vipulgoyal wrote:A triangle with three equal sides is inscribed in a circle, a point is selected inside a circle , what is the probability that the point selected is inside the triangle?
Required probability = Ratio of the areas of the triangle and the circle = (Area of the triangle)/(Area of the circle)
Now, refer to the following diagram,
Let us assume that length of the sides of the triangle = AB = BC = AC = a
and length of the radius of the circle = OC = r
Now, D is the midpoint of BC --> CD = a/2
In right angle triangle ADC, AC² = AD² + CD² ---> AD² = (AC² - CD²) = (a² - a²/4) = 3a²/4
Hence, AD = (√3/2)*a
Also, AD is the median of triangle ABC and O is the centroid.
Hence, OD = AD/3 = [(√3/2)*a]/3 = a/(2√3)
In right angle triangle ODC, OC² = OD² + CD² = (a/(2√3))² + (a/2²) = (a²/12 + a²/4) = 4a²/12 = a²/3
Hence, OC = a/√3
Hence, radius of the circle = OC = a/√3
Now, area of the triangle = (√3/4)*a²
And, area of the circle = πr² = π[a/√3]² = πa²/3
Hence, required probability = [(√3/4)*a²]/[πa²/3] = 3√3/4π
The correct answer is C.
