Geometry

Question

0

20

1

+1000

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 $BX$?

Hi6942O Jul 11, 2024

NotThatSmart · Answer 1 · 2024-07-11T02:05:14

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! :)