How do you make a model triangular pyramid?

Answer 1

I had trouble with this one as well, and this is the only comment i could find for exactly this problem, so here i am.

Treat the side of the pyramid ( not the base edge ) as a vector.

in vector geometry, we can find the projection of this vector onto the plane of the base by using the definition of the dot product

Set the origin of a 3d coordinate system as the apex of the pyramid, with an edge along an axes, for simplicity's sake

u.v= u.vcos(theta)

Where u is a vector along the side of the triangular pyramid and v is any vector in the plane of the base of the pyramid, and u,v are their respective magnitudes.

Use these to solve for the angle theta which is the angle formed between the side of the triangular pyramid and the base.

Now u.cos(theta) is the projection of u along the base of the pyramid, which coincides with the centroid of the base, directly below the apex of the pyramid, if directed properly.

We don't actually need to know the direction here, because we have enough information to go for the goal.

The lengths of the u and it's projection into the base are the hypotenuse and adjacent sides, respectively, of a right triangle, where the height of the triangular pyramid, a simple Pythagoras theorem gives us the desired answer

I hope this helped, many bothuns died to bring us this information

2 notes, first being that there is, in all likelihood, a simpler answer then this, and second, if you arent familiar with vector geometry I can try to develop a "plug and play" formula for you

I really hope this helped

Oh i was thinking it doesn't need to be equal sides the projection of any side onto the plane of the base will be directly below the apex and correct when used with the length of that side in the right triangle ( again, hypotenuse=side length, adjacent=length of projection of side onto plane of base, opposite=perpendicular height of triangular pyramid