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Given bases $a$, $b$, and the height $h$.

trapezoid

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 $y$

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  • $\begingroup$ If line $m$ parallel to the bases of the trapezoid passes through the center of mass of the trapezoid, the lines $m$ and $c$ divide the trapezoid into four regions that are not equal in the general case. Do you want the center of mass, or do you want four equal regions? $\endgroup$ – David K Mar 15 '15 at 18:55
  • $\begingroup$ I want the center of mass, I thought it is the point where all regions gets equal,Isn't it? And a few minutes ago I also found a formula for $m$, which is $m^2 = \frac{a^2+b^2}{2}$ but that didn't helped me either to find formula for $y$ $\endgroup$ – shcolf Mar 15 '15 at 20:00
  • $\begingroup$ Center of mass means it balances there. Small masses far from the pivot point can balance large masses close to the pivot point, so if the object is asymmetric (like a general trapezoid) then one end can be lighter (but longer) than the other. $\endgroup$ – David K Mar 15 '15 at 20:04
  • $\begingroup$ So you mean that I'm doing wrong approach, but then how can I find that Point? $\endgroup$ – shcolf Mar 15 '15 at 20:12
  • $\begingroup$ I posted an answer. It took a while to write it up, because I wanted it to include a derivation. $\endgroup$ – David K Mar 15 '15 at 21:05
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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.

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1190922

$x=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+a+0) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (b+0+a) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$

$y=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+0+h) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (h+h+0) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$

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  • $\begingroup$ sorry, but can you explain how did you get this? $\endgroup$ – shcolf Mar 15 '15 at 16:24
  • $\begingroup$ @shcolf \\ 1 divided the trapezoid into two triangles \ 2 found area of each \ 3 found centroid of each \ 4 combined using analytic geometry procedures, treating cases for x and y separately $\endgroup$ – Senex Ægypti Parvi Mar 15 '15 at 21:32
  • $\begingroup$ $x=\frac{a^2+ab+b^2}{3(a+b)},y=\frac{h(a+2b)}{3(a+b)}$ $\endgroup$ – Senex Ægypti Parvi Mar 15 '15 at 21:57

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