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 $10$?

Answer 1

The area of triangle $XYZ$ is $\frac12(12)(6)=36$, so $XD \cdot YD \le 80$. We can use this to write inequalities showing where point $D$ cannot be: if $XD > 10$, then the area of triangle $XYD$ is at least $\frac12(12)(10)=60$ (and similarly if $YD > 10$), and if $XD < 5$, then $YD > \frac{80}{XD} > 16$. Thus, the probability that $XD \le 10$ and $YD \le 10$ is the same as the probability that $5 \le XD \le 10$ and $5 \le YD \le 10$: this is the desired area divided by 36, or $\frac{(10-5)(10-5)}{36} = \boxed{\frac{25}{36}}$.