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Triangle $ABC$ is equilateral with side length 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?

Guest Jul 13, 2018
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The area  of the tirangle  = (1/2)side^2 * sin (60°)  = (1/2) 3^2 * √3/2 = (9/4)√3  units^2

 

We can  construct a circle at A  with a radius of 1......the area  bounded by the sides of the triangle and the circle will be  =

 

pi * (1)^2 * (60/360)  =   pi/6  units ^2

 

So....the probability that "X"  lies within one unit of "A"   is

 

[pi / 6]  / [ (9/4)√3]  =

 

pi /  6   *  4 / [9√3  ]  =

 

4 pi / [ 54 √3 ] =

 

 2 pi / [ 27√3]  ≈   .1344 ≈   13.44%

 

Here's a picture :

 

cool cool cool

CPhill  Jul 13, 2018

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