Euler’s Formula for Polyhedra - GeeksforGeeks (2024)

Euler’s formula for Polyhedra is a fundamental theorem in the field of geometry. It establishes a relationship between the number of vertices (V), edges (E), and faces (F) of a convex polyhedron. The formula is expressed as V – E + F = 2.

In this article, we have covered Euler’s Formula for Polyhedrons, its proof, related examples and others in detail.

What is Euler’s Formula for Polyhedrons?

Euler’s Formula is a basic relationship in geometry and topology, Euler’s Formula links the number of vertices (V), edges (E), and faces (F) of any polyhedron. A polyhedron is a three-dimensional solid object with flat sides and straight edges—that is, a cube, pyramid, or dodecahedron.

Named after Swiss mathematician Leonhard Euler, who originally found and verified this link in 1752, the formula formula calls for:

V – E + F = 2

where:

• V is Number of Vertices (Corner Points)
• E is Number of Edges (line segments joining the vertices)
• F is Number of Faces (flat polygonal surfaces that bound the polyhedron)

Explanation of Terms

• Vertices (V): A vertex is a point where three or more edges meet. For example, a cube has 8 vertices, while a tetrahedron (pyramid with a triangular base) has 4 vertices.

• Edges (E): An edge is a line segment that connects two vertices. In a polyhedron, edges are the “sides” of the faces. For example, a cube has 12 edges, and a tetrahedron has 6 edges.

• Faces (F): A face is a flat, polygonal surface that bounds the polyhedron. Faces are typically triangles, quadrilaterals (like squares or rectangles), or other polygons. For example, a cube has 6 square faces, and a tetrahedron has 4 triangular faces.

Euler’s Formula states that if you take the number of vertices (V), subtract the number of edges (E), and add the number of faces (F), the result will always be 2 for any polyhedron. This formula holds for any polyhedron, regardless of its shape or complexity, as long as it is a convex, closed, and finite polyhedron with no holes or intersecting faces.

The proof of Euler’s Formula is based on the idea of traversing the polyhedron and keeping track of the changes in the number of vertices, edges, and faces encountered. It involves concepts from topology and graph theory, but the formula itself is remarkably simple and elegant.

Euler’s Formula has many applications in various fields, including computer graphics, molecular modeling, and topology. It also serves as a fundamental starting point for more advanced concepts in geometry and topolog

Proof of Euler’s Formula

The proof for Euler’s formula is added below:

Theorem:

V – E + F = 2 for any convex polyhedron.

where,

• V is the verticle count
• E is the edge count
• F is the face count

Proof:

Based on the polyhedron’s face count (F), we shall prove this formula by an inductive argument.

Fundamentally, Think of a polyhedron having F = 1 face. Not a closed solid, this polyhedron has to be a single polygon. We may thus picture building a base to this polygon to produce a pyramidal polyhedron with one base face and several triangle faces intersecting at one vertex.

Here V = 1 (single vertex), E = n (where n is the base polygon’s edge count), and F = n + 1 (one base face and n triangular faces).

Using these values in Euler’s Formula yields V – E + F = 1 – n + (n + 1) = 2.

For the base case of F = 1, the formula so holds.

Step of Inductive Attraction Assume, for all polyhedra with F = k facets, where k is some positive integer Euler’s Formula holds. We wish to establish that the formula also applies for polyhedra with F = k + 1 faces.

Examine a polyhedron P with F = k + 1 faces. Beginning with a polyhedron Q with F = k faces, we can build P by cutting off one of the existing faces of Q along a closed polygonal path, therefore adding a new face F’. One calls this method “truncating” a face.

Let V_Q stand for the number of vertices, edges, and faces of Q; let E_Q stand for k, then. Likewise let V_P, E_P, and k + 1 represent the appropriate values for the polyhedron P following truncation.

We note the following changes:

• New face F’-adds one (from k to k + 1) to the count of faces.
• Number of edges rises according to the polygonal path’s count used to cut off the face. Let’s designate this rise n.
• n, the number of fresh vertices added along the polygonal path, increases the number of vertices overall.

We know from the inductive hypothesis that Euler’s Formula applies for Q: V_Q – E_Q + k = 2.

Replacing the modifications noted for P, we obtain: (V_Q + n) – (E_Q + n) + (k + 1) = V_Q – E_Q + k + 1 – n + n = 2 + 1 = 2.

Euler’s Formula likewise holds for the polyhedron P with F = k + 1 faces.

Euler’s Formula is valid for all convex polyhedra with any number of faces if the formula holds for the base case (F = 1) and the inductive step (if it holds for F = k, it also holds for F = k + 1).

By the concept of mathematical induction, therefore. This evidence shows the amazing link among the number of vertices, edges, and faces of a polyhedron that Euler’s Formula effortlessly catches.

Euler’s Formula Table

The table for Euler’s Formula is added below:

Read More,

• Arithmetic Progression
• Geometric Progression

Examples on Euler’s Formula for Polyhedra

Example 1: A cube has 8 vertices, 12 edges, and 6 faces. Verify Euler’s Formula for this polyhedron.

Solution:

Given:

• Number of vertices (V) = 8
• Number of edges (E) = 12
• Number of faces (F) = 6

To verify Euler’s Formula, we need to substitute the given values into the formula and check if the equality holds.

Euler’s Formula: V – E + F = 2

Substituting the values:

8 – 12 + 6 = 2

2 = 2 (This equality holds true)

Therefore, Euler’s Formula is satisfied for the cube.

Example 2: Find the number of faces of a polyhedron with 20 vertices and 30 edges.

Solution:

Given:

• Number of vertices (V) = 20
• Number of edges (E) = 30

To find the number of faces (F), we can rearrange Euler’s Formula as follows:

V – E + F = 2

F = 2 + E – V

Substituting the given values:

F = 2 + 30 – 20

F = 12

Therefore, the polyhedron with 20 vertices and 30 edges has 12 faces.

Example 3: If a polyhedron has 10 faces and 16 vertices, how many edges does it have?

Solution:

Given:

• Number of faces (F) = 10
• Number of vertices (V) = 16

To find the number of edges (E), we can rearrange Euler’s Formula as follows:

V – E + F = 2

E = V + F – 2

Substituting the given values:

E = 16 + 10 – 2

E = 24

Therefore, the polyhedron with 10 faces and 16 vertices has 24 edges.

Example 4: A triangular pyramid has 4 vertices, 6 edges, and 4 faces. Verify Euler’s Formula.

Solution:

Given:

• Number of vertices (V) = 4
• Number of edges (E) = 6
• Number of faces (F) = 4

To verify Euler’s Formula, we substitute the given values into the formula and check if the equality holds.

Euler’s Formula: V – E + F = 2

Substituting the values:

4 – 6 + 4 = 2

2 = 2 (This equality holds true)

Therefore, Euler’s Formula is satisfied for the triangular pyramid.

Example 4: Find the number of vertices in a polyhedron with 12 faces and 30 edges.

Solution:

Given:

• Number of faces (F) = 12
• Number of edges (E) = 30

To find the number of vertices (V), we can rearrange Euler’s Formula as follows:

V – E + F = 2

V = 2 + E – F

Substituting the given values:

V = 2 + 30 – 12

V = 20

Therefore, the polyhedron with 12 faces and 30 edges has 20 vertices.

Example 5: If a polyhedron has 8 faces and 12 vertices, how many edges does it have?

Solution:

Given:

• Number of faces (F) = 8
• Number of vertices (V) = 12

To find the number of edges (E), we can rearrange Euler’s Formula as follows:

V – E + F = 2

E = V + F – 2

Substituting the given values:

E = 12 + 8 – 2

E = 18

Therefore, the polyhedron with 8 faces and 12 vertices has 18 edges.

Example 6: A dodecahedron has 20 vertices, 30 edges, and 12 faces. Verify Euler’s Formula.

Solution:

Given:

• Number of vertices (V) = 20
• Number of edges (E) = 30
• Number of faces (F) = 12

To verify Euler’s Formula, we substitute the given values into the formula and check if the equality holds.

Euler’s Formula: V – E + F = 2

Substituting the values:

20 – 30 + 12 = 2

2 = 2 (This equality holds true)

Therefore, Euler’s Formula is satisfied for the dodecahedron.

Practice Questions on Euler’s Formula for Polyhedra

Q1. Find the number of faces of a polyhedron with 16 vertices and 24 edges.

Q2. If a polyhedron has 14 faces and 24 vertices, how many edges does it have?

Q3. A square pyramid has 5 vertices, 8 edges, and 5 faces. Verify Euler’s Formula.

Q4. A cube has 8 vertices and 6 faces. How many edges does it have?

Q5. An octahedron has 12 edges and 8 faces. How many vertices does it have?

Q6. An icosahedron has 20 faces and 30 edges. How many vertices does it have?

FAQs on Euler’s Formula for Polyhedra

How do you prove Euler’s Formula for Polyhedra?

Euler’s Formula for Polyhedra is proved using the concept of mathematical induction.

What is the Euler Characteristic of a Polyhedra?

Euler characteristic of a polyhedra is 2 for a regular polyhedron but 0 for a torus-like polyhedron.

How do you Calculate Euler’s Formula?

Euler’s Formula of a Polyhedra is, Faces + Vertices – Edges = 2.

What is the Generalised Euler’s Formula?

Generalised Euler’s Formula is: e^iφ = cos φ + i.sin φ