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Right triangle $XYZ$ has legs of length $XY = 12$ and $YZ = 6$. Point $D$ is chosen at random within the triangle $XYZ$. What is the probability that the area of triangle $XYD$ is at most $12$?

 Feb 14, 2020
 #1
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The ratio of 12 to the area of the big triangle is 1/6, so the probabiity is 1/6.

 Feb 14, 2020
 #2
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The area   of  triangle  XYZ   =(1/2)  (product of the leg lengths) =  (1/2) (12 * 6)   =  (1/2) (72)  = 36

 

So....the probability that the area of triangle XYD is at most 12 =   12 /36  =   1/3

 

 

cool cool cool

 Feb 14, 2020
 #4
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I tried, but it doesn't work.

Guest Feb 14, 2020
 #3
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+1

Never mind

 Feb 14, 2020
edited by Guest  Feb 14, 2020
edited by AnimalMaster  Feb 15, 2020

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