Bipyramids are a class of polyhedra with more faces than vertices.
A cube has 8 vertices and 6 faces. Therefore a cube has 2 more vertices than faces.
3d shapes with plane faces are called polyhedra (singular = polyhedron). The minimum number of vertices is 4, there is no maximum.
The number of vertices does not determine the number of faces. If the shape with 6 vertices was a quadrilateral based bipyramid, it would have 8 faces. A hexagonal based pyramid has 7 vertices and 7 faces. So more vertices does not necessarily imply more faces.
A cube is a geometric shape which has 6 faces and 8 vertices ie .2 more vertices than faces
A tetrahedron (4 triangular faces) is the only polyhedron that has a unique configuration of faces, edges and vertices. For any polyhedron with n (>4) faces, there is a prism with a pair of n-2 sided polygons as bases as well a pyramid whose base is an n-1 sided polygon. There are many other configurations for polyhedra with more faces. For example, there are ten [topologically] different figures with 6 faces: 3 of these are concave polyhedra.
A polyhedron is a generic term for 3 dimensional objects which are bounded by polygonal faces. They can have 4 or more vertices, 6 or more edges and 4 or more faces. The numbers of vertices (V), edges (E) and faces (F) must also satisfy the Euler characteristic: F + V = E + 2.
Oh, dude, it's like a math riddle! So, if a polyhedron has 10 more edges than vertices, we can use Euler's formula: Faces + Vertices - Edges = 2. Since we know the relationship between edges and vertices, we can substitute that in and solve for faces. So, it would have 22 faces. Math can be fun... sometimes.
3d shapes with plane faces are called polyhedra (singular = polyhedron). The minimum number of vertices is 4, there is no maximum.
A cube has 8 vertices and 6 faces. Therefore a cube has 2 more vertices than faces.
Oh, dude, that's an easy one. A shape like a cube has more faces than vertices. Like, think about it, a cube has 6 faces but only 8 vertices. So, yeah, the faces totally outnumber the vertices in that situation.
The number of vertices does not determine the number of faces. If the shape with 6 vertices was a quadrilateral based bipyramid, it would have 8 faces. A hexagonal based pyramid has 7 vertices and 7 faces. So more vertices does not necessarily imply more faces.
A cube is a geometric shape which has 6 faces and 8 vertices ie .2 more vertices than faces
An octahedron, for example. 8 faces, 6 vertices.
There are infinitely many polyhedra with one or more triangular faces.
A tetrahedron (4 triangular faces) is the only polyhedron that has a unique configuration of faces, edges and vertices. For any polyhedron with n (>4) faces, there is a prism with a pair of n-2 sided polygons as bases as well a pyramid whose base is an n-1 sided polygon. There are many other configurations for polyhedra with more faces. For example, there are ten [topologically] different figures with 6 faces: 3 of these are concave polyhedra.
No. A cube has 6 faces and 8 vertices - it has exactly 2 more vertices than faces.
There are a few families of polyhedra with identical faces. There are none whose faces have 6 or more sides. There is no special name for polyhedra whose faces are pentagons or pentagrams. A dodecahedron is an example. If coplanar faces are disallowed, the only polyhedron with quadrilateral faces are the cube and rhombohedron. There are infinitely many polyhedra with equilateral triangular faces: the tetrahedron, octahedron and icosahedron are examples.