Question 673245
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Let *[tex \LARGE x] represent the measure of the short leg.  Then *[tex \LARGE x\ +\ 12] is the measure of the long leg, and *[tex \LARGE 2x\ -\ 12] is the measure of the hypotenuse.  Use Pythagoras:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ (x\ +\ 12)^2\ =\ (2x\ -\ 12)^2]


Solve for the non-zero root.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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