Question 987219
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The sides are *[tex \Large x], *[tex \Large 2x\ +\ 6], and *[tex \Large 2x\ +\ 9].  Plug these expressions into the Pythagorean Theorem and solve for *[tex \Large x]. Then calculate *[tex \Large 2x\ +\ 6] and *[tex \Large 2x\ +\ 9].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ +\ 9)^2\ =\ (2x\ +\ 6)^2\ +\ x^2]


Note that any negative value for *[tex \Large x] would be absurd; exclude any negative root.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \