This is how I visualise the circle and the inscribed equilateral triangle ABC to look like.
Since an equilateral triangle has equal sides, the angle ABC, ACB and CAB = 60deg. O is center of the circle. The radius of the circle forms two legs, cutting the equilateral triangle ABC into 3 smaller triangles.
I started with triangle AOE. Angle OAE = 30deg (60deg / 2), angle AOE = 60deg. Thus, line AO = sqrt 3, line OE = 1, line EA = 2.
Area of triangle ABC = 1/2 * 1 * 2 * 6 = 6 ( * 6 being small triangle AOE, OEB, BOF, FOC, COD, AOD)
Area of circle = pi * r^2 = pi * (sqrt 3) ^ 2 = 3 * pi (where radius = line AO = sqrt 3)
Therefore, ratio of area of circle to area of triangle
= 3pi / 6
= pi / 2
Answer is C
I hope my working is not too confusing
triangle.JPG [ 9.65 KiB | Viewed 383 times ]
Jimmy Low, Frankfurt, Germany
GMAT Malaysia: http://gmatmalaysia.blogspot.com