Question 254586
<font face="Garamond" size="+2">


Pythagoras says that the measure of the diagonal of a square is the square root of 2 times the measure of one of the sides squared.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{s^2\ +\ s^2}\ =\ \sqrt{2s^2}]


But since the measure of a side squared is the area of a square we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{2A}]


Substitute the given area for A and do the arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>