SOLUTION: The area of a rectangle is 360 square meters. If the rectangles length is increased by 10 meters and the width is decreased by 6 meters, its area does not change. Find the perimete

Algebra ->  Rectangles -> SOLUTION: The area of a rectangle is 360 square meters. If the rectangles length is increased by 10 meters and the width is decreased by 6 meters, its area does not change. Find the perimete      Log On


   

Question 310454: The area of a rectangle is 360 square meters. If the rectangles length is increased by 10 meters and the width is decreased by 6 meters, its area does not change. Find the perimeter of the originol rectangle. (Plz help me awnser this)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is 360 square meters.
If the rectangles length is increased by 10 meters and the width is decreased
by 6 meters, its area does not change.
Find the perimeter of the original rectangle.
:
"The area of a rectangle is 360 square meters."
L * W = 360
L = 360%2FW
:
"If the rectangles length is increased by 10 meters and the width is decreased by 6 meters, its area does not change."
(L+10)*(W-6) = 360
FOIL
LW - 6L + 10W - 60 = 360
LW - 6L = -10W + 360 + 60
L(W-6) = -10W + 420
L = %28-10W+%2B+420%29%2F%28W-6%29
Replace L with 360/W
360%2FW = %28-10W+%2B+420%29%2F%28W-6%29
Cross multiply
360(W-6) = W(-10W+420)
360W - 2160 = -10W^2 + 420W
Combine like terms on the left
10W^2 + 360W - 420W - 2160 = 0
10W^2 - 60W - 2160 = 0
Simplify, divide by 10
W^2 - 6W - 216 = 0
Factor to
(W-18)(W+12) = 0
positive solution
W = 18 m is the original width
then
360%2F18 = 20 m is the length
:
Find the perimeter of the original rectangle.
2(20) + 2(18) = 76 m
:
:
Check the area when L is increased, and W is decreased
(20+10) * (18-6) = 360