Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1.

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, a2+b2

now pi/4*4=4pi/4=pi.

So your answer is pi.

Randy presses RAND 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 4p.

If I draw a square that is 1 by 1 unit. Then every possible outcome  is represented inside that square.

I mean (x,y) in tha square will represent every possible combination of the 2 random numbers.

x is one of the numbers and y is the other one.

Now I need to determine what restrictions will make this true.

Since both these numbers are between 0 and 1 it follows that 1 must be the longest side of the triangle.

Condition 1   For it to be a triangle at all, the 2 random numbers must add to more than 1

then x+y>1

y>-x+1

I have to graph that region.

Condition 2   For it to be an obtuse angled triangle x^2 +y^2 must be less than 

See the pic below to try and understand why this is true.

So $x^2+y^2<1$

I'll graph that too.

So here is my probability contour graph.

The entire sample space, (that is all possible outcomes) are in the 1 by 1 square.

The green area represents all the points that could make any triangle. with the other long side being 1

The purple area represents all the points that would NOT make an acute angled triange.

So I need to work out the are of the intersection.

$Area = \frac{\pi r^2}{4}-0.5*1*1\\ Area = \frac{\pi}{4}-0.5\\ Area = \frac{\pi-2}{4}\;units \;squared\\$

Since the are of the sample space is 1*1=1

The probabilty that I will be able to form the triangle described is    

$P (\text{That an obtuse angle with 3rd side equal 1 can be formed})=\displaystyle \frac{\pi-2}{4}$

OH this is not quite what the question asked .. You can finish it.   

I have probably just done your homework for you.

I hope you can at least learn something from my answer.

Pretty crafty, Melody......!!!

I'm going to look at this in more depth to make sure I understand it fully  !!!