Question

***15 Points***

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?


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

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

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

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

Answer 1

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

Explanation:

In every convex polyhedron, we have that:

C + V = E + 2

Where C is the number of faces, V is the number of vertices and E is the number of edges.

Then, for the truncated cube we replace E by 36 and V by 24 and get:

C + 24 =36 + 2

Solving for C, we get that the number of faces is equal to:

C = 36 + 2 - 24

C = 14

At the same way, for the truncated tetrahedron, we get that the number of faces is:

C + 12 = 18 + 2

C = 18 + 2 - 12

C = 8

So, if we compare both polyhedrons we get that the truncated cube has 6 more faces than the truncated tetrahedron.

14 - 8 = 6