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 $BX$?
A
X
5
B C
15
Since BXA = 90 then BXC also = 90
So BXC is aright triangle such that
BC^2 - CX^2 = BX^2
15^2 - 5^2 = BX^2
200 = BX^2
BX = sqrt 200 = 10sqrt 2 ≈ 14.14
+685 
