Here is the problem:
Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $[0, 1]$. What is the probability that $x, y,$ and $1$ are the side lengths of an obtuse triangle? Round your answer to the nearest tenth.
What I have gotten so far:
So by the Pythagorean inequality, I know, for an obtuse triangle that $a^{2} + b^{2} < c^{2}$, so $x^2 + y^2 < 1$. The triangle inequality tells us that $a + b > c$, which gives $x + y > 1$. I am not sure what to do with these inequalities... Hints, and ONLY hints, are appreciated! Thanks!
The area of the triangle in the first quadrant whose hypotenuse is derived from the inequality $x + y > 1\ (y = 1 - x)$ represents the number of impossible triangles $(x+y < 1)$. Area $A_1 = 0.5\cdot b\cdot h$
The area between the line $y = 1 - x$ and the circle (radius $1$) in the first quadrant derived from the inequality $x^2 + y^2 < 1\ (y = \sqrt{1 - x^2})$ represents the number of obtuse triangles. Area $A_2= 0.25\pi\cdot r^2 - A_1$.
The area between the circle and the square $x=1, y=1$ represents the number of acute triangles. Area $A_3 = 1^2 - 0.25\pi\cdot r^2$