Vertices, edges and faces

Pick up any solid with flat sides — a brick, a tent, a gemstone — and three simple things describe its shape completely: the corner points, the straight edges joining them, and the flat faces they enclose. Count all three for a cube and something surprising happens: $6$ faces, $8$ vertices, $12$ edges, and $6+8-12=2$ . Try a triangular prism, a square pyramid, any convex solid bounded only by flat polygons, and the same combination always lands on $2$ . This is not a coincidence — it is Euler's formula, one of the oldest and most useful relationships in solid geometry. This article builds the definitions, proves why the formula must hold for prisms and pyramids of any size, and shows how to use it as a fast error check whenever you sketch or count a 3-D solid.

The core insight

For any convex polyhedron, faces plus vertices minus edges always equals two, $F+V-E=2$ , so knowing any two of the three counts fixes the third.

The core idea

A polyhedron is a solid bounded only by flat polygonal faces — no curved surfaces allowed. Three building blocks describe it:

A face is one flat polygon forming part of the solid's boundary (a cube has $6$ square faces).
An edge is the line segment where exactly two faces meet (a cube has $12$ edges).
A vertex is a corner point where three or more edges meet (a cube has $8$ vertices).

These three counts are never independent. Euler's formula ties them together for every convex polyhedron:

F + V - E = 2

where: $F$ is the number of faces, $V$ is the number of vertices, and $E$ is the number of edges. Rearranged, the same relationship finds any missing count once the other two are known:

E = F + V - 2, \qquad V = E - F + 2, \qquad F = E - V + 2

Two families of solids appear over and over in this chapter. For a prism whose cross-section is an $n$ -sided polygon (an $n$ -gon), the two end faces plus $n$ rectangular side faces give $F=n+2$ ; each end contributes $n$ vertices, so $V=2n$ ; and the edges are the $n$ edges of each end plus $n$ vertical connectors, so $E=3n$ . For a pyramid on an $n$ -sided base, the base plus $n$ triangular sides give $F=n+1$ ; the base's $n$ corners plus the single apex give $V=n+1$ ; and the edges are the $n$ base edges plus $n$ slant edges up to the apex, so $E=2n$ . Both families satisfy Euler's formula for every value of $n$ , which is exactly what the derivation below proves. Counting from a solid or its net is where students actually make mistakes, so it helps to count in a fixed order every time: faces first (they are usually the easiest to see and rarely missed), then vertices (walk around the solid corner by corner), and edges last, cross-checked against $E=F+V-2$ . A net — the flat, unfolded pattern of a solid's faces — is a useful cross-check too: every face in the net becomes a face of the solid, but edges that are cut open in the net to let it lie flat get glued back together when it is folded, so a net's outer boundary edges do not equal the solid's edge count directly. This is why counting directly on the folded solid, or on a clear 3-D sketch, is safer than counting straight off a net.

Figure

A wireframe cube with its eight corner vertices marked as dots, its twelve edges drawn as lines, and its faces shaded, alongside the count six faces eight vertices twelve edges satisfying Euler's formula

Derivation AWhy Euler's formula holds for every prism and pyramid

The goal is to show, using nothing beyond algebra, that the combination $F+V-E$ equals exactly $2$ for a prism with any $n$ -sided base and for a pyramid with any $n$ -sided base, so that Euler's formula is not a lucky pattern seen in a handful of examples but a guaranteed identity for these two whole families of solids.

Step 1. Write the prism counts in terms of $n$ , the number of sides of its polygonal base. A prism has two identical $n$ -gon ends plus $n$ rectangular side faces joining them, so $F=n+2$ . Each end supplies $n$ corner vertices and the two ends are distinct sets of vertices, so $V=2n$ . The edges are the $n$ edges of the top $n$ -gon, the $n$ edges of the bottom $n$ -gon, and the $n$ vertical edges joining each top vertex to the matching bottom vertex, giving $E=3n$ .

Step 2. Substitute these three expressions into $F+V-E$ and simplify. $F+V-E = (n+2) + (2n) - (3n) = n+2+2n-3n = 2$ . The terms containing $n$ cancel exactly, leaving $2$ regardless of what value $n$ takes — this is why every prism from a triangular prism ( $n=3$ ) to a decagonal prism ( $n=10$ ) obeys Euler's formula.

Step 3. Repeat the same substitution for a pyramid on an $n$ -sided base. Here $F=n+1$ (the base plus $n$ triangular sides), $V=n+1$ (the $n$ base corners plus one apex), and $E=2n$ (the $n$ base edges plus $n$ slant edges to the apex). So $F+V-E = (n+1)+(n+1)-(2n) = n+1+n+1-2n = 2$ . Again the $n$ terms cancel, leaving $2$ .

Step 4. Read the result. Because the cancellation happens algebraically for any $n$ , Euler's formula is guaranteed for the entire prism family and the entire pyramid family at once — not just for the specific cube, triangular prism, or square pyramid a textbook happens to list. This is also why the formula is such a reliable check: any correctly counted convex, flat-faced solid must reduce the same way.

Worked example

Problem: A hexagonal prism has a hexagonal (6-sided) base. Find its number of faces, vertices and edges, verify Euler's formula, and confirm the result matches the general prism formulas.

Step 1. Faces: a prism with an $n$ -sided base has $2$ polygon ends plus $n$ rectangular side faces, so $F=n+2$ . Here $n=6$ , giving $F=6+2=8$ .

Step 2. Vertices: each of the two hexagonal ends contributes $n=6$ corner vertices, so $V=2n=2\times6=12$ .

Step 3. Edges: the two hexagonal ends together have $2\times6=12$ edges, plus $n=6$ vertical edges connecting matching vertices of the two ends, giving $E=12+6=18$ — matching the general formula $E=3n=3\times6=18$ .

Step 4. Verify Euler's formula: $F+V-E = 8+12-18 = 2$ , exactly as guaranteed. A hexagonal prism has $8$ faces, $12$ vertices and $18$ edges, and it satisfies $F+V-E=2$ .

Competitive stretch

If instead a polyhedron is known to have $F=7$ faces and $E=15$ edges, Euler's formula gives $V=2-F+E=2-7+15=10$ . Comparing with the prism pattern $F=n+2$ gives $n=5$ , and checking $V=2n=10$ and $E=3n=15$ both agree — so this is a pentagonal prism, found without drawing a single line.

Common traps

Forgetting that each edge is shared by exactly two faces and each vertex is shared by three or more edges — sloppy sketch-counting double-counts or misses lines.
Applying Euler's formula to a solid with a curved surface, such as a cylinder or cone; the formula $F+V-E=2$ is guaranteed only for flat-faced, convex polyhedra at this level.
Mixing up the prism formulas ( $F=n+2$ , $V=2n$ , $E=3n$ ) with the pyramid formulas ( $F=n+1$ , $V=n+1$ , $E=2n$ ) when substituting a value of $n$ .
Assuming that any three numbers which happen to satisfy $F+V-E=2$ must describe a real solid — the formula is a necessary check, not proof that such a polyhedron actually exists.

Ask the tutor: If you have counted a solid's faces and edges and want the vertex count checked step by step with Euler's formula, ask the AI tutor to work through your exact numbers.

Why does a vertex need at least three edges meeting there, not just two

Two edges meeting at a point would just make one bent line, not a solid corner — you need at least three faces (and therefore three edges) coming together to close off a genuine 3-D corner. Picture the corner of a cube: three edges and three faces meet there. If only two edges met, the two faces would simply continue as one flat or folded surface with no corner point worth naming, so the definition of a vertex requires three or more.

Does Euler's formula work for solids that are not prisms or pyramids, like a solid with a hexagonal face and some triangular faces mixed together

Yes, as long as the solid is convex and every face is flat — Euler's formula $F+V-E=2$ holds for any such polyhedron, not only the prism and pyramid families. The derivation above proves it specifically for prisms and pyramids because those are the shapes this syllabus asks you to count, but the formula itself is far more general; it is only the curved-surface and non-convex cases that fall outside this syllabus's guarantee.

If I count faces and vertices correctly but my edge count does not satisfy Euler's formula, what should I do

Recount the edges first, since edges are the easiest to miss or double-count on a busy sketch, especially the hidden edges at the back of a solid. A faster route is often to trust your face and vertex counts and simply compute the edges directly from $E=F+V-2$ , then check that number against your sketch — if it disagrees, you likely missed a hidden edge rather than made an arithmetic mistake.

Vertices, edges and faces

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads