# 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.

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Pentagon or equilateral triangle? Also, a center of mass is going to be a point, not a number. –  Zev Chonoles Jun 17 '13 at 20:23
the 'something' indicates the part to do with finding centre of mass of the triangle. And yes my working is trying to find the y value of the pentagon. –  maxmitch Jun 17 '13 at 20:25
The three perpendicular bisectors of an equilateral triangle intersect at a single point, which is its centroid. I presume you are wanting to break up the regular pentagon into five composite equilateral triangles; this geometric fact will give you a measurement to work with. –  RecklessReckoner Jun 17 '13 at 20:28
Ohh-kay, so now it's not a regular pentagon... –  RecklessReckoner Jun 17 '13 at 22:56

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 $$\tan(54^\circ)=\frac{1}{\sqrt{5-2\sqrt{5}}}.$$

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I don't think it is possible to have an irregular pentagon with all sides equal, right? –  Ovi Jun 17 '13 at 21:56
Here is an extreme case. Draw a line of length $2$, with a point in thw middle. That's the bottom, two of the sides. For two other sides, go up at $60^\circ$, for length $1$, and join up. There are plenty of less ectreme cases. You may be thinking of the true fact that for odd $n$, if all angles are equal, then all sides are. –  André Nicolas Jun 17 '13 at 23:25

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 $$y = \frac{64\times4 + 16\sqrt{3} \times (8 + 4/\sqrt{3})}{16\sqrt{3} + 64}$$

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