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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.

 Nov 8, 2019
 #1
avatar+448 
+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.

 Nov 8, 2019
 #2
avatar+118716 
+5

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 x2+y2<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.

 

 

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

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

 

 

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

 Nov 9, 2019
edited by Melody  Nov 9, 2019
 #6
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+1

Thank you so much Melody!!!!

Guest Nov 9, 2019
 #3
avatar+118716 
+2

I have probably just done your homework for you.

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

 Nov 9, 2019
 #4
avatar+130503 
+3

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

 

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

 

 

cool cool cool

 Nov 9, 2019
 #5
avatar+118716 
+2

Thanks Chris,

I like doing these. :)

 

I think it was Bertie who taught me how to do these :)

Melody  Nov 9, 2019
edited by Melody  Nov 9, 2019

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