+950 In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $BC = 15,$ and $CX = 5$. What is $BX$?
+1892 First, let's make some observations about the problem.
First, let's note that if \(\angle BXA = 90\), then we have \(\angle BXC = 90\)
Triangle BXA is a right triangle, so we can use Pythoagorean Thereom, we simplfy have
\(BC^2 = CX^2 + BX^2 \\ 15^2 = 5^2 + BX^2\\ 225 = 25 + BX^2 \\ 200 = BX^2\)
Square rooting both sides, we finally reach a conclusion. We finally have
\(BX = \sqrt{200} = 10\sqrt 2 \)
So our final answer is 10\sqrt2.
Thanks! :)
