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In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $AX = 5,$ and $CX = 5$. What is $BX$?

 Mar 12, 2024
 #1
avatar+128794 
+1

A

      5

            X

               5

B                 C

 

 

AB = BC

 

Triangle AXB ≈  Triangle ABC

AX / AB  = AB / AC

5 / AB = AB / 10

AB^2  = 50

AB = 5sqrt (2) = BC

 

BX / AB  = BC / AC

BX / AB = AB  / AC

BX / AB = AB / 10

10BX = AB^2

10 BX = 50

BX = 50 / 10    =  5

 

cool cool cool

 Mar 12, 2024

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