geomatry

[vipulgoyal]

A trangle with three equal sides is incribed in a circle, a point is selected inside a
circle , what is the probablity that the point selected is inside the
trangle?

3/4pi

3root2/5pi

3root3/4pi

5root3/4pi

3root3/2pi

ans not given

[Anju@Gurome]

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.

[srcc25anu]

when an equilateral triangle in inscribed in a circle, and if we drop an altitude to one of its base, the half-triangle so formed will have its sides in the ratio of 1:Root3:2
lets assume the radius of circle to be "r"
in an equilateral Triangle, centroid (point at which medians intersect) divides the median in ratio 2:1. we know radius is r so median = 3r/2. also median = height for any equilateral triangle. hence height (or side corresponding to root3 = 3r/2).now base will be side corresponding to 2 in the above ratio 1:Root3:2. therefore base = root3*r
Area of triangle = 1/2 * (root3*r)* (3r/2) => (3 root 3 * r^2) / 4
Area of circle = pie r^2
Required Probability = 3 root 3 / 4*pie
hence answer C