the area of the cardboard would be decreased by 27 cm².
What are the dimensions of the original piece of cardboard?

In the original rectangle, the length is 2 more than the width.
Let \(\displaystyle W\) = width, then \(\displaystyle W+2\) = length.
Code:
      * - - - - - - - - - - - *
      |                       |
      |                       |
    W |                       |
      |                       |
      |                       |
      * - - - - - - - - - - - *
                W + 2

The area is: \(\displaystyle \:W(W\,+\,2)\:=\:W^2\,+\,2W\) cm².

The new rectangle has its length and width decreased by 1.
Its length will be: \(\displaystyle \,(W\,+\,2)\,-\,1\:=\:W\,+\,1\)
Its width will be: \(\displaystyle \,W\,-\,1\)
Code:
      * - - - - - - - - *
      |                 |
W - 1 |                 |
      |                 |
      * - - - - - - - - *
             W + 1

Its area will be: \(\displaystyle \,(W\,+\,1)(W\,-\,1)\:=\:W^2\,-\,1\) cm².


The area of the smaller rectangle is 27 cm² less than the original rectangle.

. . \(\displaystyle W^2\,-\,1\;=\;(W^2\,+\,2W)\,-\,27\)

And there is our equation!

 
Top