Moment of Inertia of an Equilateral Triangular Plate [closed]

I was reading about moment of inertia on Wikipedia and thought it was weird that it had common values for shapes like tetrehedron and cuboids but not triangular prisms or triangular plates, so I tried working it out myself. I will post my attempt below, but for some reason I cannot find any source online that confirms or denies my solution. Please let me know if you find anything wrong with it. Thanks.

Q: What is the moment of inertia of an equilateral triangular plate of uniform density $\rho$, mass $M$, side length $L$, rotating about an axis perpendicular to the triangle's plane and passing through its center?

1. First I modeled an equilateral triangle using three lines with its center of geometry at the origin as follows: $x=\frac{1}{\sqrt{3}}y-\frac{1}{3}L \\ x=\frac{1}{3}L-\frac{1}{\sqrt{3}}y \\ y=-\frac{\sqrt{3}}{6}L$

I used the fact that the circumradius of an equilateral triangle is $\frac{\sqrt3}{3}L$ and that its height is $\frac{\sqrt{3}}{2}L$ .

1. Next, using the definition of moment of inertia ($I$) and with the help of Wolfram Alpha, I obtained the following result:

$$I=\int r^2 dm=\rho \int r^2 dA\\ =\rho \int_{-\frac{\sqrt{3}}{6}L}^{\frac{\sqrt{3}}{3}L} \int_{\frac{1}{\sqrt{3}}y-\frac{1}{3}L}^{\frac{1}{3}L-\frac{1}{\sqrt{3}}y} x^2+y^2 dxdy\\ =\frac{\rho}{16 \sqrt{3}}L^4=(\frac{4M}{\sqrt{3} L^2})(\frac{L^4}{16\sqrt{3}})\\ =\frac{1}{12}ML^2$$

closed as off-topic by sammy gerbil, ZeroTheHero, ACuriousMind♦Jun 24 '18 at 13:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, ZeroTheHero, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question. Now analyse how the moment of inertia changes when we rescale its mass and sidelength, i.e. if $i = \alpha ml^2$ and $L = 2l, M = 4m$, then $I = \alpha (4m) (2l)^2 = 16i$, where $m = mass \ of\ a \ small \ triangle$, $l = sidelength \ of\ a\ small\ triangle$ and the capital $M, L$ correspond to the larger triangle. But the moment of inertia of the big triangle can be also split into $4$ moments of inertia. Be aware that we need to use the parallel axis theorem for the $3$ triangles which enclose the central triangle. Hence, $$I = 16i = 4i + 3m(\frac{l\sqrt{3}}{3})^2.$$ Which reduces to $i = \frac{1}{12} m l^2.$