No, F + V = E + 2That's Euler's polyhedron formula (or Theorem). For a normal 3-d polyhedron to exist it must conform to that equation.
A polyhedron must have at least 4 faces, at least 4 vertices and at least 6 edges.
eighthedron eighthedron
Such a polyhedron cannot exist. According to the Euler characteristics, V + F - E = 2, where V = vertices, F = faces, E = edges. This would require that the polyhedron had only two faces.
12 vertices A prism with an n-sided base will have 2n vertices, n + 2 faces, and 3n edges.
For all polyhedra: vertices + faces = edges + 2 The given fact is: edges = vertices + 10 → vertices + faces = vertices + 10 + 2 → faces = 12
It has 6 vertices.
There is no such convex polyhedron in normal geometries because it does not satisfy the Euler characteristic. That requires that Faces + Vertices = Edges + 2
If the object is a convex polyhedron, then, by Euler's characteristics, it should have 23 faces.
The number of edges and vertices ina polyhedron will depend on the polyhedron one selects either to study, build or etc...
A polyhedron has 30 edges and 12 vertices. How many faces does it have
A polyhedron must have at least 4 faces, at least 4 vertices and at least 6 edges.
The numbers in the question do not satisfy the Euler characteristic so there cannot be such a [convex] polyhedron.
This polyhedron has 7 vertices and 12 edges.
I believe you intend to talk about a polyhedron if it is a convex polyhedron, there is a relation : F + V * E = 2 (you can experiment with current polyhedrons) the relation is not satisfied by your numbers
Edges join two vertices. Three or more edges meet at a vertex (it could be two if not in a polyhedron).
For a simply connected polyhedron,Faces + Vertices = Edges + 2
There can be no such convex polyhedron since the numbers do not satisfy the Euler characteristic which states that V - E + F = 2