Number of Hexagonal Faces in a Truncated Octahedron

To find the number of hexagonal faces in a truncated octahedron, we consider the geometry of a regular octahedron and the process of truncation:

1. Structure of a Regular Octahedron:

   • A regular octahedron has:
     • $V_0 = 6$ vertices
     • $E_0 = 12$ edges
     • $F_0 = 8$ triangular faces

2. Truncation Process:

   • A truncated octahedron is an Archimedean solid formed by slicing off (truncating) each of the vertices of a regular octahedron to a depth of one-third the edge length.
   • At each of the $6$ vertices of the octahedron, $4$ edges meet. Slicing off a vertex results in a $4$-sided polygon, creating a square face. Since there are $6$ vertices, this truncation generates:
     $\text{Number of square faces} = 6$
   • Each of the original $8$ triangular faces has $3$ corners trimmed off. Truncating the $3$ vertices of a triangle transforms it into a $6$-sided polygon (a regular hexagon). Since there were originally $8$ triangular faces, this gives:
     $\text{Number of hexagonal faces} = 8$

3. Total Faces:

   • The truncated octahedron consists of a total of $14$ faces ($8$ regular hexagons and $6$ squares), $24$ vertices, and $36$ edges.

Therefore, the number of hexagonal faces present in a truncated octahedron is 8.