SOLUTION: the perimeter of a rectangle and a square is 80 cm each. if the difference between their areas is 100 cm find the sides of the rectangle.

Question 912065: the perimeter of a rectangle and a square is 80 cm each. if the difference between their areas is 100 cm find the sides of the rectangle.
Found 2 solutions by JulietG, richard1234:
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
If the perimeter of the square is 80, then each side is 80/4, or 20.
That makes the area of the square 20*20=400
Since the difference between the rectangle and the square are is 100, the area of the rectangle equals either 500 or 300.
They're even numbers, so we'll assume that the sides are even, too.
2L + 2W = 80, therefore L + W = 40
We're looking for factors of 500 (or 300; we'll do that next if 500 doesn't work) that equal 40 when added.
Factors of 500:
1 and 500 = 501
2 and 250 =252
4 and 125 = 129
5 and 100 = 105
10 and 50 = 60
20 and 25 = 45
None of the factors of 500 fit the scenario. Let's try the other possibility, 300
Factors of 300:
1 and 300 = 301
2 and 150 = 152
3 and 100 = 103
4 and 75 = 79
5 and 60 = 65
6 and 50 = 56
10 and 30 = 40 -- we have a winner!
12 and 25 = 37
15 and 20 = 35
I'm sure there are other ways to solve this, but elimination works best for me. If you need a different way of doing it, please repost with that instruction.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Going by the above poster's solution:

Area of rectangle must be 300 (500 is impossible -- it can be proven that the maximum area of a rectangle with fixed perimeter is a square).

We want to find L, W such that L+W = 40 and LW = 300. If you notice that 30, 10 work, then you are done. However it could be that L, W are not integers (in a general case), and the easiest solution is to substitute W = 40-L, substitute that into LW = 300 to obtain L(40-L) = 300. You are left with a quadratic equation which you can solve for L.

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