What does the $a$ stand for in this picture? (And some clarification) 
(source: polygonal.de)

(source: polygonal.de)
$$ I_{triangle} = \frac{b^3h-b^2ha+bha^2+bh^3}{36} $$
$$ I_{total} = \sum^n_{k=1}(I_{triangle}+Md^2)_k $$
Source is here.
I'm trying to understand the mass moment of inertia in order to create a 2d physics engine, but I haven't found much resources to go by. One I did find was the above one, but I didn't fully understand it, and wanted to make sure about a few things before going through the effort of translating it into C++.

*

*I am correctly understanding that this is for the mass moment of inertia and not something else like the area moment of inertia right?

*What would $a$ be when the triangle is obtuse?

*Is my understanding that the mass moment of inertia is dependent on the pivot point accurate, and that if the pivot moved away from the center of mass I'd just need to adjust the 'd's to get the new mass moment of inertia value?

Alternatively, if anybody is aware of a simpler way of going about getting the mass moment of inertia for a polygon, I'd love to hear it.
 A: 
I am correctly understanding that this is for the mass moment of inertia and not something else like the area moment of inertia right?

No. The formula given are for the area moment of inertia. Although you can essentially use them to find the moment of inertia just by multiplying the answer, you get using this formula by the area mass density of the plate. (i.e. mass per unit area of the plate).

What would $a$ be when the triangle is obtuse?

$a$ is the distance between the foot of perpendicular drawn from the largest angle vertex on the opposite side and any of the endpoints of that opposite side. (You can chose any of the two vertices in each case because the expression in terms of $a$ is symmetric for $a$ and $b-a$. So we do not really need to care whether the triangle is obtuse or right or acute, because we are drawing the perpendicular from the largest angle vertex, we are sure that it is going to land on the opposite side. (not outside, as you were fearing perhaps)

Is my understanding that the mass moment of inertia is dependent on the pivot point accurate, and that if the pivot moved away from the center of mass I'd just need to adjust the 'd's to get the new mass moment of inertia value?

Yes. You would need to calculate new $d$s if you are changing the pivot. Alternatively, you can again apply the parallel axis theorem. 
A: 
What would  be when the triangle is obtuse ?

Moment of inertia in increasing order :
Acute triangle
$$ I_{acute} = \frac{\left(b^3h+bh^3\right)+bha^2-b^2ha}{36} 
\\=\frac{\left(b^3h+bh^3\right)+a b h (a - b)}{36} $$
for acute triangle $a < b$, so second term will be negative, thus $I_{acute} < I_{right}$
Right triangle
$a = 0$, so
$$ I_{right} = \frac{\left(b^3h+bh^3\right)}{36} $$
Obtuse triangle
$a \Rightarrow -a$, so
$$ I_{obtuse} = \frac{\left(b^3h+bh^3\right)+bha^2+b^2ha}{36} $$
EDIT
For better understanding of $a$ parameter - picture of all triangle types with their corresponding $a$ parameter:

