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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 $BX$?

 Jun 27, 2024
 #1
avatar+129725 
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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

 

cool cool cool

 Jun 27, 2024

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