Given a circle with an inscribed equilateral triangle with each side of the triangle having a measurement of 1

Question

Given a circle with an inscribed equilateral triangle with each side of the triangle having a measurement of 18 cm. What is the probability of selecting a point at random inside the triangle, assuming that the point cannot lie outside the circular region. Leave your answer in exact form. To receive full credit, please show your work.

Answer 1

Area of Circle/ Area of Triangle

AoT = 9*Sqrt(18^2-9^2) = 9*Sqrt(243) =~ 140.3

AoC = pi*(2*sqrt(27))^2 = pi*4*27 =~ 339.29

140.3/339.29 =~ 41.4%

Answer 2

The circumradius, r, of the equilateral triangle = abc/(4A) where A is area and a,b,c are the 3 sides, so r = 18^3/(4A)
But A = s^2sqrt(3)/4 so r= 18^3/(4 *18^2sqrt(3)/4) = 18/sqrt(3)
Area of circle = pir^2 = (6sqrt(3))^2pi = 108pi
Area of triangle = 18^2sqrt(3)/4 = 81sqrt(3)
So probability is 81sqrt(3)/108pi = 3sqrt(3)/(4pi)

Answer 3

each and every attitude of the inscribed triangle is 60 degrees, if we draw a line from one the vertexes to the middle of the circle we may be able to bisect the attitude. Now drop a perpendicular from the middle to between the elements of the triangle and we may be able to bisect that section. we've a suitable triangle with an attitude of 30 degrees and a aspect of 9 meters The hypotenuse of the right triangle is the radius cos 30 deg = 9/radius 0.86603 = 9/radius radius = 10.3923 m Now all individuals understand that the circumference = 2?r C = 2?*10.3923 = sixty 5.2968 The triangle vertexes are the top aspects of three equivalent arcs of 21.seventy seven meters. .