Co-relating moments of inertia with geometry.

Let the moment of inertia of the triangle of side $a$ is $I$ .

We know that the moment of inertia is proportional to the mass and square of side length. $I \propto Ma^2$ Also, mass is proportional to the area or square of side $M \propto a^2$ $I \propto a^4$

Let us draw two triangles of side length $2a$ over the top and bottom of the hexagon. The geometry will look like as shown,

The moment of inertia of anyone orange triangle will be $16I$ because the side length is $2a$ and $I \propto a^4$

Now we can break this geometry into two equilateral triangles of side length $4a$ and moment of inertia $256I$ each.

Thus, the moment of inertia of the hexagon will be the total moment of inertia minus the moment of inertia of the orange triangles.

Putting the values we get, Moment of inertia of the hexagon = $2 \times 256 I - 2 \times 16 I = \boxed{480 I}.$

If the moment of inertia of a triangle of side $a$ about the given axis is $\mathcal{I}$ , then the moment of inertia of a triangle of side $2a$ about the corresponding axis is $16\mathcal{I}$ (the mass has been multiplied by $4$ and the dimensions doubled). Similarly the moment of inertia of a triangle of side $4a$ about the corresponding axis is $256\mathcal{I}$ . Thus the moment of inertia of the top half of a hexagon of side $2a$ about the given axis (which can be seen as a triangle of side $4a$ with a triangle of side $2a$ removed is $(256-16)\mathcal{I} = 240\mathcal{I}$ . Doubling up, the moment of inertia of a hexagon of side $2a$ is $480\mathcal{I}$ .

Alternatively, suppose that the moment of inertia of an triangle of mass $m$ and side $a$ is $J(a)$ . Then, by symmetry and the parallel axes theorem, $16J(a) \; = \; J(2a) \; = \; 2J(a) + 2\big[J(a) + m(\tfrac12a)^2\big] \; = \; 4J(a) + \tfrac12ma^2$ and hence $J(a) = \tfrac{1}{24}ma^2$ . Now splitting the hexagon of side $2a$ into $24$ triangles of side $a$ and applying the parallel axes theorem many times, we see that the moment of inertia of the hexagon is $4J(a) + 8\big[J(a) + m(\tfrac12a)^2\big] + 8\big[J(a) + ma^2\big] + 4\big[J(a) + m(\tfrac32a)^2\big]$ which is equal to $24J(a) + 2ma^2 + 8ma^2 + 9ma^2 \; = \; 24J(a) + 19ma^2 \; = \; 20ma^2 \; = \; 480J(a)$ making the answer $\boxed{480}$ again.

Although this problem may be solved by simply considering proportions, it also serves as a good exercise in calculating moments of inertia.

The moment of inertia of an equilateral triangular plate with side-length $a$ and thickness $t$ is given by integrating the squared distance between the axis of rotation (let's call it the $y$ -axis) and each mass element ${\rm d}m = \rho \,{\rm d}V$ .

The density is $\rho = m/V = m \left( \frac{1}{2} a \cdot \frac{\sqrt3 a}{2} \cdot t \right)^{-1} = \frac{4 \sqrt{3} m}{3 a^{2} t} \,.$ Thus, the moment of inertia for the triangular plate may be integrated as $I = \int_{m} r^2 - y^2 \,{\rm d}m = \int_{V} (r^2-y^2 )\rho\,{\rm d}V = \frac{4 \sqrt{3} m}{3 a^{2} t} \int\limits_{-t/2}^{t/2} \int\limits_{0}^{\sqrt3a/2} \int\limits_{-a/2 +y/\sqrt3}^{\ a/2-y/\sqrt3} (x^2+z^2) \,{\rm d}x\,{\rm d}y\,{\rm d}z = m \left( \frac{a^2}{24} + \frac{t^2}{12} \right)\,.$ Assuming the thickness to be negligible ( $t \ll a$ ), $I$ simplifies to $I = \frac{1}{24}ma^2\,.$ Since we may add and subtract moments of inertia about the same axis, we note that the hexagon is equivalent to two large triangles of side length $4a$ , less two triangles of side length $2a$ . Since the mass of these triangles scale (for a fixed thickness) with the length squared, their masses are respectively $16m$ and $4m$ .

The moment of inertia of the hexagon is then $I_{\rm hex} = 2(I_{4a} - I_{2a}) = \frac{1}{12} \left[ (16m)(4a)^2 - (4m)(2a)^2 \right]=20ma^2\,.$ And the ratio desired is $k = \frac{I_{\rm hex}}{I} = 20\cdot 24 = \boxed{480}\,.$