Question 1099127
.
<U>1.  &nbsp;&nbsp;&nbsp;&nbsp;Non-standard and unexpected solution</U>


<pre>
Let L and W be the dimensions of the expanded rectangle.


Then its perimeter is 2*(L+W),  hence L+W = {{{440/2}}} = 220 ft

and the average between L and W is  {{{(L+W)/2}}} = {{{220/2}}} = 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) = {{{12100 - x^2}}}

which implies  {{{x^2}}} = 12100 -7200 = 4900.


Hence, x = {{{sqrt(4900)}}} = 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.
</pre>

Solved.


Did you noticed that I solved the problem MENTALLY ?



<U>2.  &nbsp;&nbsp;&nbsp;&nbsp;Another mental solution</U>


<pre>
The area of 7200 ft^2 is the product of two numbers, whose sum is  {{{440/2}}} = 220, as we saw in the <U>Solution 1</U>.


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  {{{180/3}}} = 60 ft,

and you got the same answer.
</pre>


<U>3. &nbsp;&nbsp;&nbsp;&nbsp;Algebraic solution</U>


<pre>
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, {{{180/3}}} = 60 ft  is your answer to the problem's question.
</pre>