Inside a regular hexagon ABCDEF is drawn triangle ACE. What is the ratio of the area of triangle ACE to that of the hexagon?
+128399 Let the side of the hexagon = S
Its area = 6(1/2)S^2 sin60“ = 3S^2sin 60°
We can find the side of the triangle using the Law of Cosines
This is
sqrt [ S^2 + S^2 - 2S^2 cos (120°)] = sqrt [ 2S^2 + S^2] = S√3
The area of this equilateral triangle is (1/2)(S√3)^2 sin60° = (3/2)S^2sin60°
So the ratio is
(3/2)S^2 sin 60°
_____________ = (3/2) / 3 = 1 / 2
3S^2 sin 60°
