SOLUTION: "The perimeter of a rectangle is 36 m. The area is 65 m^2. Find the dimensions of the rectangle." I'm supposed to use the Quadratic formula in this somehow, but I don't really u

Question 731276: "The perimeter of a rectangle is 36 m. The area is 65 m^2. Find the dimensions of the rectangle."
I'm supposed to use the Quadratic formula in this somehow, but I don't really understand how I get A, B, and C from this. I tried to do it, and I confused myself halfway through it. Thank you for your help!
Found 2 solutions by richwmiller, josgarithmetic:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
2l+2w=36
l=65/w
2*(65/w)+2w=36
65/w+w-36=0
65+w^2-36w=0
w^2-18w+65=0
factor
(w-5)*(w-13)=0
w = 5
width is shorter than length
l=13
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
What are A, B, and C supposed to be?

Just use what you know about a rectangle and the given information about the problem.

Let's say w = width of the rectangle and h = height of the rectangle.
Perimeter is 2w+2h=36.
Area is wh=65.

Pick one of the equations, any one of them. Solve for one of the variables, w or h; it does not matter which. Substitute that formula into the other equation. Now you should have what will be a quadratic equation. You may need to simplify it before using it.

A possible way to go,

-------------
, to continue starting with area relationship
, substituting what was found for h

You can certainly use the general solution to quadratic formula if you want to but that polynomial seems factorable. I will use quadratic formula solution.

OR

That was for w. You can then use either the perimeter equation or the area equation to find h.

postnote: You probably meant Ax^2+Bx+C=0 when you mentioned some confusion about "A, B, and C".
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