According to the Euler characteristic which applies to all simply connected polyhedra,# edges + 2 = # vertices + # faces.
So the answer is 2 fewer.
An octahedron, for example. 8 faces, 6 vertices.
There are infinitely many sets. For example, a cube, cuboid, parallelepiped, rhombohedron and their less regular counterparts all have 6 quadrilateral faces, 12 edges and 8 vertices. There are similar sets for polyhedra with a different number of faces.
None. Using Euler's formula v - e + f = 2, where v is vertices, e is edges, and f is faces, we see that for your question f = 3. No solid figure can have less than 4 faces (a tetrahedron).
No. A cube has 6 faces and 8 vertices - it has exactly 2 more vertices than faces.
A nonagon (9 sides) for example.
A cube
6 faced more or less.. !! Counting faces and one of the bottom = 7 .!! See yaa kisees . ;D hope my ansewer help u a lot.!
There's no least, and there's no most. There's only one set of numbers: 12 edges and 8 vertices. That's it. No more, no less, no least, no most.
None.A polygon is made up of straight line edges between its vertices. There are the same number of edges as vertices in a polygon.In the case of a polygon, it is convex if all interior angles are less than 180o.
There is no such figure.
A rhombus is a two-dimensional shape, so it does not have any vehicles. However, it does have edges and faces. A rhombus is a quadrilateral with four sides of equal length. Each pair of opposite sides is parallel and the opposite angles are equal. A rhombus has the following properties: It has four edges or sides of equal length. It has four vertices (corners) where the edges meet. It has four angles, each of which measures less than 180 degrees. It has two diagonals that bisect each other at right angles. Since a rhombus is a flat shape, it has only one face, which is a quadrilateral with four sides. Therefore, a rhombus has 4 edges, 1 face, and 0 vehicles. Hope this helps you!
It is a sphere, a cylinder and a cone that have less than 3 vertices!