Question 334126
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Let *[tex \Large x] represent the measure of the short leg. Then *[tex \Large x\ +\ 4] must represent the measure of the long leg, and *[tex \Large x\ +\ 8] must represent the measure of the hypotenuse.

Use Pythagoras:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 8)^2\ =\ (x\ +\ 4)^2\ +\ x^2]

Expand the two squared binomials, collect like terms, and solve the resulting quadratic for *[tex \Large x], the measure of the short leg.

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