polyhedrons need flat face and edges, corners which cylinder cones don't have.
Two polyhedrons have 18 edges: truncated tetrahedron and hexagonal prism.
I can just think of the pentagonal pyramid.
Yes. to add to that a vertex must be connected to at least 3 edges to be 3d, an edge is always connected to 2 vertexes, so the closest the two can ever be is vetexes x 3 = edges x 2, but when working with any platonic solid you can follow this: vertexes x (faces / vertexes) x [edges on one side] = edges x 2 or vertexes x [faces meeting at one vertex] = edges x 2 when working with any other polyhedron [vertexes with x amount of faces] x (x) + [vertexes with y amount of faces] x (y) ...{and so on} = edges x 2
A prism with an n-sided base will have 2n vertices, n + 2 faces, and 3n edges.
A cuboid douses have 8 verticess and 12 edges.
polyhedrons need flat face and edges, corners which cylinder cones don't have.
4 edges A rectangle has four edges. A rectangle has 4 sides but no edges which are normally applicable to polyhedrons.
polyhedrons need flat face and edges, corners which cylinder cones don't have.
Two polyhedrons have 18 edges: truncated tetrahedron and hexagonal prism.
I can just think of the pentagonal pyramid.
Add the number of faces of the shape to the number of vertices, and then subtract 2 to give you the number of edges. This works for most polyhedrons. I hope I have helped :) ALSO, take the number of vertices, divide that by two, then add that answer to the number of vertices... that will give you the number of edges unless it is a pyramid. Here is one that is GURANTEED TO WORK BECAUSE I HAVE TRIED IT. Here is is: Edge= 2 times(Vertice-1)
No. There must be at least three but theyre can be more.
No. For example, a cube is a polyhedron and 3 edges meet at each vertex.
you take face, than add the vertice, and subtract 2 from it this works for almost al polyhedrons but it doesn't work for a cylinder
pentagonal antiprism, pentagonal deltohedron(not deltahedron), decagonal pyramid, and many more.
A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edges.