Question 732957
<font face="Times New Roman" size="+2">


Let *[tex \LARGE w] represent the width.  Then *[tex \LARGE 2w\ +\ 2] must represent the length.


The square of the width plus the square of the length equals the square of the diagonal -- Pythagoras said it, I believe it, that settles it.


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ \left(2w\ +\ 2\right)^2\ =\ 169]


Solve the quadratic for *[tex \LARGE w], discarding the negative root.  Or save yourself some work by remembering that 13, 12, 5 is a Pythagorean triple.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>