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 ?

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) (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.

-

You can use the given values any time you want, and then find the other variable using the earlier relation found before the substitution.... .

Keeping with a general solution of y,

If you use given values right in the quadratic equation,

, 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) (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)

(W - 5)(W - 4) = 0
W, or width = OR
Note that:
If width = 5, then length = 4
If width = 4, then length = 5
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