+279 In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 1, and median $\overline{BN}$ has a length of 1. What is the length of the hypotenuse of the triangle?
+2 For this triangle to exist we must have a triangle of two equal side lenghts. In this case we only have a 45, 45, 90 triangle. This gives us a neat diagram with that triangle with it's medians. Now we have a little equation by setting the legs as x. \(x^2+(x/2)^2=1 \) is our equation but setting x^2 as y we simplify into \(y+y/4=1\) solve it to get y=4/5 so x= sqrt 4/5.
