SOLUTION: the length of a rectangle is 2in longer than its width. if the perimeter is 36in, what is the area?

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Question 862466: the length of a rectangle is 2in longer than its width. if the perimeter is 36in, what is the area?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is k inches more than its width. If perimeter p is known, what is area A?

Let L = length
Let w = width
k is a described constant, as is p.
A is a variable, unknown, but calculable, for the area.
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L=w%2Bk.

2w%2B2L=p
2w%2B2%28w%2Bk%29=p
4w%2B2k=p
4w=p-2k
highlight%28w=%28p-2k%29%2F4%29
Use w to find L.
L=w%2Bk
L=%28p-2k%29%2F4%2Bk------this form may be good enough...
L=%28p-2k%29%2F4%2B4k%2F4
L=%28%28p-2k%29%2B4k%29%2F4
highlight%28L=%28p%2B2k%29%2F4%29 -------this form seems better.

Now, you want Area A, and you begin forming highlight_green%28A=w%2AL%29.
A=%28%28p-2k%29%2F4%29%28%28p%2B2k%29%2F4%29
highlight%28highlight%28A=%28p%5E2-%284k%5E2%29%29%2F16%29%29

The solution is generalized. You should now appreciate the use of purely solving all in symbols. You have a formula for the area which was asked. What would be important is that the values for p and k make sense in the described situation. Their actual values are not too important; only that they make sense in the situation is important. In your example, p=36 and k=2. Use them and compute A.