Question

Can a polyhedron have 10 faces, 20 edges and 15 vertices?

Answer 1

Hint: We know that Euler’s formula holds for every polyhedron. Here, we are given that the number of faces is 10, the number of edges is 20 and the number of vertices is 15. If the given conditions satisfy the Euler’s formula, then such a polyhedron is possible otherwise not.

Complete step by step solution: We know that a polyhedron is a three-dimensional closed figure with straight edges, flat surfaces and sharp vertices.
Also, it is known that Euler’s formula holds true for every polyhedron.
Euler’s formula gives us the relationship between the number of faces, number of edges and number of faces of a polyhedron.
So, here, we can apply the Euler’s formula on the given number of faces, edges and vertices to check if it forms a polyhedron.
Euler’s formula states that the summation of the number of faces and vertices is exactly two more than the number of edges, which can be written as
$F + V = E + 2$ or
$F + V - E = 2$, where $F$ represents the number of faces, $V$ represents the number of vertices and $E$ represents the number of edges.
We are given that the number of faces is 10, the number of edges are 20 and number of vertices are 15.
On substituting the values in the Euler’s formula, we will get,
$
  10 + 15 - 20\mathop = \limits^? 2 \\
  25 - 20\mathop = \limits^? 2 \\
  5 \ne 2 \\
$

Since the Euler’s formula does not hold true for the given number of faces, edges and vertices, therefore, there does not exist any polyhedron with 10 faces, 20 edges and 15 vertices.

Note: Euler’s formula does not work for the shapes with holes or when the opposite faces intersect together at a common point. Euler’s formula is very helpful to determine if a polyhedron is possible when the number of edges, faces and vertices are given.