SOLUTION: A given rectangle has a length 2 feet longer than it's width. If the width is shortened by 2 feet and the length is doubled,a new rectangle is formed. The perimeter of the new rect

Algebra ->  Test -> SOLUTION: A given rectangle has a length 2 feet longer than it's width. If the width is shortened by 2 feet and the length is doubled,a new rectangle is formed. The perimeter of the new rect      Log On


   

Question 1090190: A given rectangle has a length 2 feet longer than it's width. If the width is shortened by 2 feet and the length is doubled,a new rectangle is formed. The perimeter of the new rectangle is 12 more than the perimeter of the original rectangle. What are the dimensions of the original rectangle?
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Original rectangle dimensions, w+2 and w;
Perimeter p=2%28w%2B2%29%2B2w.

Second rectangle dimensions, 2(w+2) and w-2 for length and width;
Perimeter 2%2A2%28w%2B2%29%2B2%28w-2%29=p%2B12.

-

STEPS
p=2w%2B4%2B2w
p=4w%2B4
-
4w%2B8%2B2w-4=p%2B12
6w%2B4=p%2B12
6w-p=8
6w-%284w%2B4%29=8
6w-4w-4=8
2w-4=8
w-2=4
highlight%28w=6%29-------width, original rectangle
-
w%2B2=highlight%288%29------length, original rectangle