Problem Solving

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

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

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.

Here it is. Enjoy my supreme art skill