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