SOLUTION: If the length of a rectangle is increased by 3times then the perimeter of a rectangle is 440meters and the area is 7200meter^2 .what is the real length of a rectangle

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.

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.

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.