In triangle \(ABC, \angle B= 90^\circ\) Point \(X\) is on \(\overline {AC}\) such that \(\angle BXA=90^\circ, AX = 12\) and \(CX = 4\). What is \(BX\)?
Triangle ABC is a right triangle, so by the Pythagorean Theorem, AC=122+42=13. Then by the Pythagorean Theorem in triangle ABX, BX^2 = 12^2 - 4^2 = 144 - 16 = 128, so BX= sqrt(128) = 8*sqrt(2).
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