A equilateral triangular lamina, has a moment of Inertia of I if the axis of rotation passes through the centroid of the triangle, perpendicular to it's plane. What is the moment of inertia of a regular hexagon(Again, through it's geometrical centre, perpendicular to the plane), provided that, the length of the side of the triangle is equal to the sides of the regular hexagon. (I have assumed the side of the eq. triangle to be a. And m to be the mass of the eq. triangle. M of the hexagon M=6m) The equations, I have used to solve this, i)I' = I + Md^2 ii)The cosine rule. iii)Basic Trigonometry. iv)The radius of gyration is directly proportional to the length of the side of the triangle. I did not receive any answer which had only I in it. The solution provided in the book, is something I did not understand. It's given that it could be taken as two sides of eq. triangle of sides 2a. Hence using the the fourth point. (Since the hexagon is like 6 eq. triangle) The gave the moment of inertia as 32 I. Then the subtracted 2I, as they remove the two extra triangle due to the us adding two extra triangles, while constructing. But is it not fundamentally flawed, doing such a thing. By the parallel axis theorem their moment of inertia should be 2(I + Md^2) where d = a + 3^0.5/2 a The answer provided is 30I. Thank you.