<PRE><B>Question 710129</B>
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The perimeter of a square with side measure *[tex \LARGE s] is *[tex \LARGE 4s].  The area of such a square is *[tex \LARGE s^2].


A square with one-half the perimeter would have a perimeter of *[tex \LARGE \frac{4s}{2}], hence the measure of one side would be *[tex \LARGE \frac{s}{2}].  And the area of this smaller square is *[tex \LARGE \left(\frac{s}{2}\right)^2].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s^2\ =\ c\left(\frac{s}{2}\right)^2]


Solve for *[tex \LARGE  c]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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