SOLUTION: a rectangle has whole number sides.The length of the rectangle is tripled and the width is doubled.the new area is 204 cm squared.the new perimeter is 110 cm.what was the old peri

Algebra ->  Rectangles -> SOLUTION: a rectangle has whole number sides.The length of the rectangle is tripled and the width is doubled.the new area is 204 cm squared.the new perimeter is 110 cm.what was the old peri      Log On


   

Question 724205: a rectangle has whole number sides.The length of the rectangle is tripled and the width is doubled.the new area is 204 cm squared.the new perimeter is 110 cm.what was the old perimeter?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x= the length of one of the sides of the new, enlarged rectangle
y= the length of the other side of the new, enlarged rectangle
The new area is xy=204 .
The new perimeter is 2%28x%2By%29=100 --> x%2By=55
I could set up to solve system%28xy=204%2Cx%2By=55%29, end up with a quadratic equation, and solve it by factoring. The hint that the length of the sides were whole numbers tells me that the factoring approach would work.
Finding two numbers that add up to 55 and multiply to yield 204 is a shorter way to do that.
Since 51%2B4=55 and 51%2A4=204 the new rectangle sides measure 51cm and 4cm
Since 51cm=3%2A17cm and 4cm=2%2A2cm , it is obvious that the original length and width were 17cm and 2cm.
(The original length and width could not have been 4cm%2F3=4%2F3cm and 17cm%2F2=17%2F2cm respectively, because that would gives us a length shorter than the width and no whole number side lengths).
With a length of 17cm and a width of 2cm , the perimeter of the original rectangle is
2%2817cm%2B2cm%29=highlight%2838cm%29