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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!

Max0815
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    Draw the region in $\mathbb{R}^2$ where $x,y\ge0$, $x^2+y^2\le1$, and $x+y\ge1$. What is the area? – robjohn Sep 23 '18 at 19:09
  • Sorry, but I am having trouble understanding the r^2 thing. How would I graph this if you don't have graph paper? Is there an algebraic way of solving for the area?Also, can you please reply in the answers? – Max0815 Sep 23 '18 at 20:03

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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 $A_3 = 1^2 - 0.25\pi\cdot r^2$

Phil H
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