Question 198711
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Let *[tex \Large x] represent the measure of the longer leg, then *[tex \Large x + 2] is the measure of the hypotenuse and *[tex \Large x - 2] is the measure of the short leg


If we can believe our good friend Pythagoras,


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


Square both binomials and collect like terms and then solve the very simple resulting quadratic.  One of the roots won't make any sense, so toss it.  The other root is the measure of the long side.  Add 2 to get the hypotenuse.


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