SOLUTION: The width of a rectangle is 5 times its length. If the width of the rectangle is doubled, then doubled again, the new perimeter is which percent of the original perimeter?
Algebra ->
Rectangles
-> SOLUTION: The width of a rectangle is 5 times its length. If the width of the rectangle is doubled, then doubled again, the new perimeter is which percent of the original perimeter?
Log On
Question 364453: The width of a rectangle is 5 times its length. If the width of the rectangle is doubled, then doubled again, the new perimeter is which percent of the original perimeter?
You can put this solution on YOUR website! The width of a rectangle is 5 times its length. If the width of the rectangle is doubled, then doubled again, the new perimeter is which percent of the original perimeter?
..
let length be x
width = 5x
perimeter = 2*(L+W)
=2*6x
=12x
..
2*2*5x= 20x
length = x
...
perimeter = 2*(20x+x)
=42x
Ratio = 42x/12x
2.5 *100 = 250 %
...
m.ananth@hotmail.ca
