Verify Euler's formula F+V-E=2 for a cube and tetrahedron

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Question

Verify Euler’s formula F+VE=2F + V - E = 2 for: (a) A cube (b) A tetrahedron

where FF = number of faces, VV = number of vertices, and EE = number of edges.

Solution — Step by Step

Euler’s formula for polyhedra states:

F+VE=2F + V - E = 2

This formula holds for all convex polyhedra — three-dimensional shapes where all faces are flat polygons, all vertices are points where edges meet, and the shape is “convex” (no dents). It was discovered by Leonhard Euler in the 18th century.

A cube has:

  • Faces (F): 6 (top, bottom, front, back, left, right — all squares)
  • Vertices (V): 8 (the 8 corner points)
  • Edges (E): 12 (4 on top, 4 on bottom, 4 vertical edges connecting them)

Apply Euler’s formula:

F+VE=6+812=1412=2F + V - E = 6 + 8 - 12 = 14 - 12 = \boxed{2} \checkmark

The formula is satisfied.

A tetrahedron (triangular pyramid) is the simplest polyhedron — 4 triangular faces:

  • Faces (F): 4 (all equilateral triangles in a regular tetrahedron)
  • Vertices (V): 4 (one at each apex of the triangles)
  • Edges (E): 6 (each triangle shares its edges with an adjacent face)

Apply Euler’s formula:

F+VE=4+46=86=2F + V - E = 4 + 4 - 6 = 8 - 6 = \boxed{2} \checkmark

The formula is satisfied.

PolyhedronFVEF+V-E
Tetrahedron4462
Cube68122
Square pyramid5582
Triangular prism5692
Octahedron86122

No matter what convex polyhedron you choose, F+VEF + V - E always equals 2. This is remarkable — the shape’s topology (the way faces, edges, and vertices are connected) always yields this invariant.

Why This Works

Euler’s formula is a topological theorem — it’s about the connectivity structure of the polyhedron, not its specific geometry. A cube and a sphere are topologically equivalent (you can continuously deform one into the other without tearing), and Euler’s formula holds for both.

The number 2 represents the Euler characteristic of the sphere. Any polyhedron that is topologically equivalent to a sphere has Euler characteristic 2.

For shapes with “holes” (like a torus/donut), the Euler characteristic is 0. A donut-shaped polyhedron would satisfy F+VE=0F + V - E = 0.

Alternative Method

For a quick check without memorising the counts, use these rules for common shapes:

  • Prism with nn-gon base: F=n+2F = n+2, V=2nV = 2n, E=3nE = 3n. Check: (n+2)+2n3n=2(n+2) + 2n - 3n = 2
  • Pyramid with nn-gon base: F=n+1F = n+1, V=n+1V = n+1, E=2nE = 2n. Check: (n+1)+(n+1)2n=2(n+1) + (n+1) - 2n = 2

Common Mistake

Students sometimes miscount edges of a cube. A systematic way: the 12 edges of a cube come from 4 on the top face + 4 on the bottom face + 4 vertical edges connecting corresponding vertices. Do NOT count each edge twice — each edge is shared by exactly 2 faces. If you count by faces (each face has 4 edges, 6 faces → 6×4=246 \times 4 = 24) and divide by 2 (each edge is shared): 24/2=1224/2 = 12. This systematic method prevents undercounting or overcounting.

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