Did get the same result, though used a different method..can someone verify?
I) Calculate height of the triangle - dropping an altitude to the base, the right triangle would be in the ratio of 1:sqrt(3):2. Since we have an equilateral triangle, the median from A would create a right triangle with one side (hyp) as 6, base as 3 (6/2) and therefore, height as 3*sqrt(3)
II)For an inscribed circle in an equilateral triangle, the center of the circle would coincide with the intersection of the medians of the triangle, the medians dividing themselves in a ratio of 1:2. So, the shorter median would be 1/3rd of the total median length. This would also be the radius of the circle = [3*sqrt(3)]/3 (from I) = sqrt(3)
III) Area of circle = pi*[sqrt(3)]^2 = pi*3
IV) Area of Triangle = 1/2*b*h = [1/2]*[6]*[3*sqrt(3)] = 9*sqrt(3)
V) Subtracting the two, Area of Tri - Area of Cir = 9*sqrt(3) - pi*3 = 3 [3sqrt(3) - pi]
VI) Since all the figures are symmetric (Equilateral traingle and the circle inscribed, the three areas created by the vertices of the triangle and the minor arc of the circle would be the same. Hence, for one area, we divide this figure by three
= [3{3sqrt(3)} - pi] / 3 = 3sqrt(3) - pi
Makes sense? Or did I go wrong somewhere?
