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1. Non-standard and unexpected solution
Let L and W be the dimensions of the expanded rectangle.
Then its perimeter is 2*(L+W), hence L+W = = 220 ft
and the average between L and W is = = 110 ft.
Since 110 ft is the average, L = 110 + x and W = 110 - x, where x is an unknown deviation.
Then the area = 7200 = L*W = (110+x)*(110-x) =
which implies = 12100 -7200 = 4900.
Hence, x = = 70 ft.
Therefore, L = 110 + 70 = 180 (I assume that L is the larger dimension !), and the length of the original rectangle was/is 180/3 = 60 ft.
Solved.
Did you noticed that I solved the problem MENTALLY ?
2. Another mental solution
The area of 7200 ft^2 is the product of two numbers, whose sum is = 220, as we saw in the Solution 1.
Try to find decomposition of the integer 7200 into the product of two factors that sum up to 220:
7200 = 10*720 = 20*360 = 30*240 = 40*180 . . . ! O-o-o-ps ! That's it !
So, you just found the length of the expanded rectangle.
It is 180 ft.
Hence, the length of the original rectangle was = 60 ft,
and you got the same answer.
3. Algebraic solution
You have
L + W = 220 (1) and
LW = 7200. (2)
From (1), express W = 220-W and substitute into (2). You will get
L*(220-L) = 7200, or
L^2 - 220*L + 7200 = 0.
Solve this quadratic equation by any method you know.
For example, using factoring for the left side
(L-180)*(L-40) = 0.
Then for L (for the larger dimension) you obtain L = 180 ft.
Finally, = 60 ft is your answer to the problem's question.