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by nidhis.1408 » Wed Nov 28, 2012 2:05 pm
A triangle with three equal sides is inscribed inside a circle. A point is selected at random inside the circle. What is the probability that the point selected is inside the triangle?

1. 3/4pi
2. 3√2/5pi
3. 3√3/4pi
4. 5√3/4pi
5. 3√3/2pi
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by The Iceman » Wed Nov 28, 2012 11:33 pm
nidhis.1408 wrote:A triangle with three equal sides is inscribed inside a circle. A point is selected at random inside the circle. What is the probability that the point selected is inside the triangle?

1. 3/4pi
2. 3√2/5pi
3. 3√3/4pi
4. 5√3/4pi
5. 3√3/2pi
if r is the radius of the circle, then length of each side of the triangle is √3r.

So the required probability = area of triangle / area of the circle = 3√3/4pi
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by Bill@VeritasPrep » Wed Nov 28, 2012 11:41 pm
I assigned values to the sides of the triangles, then determined the area of the circle.

If the triangle's side length is 6, its area is 9sqrt3.

The circle must have a radius of 6/(sqrt3), and we can then find the area of 12pi.

P(point in triangle) = 9sqrt3/12pi = 3sqrt3/4pi.

I'll attach a diagram in a second.
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by Bill@VeritasPrep » Wed Nov 28, 2012 11:46 pm
Here it is. Enjoy my supreme art skill :D
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