SOLUTION: A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square cent
Algebra ->
Customizable Word Problem Solvers
-> Geometry
-> SOLUTION: A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square cent
Log On
Question 201345: A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the demensions of the original rectangle? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the demensions of the original rectangle?
.
Let w = width of original rectangle
then
4w = length of original rectangle
.
(w+2)(4w+5) = w(4w) + 270 (if you don't see this, write back)
4w^2+5w+8w+10 = 4w^2 + 270
4w^2+13w+10 = 4w^2 + 270
13w+10 = 270
13w = 260
w = 20 cm (width of original rectangle)
.
Length:
4w = 4(20) = 80 cm (length of original rectangle)
