Question

A truncated cube is a convex polyhedron with 36 edges and 24 vertices. A truncated tetrahedron is a convex polyhedron

with 18 edges and 12 vertices.

How do the number of faces of a truncated cube and a truncated tetrahedron compare?

The truncated cube has 6 more faces than the truncated tetrahedron.

The truncated cube has 8 more faces than the truncated tetrahedron.

The truncated cube has 12 more faces than the truncated tetrahedron.

The truncated cube has 18 more faces than the truncated tetrahedron.

Answer 1

Answer A:The truncated cube has 6 more faces than the truncated tetrahedron.

Explanation:

Answer 2

The truncated cube has 6 more faces than the truncated tetrahedron.

Explanation:

==>Given:

truncated cube=>

edges (E) = 36

vertices (V) = 24

faces (F) = ? (unknown)

According to Euler's polyhedron formula, Vertices (V) - Edges (E) + Faces (F) = 2, let's find how many faces the truncated cube has

=> 24 - 36 + F = 2

- 12 + F = 2

F = 2 + 12

F = 14

truncated tetrahedron=>

edges (E) = 18

vertices (V) = 12

faces (F) = ? (unknown)

Using V - E + F = 2, find F

=>12 - 18 + F = 2

- 6 + F = 2

F = 2 + 6

F = 8

Comparing the faces of the truncated cube (14) and the truncated tetrahedron (8), the truncated cube has 6 more faces than the truncated tetrahedron (14 - 8 = 6)