SOLUTION: the perimeter of a rectangle os 18ft, and the area of the rectangle is 20 sq ft. what is the width of the rectangle ?

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Question 922832: the perimeter of a rectangle os 18ft, and the area of the rectangle is 20 sq ft. what is the width of the rectangle ?

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Try doing this in general.

x and y are the dimensions.
Let A be the given area.
Let p be the given perimeter.


system%282x%2B2y=p%2Cxy=A%29

2x=p-2y
x=%28p-2y%29%2F2
-
%281%2F2%29%28p-2y%29y=A
%28p-2y%29y=2A
py-2y%5E2-2A=0
highlight_green%282y%5E2-py%2B2A=0%29

You can use the given values any time you want, and then find the other variable using the earlier relation found before the substitution.... highlight%28x=%28p-2y%29%2F2%29.

Keeping with a general solution of y,
highlight%28y=%28p%2B-+sqrt%28p%5E2-4%2A2%2AA%29%29%2F%282%2A2%29%29

If you use given values right in the quadratic equation,
2y%5E2-18y%2B2%2A20=0
y%5E2-9y%2B20=0
%28y-5%29%28y-4%29=0, seems faster than using the general solution to a quadratic equation;
y=5 OR y=4, and correspondingly, x=4 or x=5.

Dimensions are 4 feet by 5 feet.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
the perimeter of a rectangle os 18ft, and the area of the rectangle is 20 sq ft. what is the width of the rectangle ?

Let width and length, be W and L, respectively
Then: 2(W + L) = 18_____2(W + L) = 2(9)____W + L = 9_____L = 9 - W ----- eq (i)
Also, LW = 20 ------ eq (ii)
W(9 - W) = 20 ------- Substituting 9 - W for L in eq (ii)
9W+-+W%5E2+=+20
W%5E2+-+9W+%2B+20+=+0
(W - 5)(W - 4) = 0
W, or width = highlight_green%285%29 OR highlight_green%28W+=+4%29
Note that:
If width = 5, then length = 4
If width = 4, then length = 5