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Re: center of gravity formula
Posted:
May 19, 1998 9:41 PM


Imrich Miklosko wrote:
> We have an isosceles triangle with 2 sides equal to 100 cm. We have > figured out the third side to be equal to 1.7453292. We need to know > the formula to figure out the center of gravity of this triangle. I > know how to draw the intersection point of the three angles, now I > need to find out the distances of the intersection points. ___x___ ___ > \ /  r\ /r  \ /  V \   Any advice would be appreciated. Thanks > in advace.
That's a mighty skinny triangle! The center of gravity is simply the point where the 3 medians meet. By the way, that point (the centroid) is also 2/3 of the way from any vertex to the midpoint of the opposite side. I hope this helps.



