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 $A$?
+128577 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%
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