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 $12$?
The ratio of 12 to the area of the big triangle is 1/6, so the probabiity is 1/6.
+129918 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
