Center of mass of Right angle Trapezoid Given bases $a$, $b$, and the height $h$.

Get the $M(x,y)$ coordinates formula from point $O(0,0)$, where $M$ is center of mass. Wiki has a formula for $M(y) = \frac{h}{3}\frac{2a+b}{a+b}$. And I'm interested in how to find that formula (also for $M(x)$).
The median $c$ is dividing the Trapezoid into 2 equal surfaces. I need to draw $m$ in right place $y$, so all 4 surfaces stay equal each other.
Here's my try:
$y(\frac{a+m}{2})=(h-y)(\frac{m+b}{2})$
$\frac{a+m}{2}=(\frac{h}{y}-1)(\frac{m+b}{2})$
$y=\frac{h(m+b)}{a+2m+b}$
on the other look
$2y(\frac{a+m}{2})=h(\frac{a+b}{2})$
$y=\frac{h(a+b)}{2(a+m)} = \frac{h(m+b)}{a+2m+b}$
From the last equality we have: $m^2 = \frac{a^2+b^2}{2}$
But I still cant get that formula back for $y$
 A: The formula for the distance from the base $b$ to the center of mass of a trapezoid
is 
$$\bar y = \frac{b+2a}{3(a+b)} h.$$
You can find this in many on-line sources such as Wolfram Mathworld.
You can prove this by integrating $\int_0^h y(b + (a-b)\frac yh) dy$
and dividing by the area of the trapezoid. But here's a derivation without calculus,
using the fact that the distance from a side of a triangle to the triangle's
centroid is $\frac13$ the height of the triangle.
Let $T$ be a trapezoid with bases $a$ and $b$.
For the case $a < b$, extend the non-parallel sides of the trapezoid
until they intersect. The base $b$ of the trapezoid and the two extended
sides form a triangle $B$; the base $a$ divides this triangle into two pieces,
one of which is $T$ and the other of which is a triangle which we'll call $A$.
If the height of the trapezoid $T$ is $h$, the height of $B$ is $\frac{b}{b-a}h.$
The centroid of $B$ is at a distance $\frac{b}{3(b-a)}h$ from base $b$.
But another way to find the centroid of $B$ is to balance the two figures
$A$ and $T$ that compose $B$.
The triangle $A$ has height $\frac{a}{b-a}h$, so its area is
$\frac{a^2}{2(b-a)}h$ and its centroid is a distance $\frac{a}{3(b-a)}h + h$
from the base $b$ of the trapezoid.
Trapezoid $T$ has area $\frac{a+b}{2} h$ and a centroid at the unknown
distance $\bar y$ from base $b$.
To "balance" the two regions, we take a weighted average 
of the distance of their centroids from base $b$.
The "weights" in this average are just the areas of the two regions.
This weighted average is the same as the distance of the centroid of $B$
(the combined figure) from base $b$.
That is,
$$\begin{eqnarray}
\frac{b}{3(b-a)}h
 &=&\frac{\mathop{Area}(A) \cdot \left(\frac{a}{3(b-a)}h + h\right)
       + \mathop{Area}(T) \cdot \bar y}{\mathop{Area}(A) + \mathop{Area}(T)}\\
 &=&\frac{\frac{a^2}{2(b-a)}h \cdot \left(\frac{a}{3(b-a)} + 1\right)h
       + \frac{a+b}{2} h \cdot \bar y}{\frac{a^2}{2(b-a)}h + \frac{a+b}{2} h}\\
\end{eqnarray}$$
Solve for $\bar y$. This looks messy, but it can be simplified if you
realize that
$$\frac{\mathop{Area}(T)}{\mathop{Area}(A)} = \frac{b^2 - a^2}{a^2}.$$
If you divide both the numerator and denominator on the 
right-hand side of the weighted average by $\mathop{Area}(A)$, you get
$$
\frac{b}{3(b-a)}h
 = \frac{ \frac{3b - 2a}{3(b-a)} h
       + \frac{b^2 - a^2}{a^2} \cdot \bar y}{1 + \frac{b^2 - a^2}{a^2}}
$$
After you finish collecting all the terms in
$a$, $b$, and $h$ on the left side of this equation, and factor
$(b-a)^2$ out of $b^3 - 3ba + 2a^3$, it all simplifies to the formula
for $\bar y$ shown above.
A: 1190922   
$x=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+a+0) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (b+0+a) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix}      + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$   
$y=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+0+h) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (h+h+0) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix}      + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$   
