Centre of Mass of a Triangle

Result

The CM of a uniform triangular lamina of height $h$ lies at $\frac{h}{3}$ from the base:

$y_{cm} = \frac{h}{3}$

This point is the centroid of the triangle — the intersection of the three medians.

Derivation

Place the triangle with base $b$ along the $x$-axis and apex at height $h$.

At height $y$, by similar triangles, the width of the horizontal strip is:

$w(y) = b\left(1 - \frac{y}{h}\right)$

Mass element (strip of width $w$, height $dy$, surface density $\sigma$):

$dm = \sigma \cdot w(y) \, dy = \sigma b\left(1 - \frac{y}{h}\right) dy$

Total mass: $M = \sigma \cdot \frac{1}{2}bh$

$y_{cm} = \frac{1}{M}\int_0^h y \, dm = \frac{\sigma b}{\sigma bh/2} \int_0^h y\left(1 - \frac{y}{h}\right) dy$

$= \frac{2}{h}\int_0^h \left(y - \frac{y^2}{h}\right) dy = \frac{2}{h}\left[\frac{h^2}{2} - \frac{h^2}{3}\right] = \frac{2}{h} \cdot \frac{h^2}{6} = \frac{h}{3}$

$\boxed{y_{cm} = \frac{h}{3}}$

The centroid property

The CM of a triangle is also its centroid — the point where the three medians intersect. Each median divides the triangle into two triangles of equal area.

The centroid divides each median in ratio $2:1$ from the vertex. So the CM is $\frac{1}{3}$ of the way from the base and $\frac{2}{3}$ of the way from the apex.

For any triangle, regardless of shape

This result holds for any triangle — equilateral, isoceles, right-angled, or scalene — as long as it is uniform.

The CM coordinates from any vertex:

$x_{cm} = \frac{x_1 + x_2 + x_3}{3}, \quad y_{cm} = \frac{y_1 + y_2 + y_3}{3}$

The centroid is the average of the three vertex coordinates.