General formula for centre of mass of polygons

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SUMMARY

The center of mass for a triangle is calculated using the formula ##\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})##. For a planar four-sided figure (quadrilateral) ABCD, the position vector of the center of mass is ##\frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})##. This formula generalizes to n-sided figures, confirming that the center of mass can be determined similarly for any polygon. In three-dimensional space, the center of mass for an arbitrary triangular pyramid also follows the same principle, maintaining the formula ##\frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})##.

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The position vector of the center of mass of a triangle is ##\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})##.

Screen Shot 2016-07-16 at 4.08.19 am.png


Is the position vector of the center of mass of a planar four-sided figure ABCD ##\frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})##? Does this generalise to n-sided figure?

How about a 3D figure ABCD in the form of an arbitrary triangular pyramid (whose faces need not be equilateral)? Is the center of mass also ##\frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})##? Does this generalise to n-dimensonal figure?
 
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