Geometric Probability Problem, Random Numbers $0$-$1+$ Triangles. Randy presses RANDOM on his calculator twice to obtain two random numbers between $0$ and $1$. Let $p$ be the probability that these two numbers and $1$ form the sides of an obtuse triangle. Find $p$.
At first, I thought that the answer would be $.25$, because the probability of both of the numbers being above $.5$ is $.5$. But I realized that this is not true, for cases such as $.99, .99, 1,$ etc. This problem has me a bit stumped. I'm fairly certain I need to use a graph, but I don't know how.

I bet they are looking for you to use the Pythagorean Theorem. Since the random number will be less than 1, you can assume 1 is the largest side length. For obtuse triangles, $a^2+b^2<c^2$. In this case, because c is 1, you can simplify this to $a^2+b^2<1.$ This is a two-variable probability problem, so the easiest way to do it is integration, but you needn't integrate. If you set $a=x$ and $b=y$, you get the unit circle equation $x^2+y^2<1$. The area of this circle inside the parameters $0<x<1$ and $0<y<1$ is $[\pi(r^2)]/4$, or $\pi/4$. The total area is the unit square. Thus, the probability of an obtuse triangle is $(\pi/4)/1$, or just $\pi/4.$
