Prove that the static moment of a triangle with base

Prove that the static moment of a triangle with base $a$, with respect to that base, depends only on the height of the triangle (and not on where the apex sits horizontally).

Step 1 · General triangle

Let $A=(0,0),\,B=(a,0),\,C=(c,h)$ for arbitrary $c\in\mathbb R,\,h>0$. The left edge $AC$ gives $x_L(y)=\frac chy$; the right edge $BC$ gives $x_R(y)=a+\frac{c-a}hy$.

Step 2 · The width is independent of $c$

$$ w(y) = x_R(y)-x_L(y) = a+\frac{c-a}hy-\frac chy = a-\frac ahy = a\left(1-\frac yh\right), $$

with every occurrence of $c$ cancelling exactly.

Step 3 · Static moment

$$ M = \varrho\int_0^hy\cdot w(y)\,dy = \varrho a\int_0^h\left(y-\frac{y^2}h\right)dy = \varrho a\left(\frac{h^2}2-\frac{h^2}3\right) = \frac{\varrho ah^2}6. $$

This expression involves only $a$ and $h$, with no dependence on $c$ — proving the claim.

Additional Notes

• This derivation matches the source exactly.
• Equivalently, $M=m\bar y$ with $m=\frac{ah}2$ (area) and $\bar y=\frac h3$ — and the fact that every triangle's centroid sits at exactly $\frac13$ of the height above its base, regardless of the apex's horizontal position, is itself the geometric heart of why $M$ only depends on $h$.
• Verified numerically for several skewed apex positions (e.g. $a=2.4,\,h=1.8$, with $c=0,\,1,\,-3,\,5$): every case gives $M\approx1.296$, matching $\frac{ah^2}6$ exactly and confirming the independence from $c$.