SOLUTION: The length of a certian rectangle is 40m greater thans its width. If the width were reduced by 20m and the lenght increased by 80m,the perimeter of the new rectangle would be twice

Question 110518: The length of a certian rectangle is 40m greater thans its width. If the width were reduced by 20m and the lenght increased by 80m,the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The length of a certain rectangle is 40m greater than it's width. If the width were reduced by 20m and the length increased by 80m,the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
:
Let x = original width
:
It says the length is 40 m greater, therefore
(x+4) = original length
:
If the width is reduced by 20:
(x-20) = new width
:
If the length were increased by 80:
(x+4) + 80
(x+84) = the new length
:
New perimeter = twice old perimeter:
2(x-20) + 2(x+84) = 2[2x + 2(x+4)]
:
2x - 40 + 2x + 168 = 2[2x + 2x + 8]
:
2x + 2x + 168 - 40 = 2(4x + 8)
:
4x + 128 = 8x + 16
:
128 - 16 = 8x - 4x
:
4x = 112
:
x = 28 m is the original width
Then
28 + 4 = 32; original length
:
:
Check solution by finding the perimeters:
2(28) + 2(32) = 120
2(8) + 2(112) = 240, which is twice as long
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