Moment of Inertia (triangular plate) I want to generalize the formula for the MOI of a triangular plate (sides $a,b,c$) about an axis passing through mid point of one sides and perpendicular to it's plane . 

The mass of plate $M$ is uniformly distributed on it's area. 
I can use parallel axis theorem if I know the MOI about an axis passing through COM of the plate, but I don't even know that. Also there is not symmetry in this, so taking small strips of masses to integrate will also not help.So, I was unable to start
Please Help.
If you solve by integration then please help me get the integration term too.
 A: It's actually not too hard to calculate the moment of inertia (MOI) of a right triangle.  And you can make your triangle out of a large right triangle minus a smaller right triangle.  So your MOI is just the MOI of the bigger triangle minus the MOI of the smaller one.
Step 1:
Extend line $b$ (move vertex $C$) until you have a right triangle.  We'll calculate the MOI relative to vertex $A$.  I agree with @fibonatic that it's easiest to use polar coordinates.  So we have 
\begin{equation}
I_{ABC'} = \int_0^\alpha \int_0^{R(\theta)} (\rho_A\, r^2) r\, dr\, d\theta~.
\end{equation}
Here, $\alpha$ is the angle $\angle B A C$, which you can calculate using trig.  Also, $R(\theta)$ is the length of a line $Aa$ that goes from vertex $A$ to line $BC=a$, where $\theta$ is the angle of that line above line $b$.  A little simple trig says that $R(\theta) = b/\cos\theta$.
\begin{align}
I_{ABC'} &= \rho_A\, \int_0^\alpha \int_0^{b/\cos\theta} r^3\, dr\, d\theta 
\\
&= \rho_A\, \int_0^\alpha \left. \frac{r^4}{4} \right|_0^{b/ \cos\theta}\, d\theta 
\\
&= \frac{\rho_A\, b^4}{4}\, \int_0^\alpha \cos^{-4}\theta\, d\theta
\\
&= \frac{\rho_A\, b^4}{24}\, \frac{3 \sin\alpha+\sin (3 \alpha
   )} {\cos^3\alpha}
\end{align}
[I'm terrible at integrals, so I just looked up the answer for $\cos^{-4}$.  But I think it's easy enough to convince you that you can handle any reasonable MOI problem.]
Step 2:
Use this formula and the parallel-axis theorem to get the MOI of the smaller right triangle that you need to remove from the bigger one to get your triangle.
Step 3:
Subtract the result of step 2 from step 1 to get the MOI of your triangle about the vertex $A$.
Step 4:
Use the parallel-axis theorem to get your MOI relative to whatever origin you want.
A: Moment of Inertia is defined as:
$$
I={\sum}mr^2
$$
which in this case can be rewritten into an integral:
$$
I=\rho\int_A{r^2dA}
$$
Since the shape of the triangle can't be described by one formula, you would have to split the integral into multiple sections. And I will use polar coordinates, in which case $dA=rd\theta dr$:
$$
I=\rho\left(\int_{\theta_1}^{\theta_2}\int_0^{r(\theta)}r^3drd\theta+\int_{\theta_2}^{\theta_3}\int_0^{r(\theta)}r^3drd\theta\right)
$$
with $\theta_1=0$ radians, $\theta_2$ is equal to the angle between $AOB$ which is equals: $\theta_2=\pi-\alpha-\frac{\beta}{2}$, with $\alpha$ the angle between $BAC=\cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)$ and $\beta$ the angle between $ABC=\cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)$. And $\theta_3=\pi$ radians. For the first integral the range to which $r$ has to integrated equals: $r(\theta)=\frac{1/2b\sin\alpha}{\sin(\pi-\alpha-\theta)}$. For the second integral the range to which $r$ has to integrated equals: $r(\theta)=\frac{1/2b\sin\gamma}{\sin(\pi-\gamma-(\pi-\theta))}=\frac{1/2b\sin\gamma}{\sin(\theta-\gamma)}$, with $\gamma=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)$.
I hope this helped, and if you are still not able to solve this, I will try to add the solution to this equation.
Edit: here is the expression after solving the integral:
$$
I=\frac{\rho b^4}{192}\left(\sin\gamma\cos\gamma(2-\cos(2\gamma))-\frac{\sin^4\gamma\cos(\gamma-\theta_2)(2-\cos(2\gamma-2\theta_2))}{\sin^3(\gamma-\theta_2)}+\sin{\alpha}\cos{\alpha}(2-\cos(2\alpha))-\frac{\sin^4\alpha\cos(\alpha-\theta_2)(2-\cos(2\alpha-2\theta_2))}{\sin^3(\alpha-\theta_2)}\right)
$$
A: Now I have got a method to get it directly.And again it came out to be an easy problem.
The answer comes out to be  $$I= \dfrac{m}{12}(a^2+c^2)$$
See if we add another such plate along it's side $AC$ , then it comes out to be a parallelogram plate , whose MOI is known, same as rectangular plate
So, by symmetry arguments , both the triangular plates have the same MOI.



