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
90 X
5
90 B 15 C
If BXA = 90 then BXC = 90
Then triangle BXA is right and by the Pythagorean Theorem
BC^2 = CX^2 + BX^2
15^2 = 5^2 + BX^2
225 = 25 + BX^2
200 = BX^2
BX = sqrt(200) = 10sqrt (2)
+1234 
