<PRE><B>Question 1194689</B>
&lt;br&gt;
An &quot;irregular field that is in the shape of two squares, side by side&quot; suggests that the side lengths of the two squares are different; so the field looks something like this:&lt;br&gt;&lt;pre&gt;

  +-----------+
  |           |
  |           +-----+
  |           |     |
  |           |     |
  +-----------+-----+&lt;/pre&gt;&lt;br&gt;
or perhaps this:&lt;br&gt;&lt;pre&gt;
  +-----------+
  |           +-----+
  |           |     |
  |           |     |
  |           +-----+
  +-----------+&lt;/pre&gt;&lt;br&gt;

Let x and y be the side lengths of the larger and smaller square, respectively.&lt;br&gt;
Then the combined area of the two squares is&lt;br&gt;
{{{x^2+y^2}}}&lt;br&gt;
The perimeter of the field is&lt;br&gt;
{{{3x+3y+(x-y) = 4x+2y}}}&lt;br&gt;
So we need to solve a pair of simultaneous equations:&lt;br&gt;
(1) {{{x^2+y^2=17396}}}
(2) {{{4x+2y=572}}}&lt;br&gt;
You can solve the pair of equations by solving (2) for y and substituting in (1); you will get an ugly quadratic equation whose solutions you would probably need a calculator for.&lt;br&gt;
Since a calculator was going to be necessary, I solved the problem by a different path.&lt;br&gt;
I used the TABLE feature of a TI-83 calculator to find integer solution(s) of&lt;br&gt;
{{{x^2+y^2=17396}}} --&gt; {{{y=sqrt(17396-x^2)}}}&lt;br&gt;
and found there was only one: x=100 and y=86.&lt;br.
And that satisfied the condition that the perimeter 4x+2y is 572.&lt;br&gt;
From there, I can&#39;t answer the question that was asked.  If the field looks like the first diagram above, the longest side of the field is 100+86=186 yards; if it looks like the second diagram, the longest side of the field is 100 yards.&lt;br&gt;
ANSWER: The longest side of the field is either 100 yards or 186 yards&lt;br&gt;
</PRE>