Centre of mass of a rod with infinitesimal width The following is a thought that I am struggling to comprehend, most likely because of some logical flaw regarding infinitesimal objects.
Consider a rod of uniform density, length $L$ and thickness $T$. The horizontal distance from the side of the rod to the centre of mass of the rod is denoted $\overline{x}_{rod}$.
It is known that: $$\overline{x}_{rod} = \frac{L}{2}$$
Note how this distance does not depend on $T$. Hence, even as $T \to 0$, 
 $\overline{x} = \frac{L}{2}$.
I would conclude that this is therefore the centre of mass for a rod of infinitesimal thickness.
However, now consider a right-angled triangular body with a base of $B$ and a height of $H$. The horizontal distance from the  to the point-end of the body to its centre of mass is denoted $\overline{x}_{triangle}$.
It is known that: $$\overline{x}_{triangle} = \frac{2}{3}B$$
Note how this distance does not depend on $H$. Hence, even as $H \to 0$, 
 $\overline{x} = \frac{2}{3}B$.
This is where my confusion comes in. A triangle with an infinitely small $H$ is essentially a rod of infinitely small $T$ – is it not? Which is then the correct distance to the centre of mass of an infinitely thin rod?
My one thought is that regardless of how infinitely small $H$ is, it is still infinitely larger than the height on the other side of the triangle, which is $0$. Hence, the two objects, the rod and the triangle, are not the same.
Or perhaps because it is nonsensical to calculate the centre of mass of an object with infinitesimal mass? 
However, there still seems to be a sudden jump between $\overline{x} = \frac{2}{3}B$ and $\overline{x} = \frac{L}{2}$.
How is this possible? What am I missing?
 A: The mistake is in the line ``A triangle with an infinitely small $H$ is essentially a rod of infinitely small $T$''. They might appear to be the same, and in some aspects they are, but in others they are quite different. In fact, you already know a difference, as you pointed out the different locations of the centers of mass. The correct conclusion to draw from this observation is therefore that the limiting rod for a rectangle and that of a triangle are different objects, and hence there is no jump in the location of the center of mass.
Considering the triangle (in the 2D plane) with vertices $(0,0)$, $(B,0)$, and $(0,H)$, the location of its center of mass is at $(\frac{B}{3},\frac{H}{3})$.
Likewise for the triangle with vertices $(B,H)$, $(0,H)$, and $(B,0)$, the location of its center of mass is at $(\frac{2 B}{3},\frac{2 H}{3})$.
Whereas for the rectangle with vertices $(0,0)$, $(B,0)$, $(B,H)$, $(0,H)$, the centre of mass is located at $(\frac{B}{2},\frac{H}{2})$.
This is true regardless of the values of $B$ and $H$. Note that the centers of mass of either triangle are never coinciding with that of the rectangle. In fact the $x$-coordinate does not depend on the value of $H$. The center of mass of the combined triangles, however, which is the average of their two centers of mass locations, does correspond to that of the rectangle.
A: graphical hint:

.. independently from H
