How to find the centre of mass of an equilateral triangle? I have a pentagon lamina with all the sides length 8cm to find the centre of mass of the pentagon do you do:
$$\frac{4+8+ \frac{1}{3}*something?}{8^2+16\sqrt3}$$
A uniform lamina ABCDE consists of a square ACDE and an equilateral triangle ABC. Find the distance of the centre of mass from AC.
 A: Specifying the sides is not enough. You probably mean that also the angles are all equal, so we are dealing with the regular pentagon of side $8$. 
The centre of mass will be at the centre of the circle that the regular pentagon is inscribed in. 
To describe the location of the centre, imagine that $AB$ is one of the edges, and let $O$ be the centre of the circle. Let $M$ be the midpoint of $AB$. To get to $O$, we draw the perpendicular bisector of $AB$, and go up a distance $d$ from $M$, where $d=4\tan(54^\circ)$. 
Remark: One can get an explicit expression for the required $\tan(54^\circ)$ in terms of square roots if that is desired.  It turns out that
$$\tan(54^\circ)=\frac{1}{\sqrt{5-2\sqrt{5}}}.$$  
A: Ler $A$ be the center of mass of the triangle, $B$ of the square, $C$ of the lamina. Then (we are calculating a barycentre), $$C = \frac{\text{Area(triangle)}A + \text{Area(square)}B}{\text{Area(lamina)}}$$.
$$B = (4,4)$$ $$C = (4, 8+4 \tan(\pi/6)) = (4, 8 + 4/\sqrt{3})$$
$$Area(square) = 64$$
$$\text{Area(triangle)} = 8 \times 4\tan(\pi/3) = 16\sqrt{3}$$
If we put $C=(x,y)$, $x=4$ and $$y = \frac{64\times4 + 16\sqrt{3} \times (8 + 4/\sqrt{3})}{16\sqrt{3} + 64}$$
