SOLUTION: How do you find the length and width of a rectangle if the perimeter=38 and the area=90 square meters?

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Question 9674: How do you find the length and width of a rectangle if the perimeter=38 and the area=90 square meters?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You could try this:
A = L X W = 90 m^2 This is the area.
P = 2(L+W) = 38 m This is the perimeter. Solve this for L and substitute in the above equation, then solve for W.
2(L+W) = 38
L+W = 19
L = 19-W
%2819-W%29W+=+90 Simplify and solve for W
19W+-+W%5E2+=+90
W%5E2+-+19W+%2B+90+=+0 Solve this quadratic by factoring.
%28W-9%29%28W-10%29=+0 Apply the zero product principle.
W-9 = 0 or W-10 = 0
If W-9 = 0, then w = 9
If W-10 = 0, then W = 10
These two numbers are actually the length and the width of the rectangle.
L = 10 m and W = 9 m.
Check:
A = LW = 10(9) = 90 m^2
P = 2(L+W) = 2(10+9) = 2(19) = 38 meters.