+0  
 
0
1
1
avatar+104 

In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $AX = 15,$ and $CX = 6$. What is $BX$?

 Jun 4, 2024

Best Answer 

 #1
avatar+104 
+1

I solved it.

 

Let x represent BX.

15^2 + x^2 (AB) + 6^2 + x^2 (BC) = 21^2 (AC)

Solving this equation gives us x = 3sqrt(10)

Therefore, BX = 3sqrt(10)

 Jun 4, 2024
 #1
avatar+104 
+1
Best Answer

I solved it.

 

Let x represent BX.

15^2 + x^2 (AB) + 6^2 + x^2 (BC) = 21^2 (AC)

Solving this equation gives us x = 3sqrt(10)

Therefore, BX = 3sqrt(10)

SilviaJendeukie Jun 4, 2024

2 Online Users

avatar