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10 November, 18:21

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.

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  1. 10 November, 19:47
    0
    The truncated cube has 6 more faces than the truncated tetrahedron.

    Step-by-step 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)
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