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It has 6 vertices.

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Q: If a convex polyhedron has 12 edges and 8 faces then how many vertices does it have?
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Related questions

What has 4 faces 9 edges 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


How many faces if there is 16 vertices and 37 edges?

If the object is a convex polyhedron, then, by Euler's characteristics, it should have 23 faces.


A polyhedron with 12 vertices and 30 edges has how many faces?

A polyhedron has 30 edges and 12 vertices. How many faces does it have


What is the name of a shape that has 5 faces 8 edges 6 vertices?

There is not a polyhedron with the given number of faces, edges and vertices.


Why is a sphere not a polyhedron?

A sphere is not a polyhedron because it has no edges, no vertices and no flat faces The word 'polyhedron' means many faces.


What polyhedron have some a face but no edges and no vertices?

A polyhedron must have at least 4 faces, at least 4 vertices and at least 6 edges.


If a polyhedron has 5 faces and 6 vertices how many edges are there?

It is a triangular prism that has 5 faces, 6 vertices and 9 edges


If a polyhedron has 10 more edges than vertices how many faces does it have?

For all polyhedra: vertices + faces = edges + 2 The given fact is: edges = vertices + 10 → vertices + faces = vertices + 10 + 2 → faces = 12


Euler's rule for a polyhedron?

For a simply connected polyhedron,Faces + Vertices = Edges + 2


What 3d shape has 4 faces 8 edges and 5 vertices?

The numbers in the question do not satisfy the Euler characteristic so there cannot be such a [convex] polyhedron.


Is it possible for convex polyhedron to have 6 faces 8 vertices and 10 edges?

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.


Polyhedron has 10 vertices and 10 edges how many faces?

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.