Platonic Solids

Meet the five Platonic solids, count their faces, edges and vertices, and find out why there can never be a sixth.

Advanced30 min18 problems
Practice Problems

Definition

A Platonic solid (also called a regular polyhedron) is a solid that meets three conditions:
  1. 1.Every face is the same regular polygon (all sides and all angles equal).
  2. 2.The same number of faces meet at every vertex.
  3. 3.It is convex: no dents, every corner points outward.
There are exactly five of them:
  • Tetrahedron: 4 equilateral triangles, 3 at each vertex. V=4V = 4, E=6E = 6, F=4F = 4
  • Cube: 6 squares, 3 at each vertex. V=8V = 8, E=12E = 12, F=6F = 6
  • Octahedron: 8 equilateral triangles, 4 at each vertex. V=6V = 6, E=12E = 12, F=8F = 8
  • Dodecahedron: 12 regular pentagons, 3 at each vertex. V=20V = 20, E=30E = 30, F=12F = 12
  • Icosahedron: 20 equilateral triangles, 5 at each vertex. V=12V = 12, E=30E = 30, F=20F = 20
All five satisfy Euler's formula V−E+F=2V - E + F = 2. The reason there are only five is the angle sum at a vertex: the face angles that meet at a corner must add up to less than 360°360°, otherwise there is no corner: at exactly 360°360° the faces lie flat, and above 360°360° they do not fit around the point.
(faces at a vertex)×(one face angle)<360°\text{(faces at a vertex)} \times \text{(one face angle)} < 360°

Try it now

Which of these solids is NOT a Platonic solid?

Worked Examples

An octahedron has 8 faces, each an equilateral triangle, and 4 faces meet at every vertex. Find the number of edges and vertices.
1

Count all triangle sides

8 faces×3 sides=248 \text{ faces} \times 3 \text{ sides} = 24 → 24 sides in total
2

Each edge is shared by 2 faces

E=24÷2E = 24 \div 2 → E=12E = 12
3

Count all triangle corners: each vertex collects 4 of them

8×3=248 \times 3 = 24 corners, V=24÷4V = 24 \div 4 → V=6V = 6
4

Check with Euler's formula

V−E+F=6−12+8V - E + F = 6 - 12 + 8 → 22, correct

Common Mistakes

Calling any solid with identical faces a Platonic solid

Why it's wrong: The triangular bipyramid (two tetrahedra glued together) has 6 identical equilateral triangles, but 3 faces meet at two of its vertices and 4 faces meet at the other three.

Correct: Check both conditions: the same regular polygon for every face AND the same number of faces at every vertex.

Counting the edges of a cube as 6×4=246 \times 4 = 24

Why it's wrong: Every edge belongs to two faces, so counting the sides of every face counts each edge twice.

Correct: Divide by 2: E=6×42=12E = \frac{6 \times 4}{2} = 12. In general E=F×n2E = \frac{F \times n}{2}, where nn is the number of sides of one face.

Thinking 6 triangles or 4 squares at a vertex make a new solid

Why it's wrong: 6×60°=360°6 \times 60° = 360° and 4×90°=360°4 \times 90° = 360°: the faces fill the full turn around the point and lie flat, like floor tiles.

Correct: The angle sum at a vertex must be strictly less than 360°360°.

Thinking every gaming die is a Platonic solid

Why it's wrong: The ten-sided die (d10) has kite-shaped faces, which are not regular polygons, and its vertices do not all look the same.

Correct: Only d4, d6, d8, d12 and d20 are Platonic solids. The d10 is a fair die anyway because all its faces are identical in shape and position.

Interactive Visual

The five Platonic solids: tap to count the faces

Count them

Tap the solid or Start counting: every face lights up one by one.

Faces

4

Edges

6

Vertices

4

Face shape

Equilateral triangle

Faces at each vertex

3

Volume

V = (√2/12)a³

7.54 units³

Surface Area

SA = √3a²

27.71 units²

Nets of the five Platonic solids

Net

8 identical faces

Faces

8

Edges

12

Vertices

6

Face shape

Equilateral triangle

Faces at each vertex

4

Volume

V = (√2/3)a³

30.17 units³

Surface Area

SA = 2√3a²

55.43 units²

3D Shape Viewer

Faces

4

Edges

6

Vertices

4

Face shape

Equilateral triangle

Faces at each vertex

3

Euler's formula: V − E + F = 4 − 6 + 4 = 2

Volume

V = (√2/12)a³

V = √2/12 · 3³ ≈ 3.18

3.18 cm³

Surface Area

SA = √3a²

SA = √3 · 3² ≈ 15.59

15.59 cm²

Interactive Sandbox

Expression Calculator

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History

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which of these solids is NOT a Platonic solid?

Why It Matters

Platonic solids have fascinated people for more than 2000 years, and they still show up all around you:
  • Games: the dice d4, d6, d8, d12 and d20 are exactly the five Platonic solids. Because every face is the same and sits the same way, each number has the same chance.
  • Biology: many viruses (such as the herpes virus and the adenovirus) build their protein shell in the shape of an icosahedron, because 20 identical triangles make a strong, closed shell from many copies of the same small piece.
  • Chemistry and crystals: a methane molecule has the shape of a tetrahedron, table salt and pyrite grow as cubes, and fluorite and diamond often grow as octahedra.
  • Architecture: geodesic domes start from an icosahedron whose faces are divided into smaller triangles.
  • History: Plato linked four solids to fire, earth, air and water and the dodecahedron to the whole universe. Euclid proved that there are only five, and Johannes Kepler (1596) tried to explain the orbits of the planets by nesting the solids inside each other.
The proof that there are only five is one of the oldest and most beautiful arguments in mathematics, and you can follow it with nothing more than angles.

Real World Applications

Fair Dice for Board Games

Role-playing games use all five Platonic solids as dice. Because every face is the same regular polygon and every vertex looks the same, no face is more likely to land on top.

Example:

On a d20 (icosahedron) each number has probability 120=0.05\frac{1}{20} = 0.05, so rolling 15 or higher (15 to 20, six numbers) has probability 620=0.3\frac{6}{20} = 0.3.

1Try It Yourself

You roll a fair 12-sided die (a dodecahedron) numbered 1 to 12.

What is the probability of rolling a number greater than 9?

Step 1: Write the mathematical expression

Favorable outcomes divided by all outcomes:

Virus Shells

Many viruses wrap their genes in a shell (capsid) shaped like an icosahedron. Nature uses it because 20 identical triangles close up into a strong shell built from many copies of the same small protein.

Example:

An icosahedral capsid has 12 vertices where 5 triangles meet. Scientists use these 12 points of five-fold symmetry to describe the structure.

2Try It Yourself

A science museum builds a giant model of a virus capsid from 20 triangular panels joined by metal struts along the edges.

How many struts (edges) does the model need?

Step 1: Write the mathematical expression

Count all panel sides, then remember each strut is shared:

Paper Lamps and Models

Designers fold lamps, gift boxes and decorations from nets of Platonic solids. The surface area tells them how much paper or material they need.

Example:

A cube lamp with 20 cm edges needs 6×202=24006 \times 20^2 = 2400 cm² of paper (plus the glue tabs).

3Try It Yourself

You fold an octahedron lamp from paper. Each edge is 10 cm long.

How much paper do the 8 faces cover? Round to one decimal place.

Step 1: Write the mathematical expression

Use SA=23 a2SA = 2\sqrt{3}\,a^2:

Key Takeaways

  • 1A Platonic solid has identical regular polygon faces, the same number of faces at every vertex, and no dents (convex)
  • 2There are exactly five: tetrahedron (4 triangles), cube (6 squares), octahedron (8 triangles), dodecahedron (12 pentagons), icosahedron (20 triangles)
  • 3Only five exist because the face angles at a vertex must add up to less than 360°360°: 3, 4 or 5 triangles, 3 squares or 3 pentagons
  • 4Edges and vertices follow from the faces: E=F×n2E = \frac{F \times n}{2} and V=F×nkV = \frac{F \times n}{k} (nn sides per face, kk faces per vertex), and V−E+F=2V - E + F = 2
  • 5Duals swap VV and FF: cube and octahedron, dodecahedron and icosahedron; the tetrahedron is its own dual
  • 6Surface area for edge aa: tetrahedron 3a2\sqrt{3}a^2, cube 6a26a^2, octahedron 23a22\sqrt{3}a^2, icosahedron 53a25\sqrt{3}a^2

Frequently Asked Questions

The Greek philosopher Plato described them around 360 BC in his dialogue Timaeus. He linked the tetrahedron to fire, the cube to earth, the octahedron to air, the icosahedron to water and the dodecahedron to the universe. They were known before him, and Euclid later proved that there are exactly five.
The Greek philosopher Plato described them around 360 BC in his dialogue Timaeus. He linked the tetrahedron to fire, the cube to earth, the octahedron to air, the icosahedron to water and the dodecahedron to the universe. They were known before him, and Euclid later proved that there are exactly five.
No. Its faces are kites, not regular polygons, and some vertices have 3 faces while the two tips have 5. It is still fair, because all its faces are identical and placed the same way.
The tetrahedron itself. It has V=4V = 4 and F=4F = 4, so swapping vertices and faces gives a tetrahedron again (turned upside down).
Yes. In four dimensions there are six regular shapes (called polytopes), and in five or more dimensions there are only three. Our 3D world has five.

Glossary

Platonic solid (regular polyhedron)
A convex polyhedron whose faces are identical regular polygons, with the same number of faces meeting at each vertex
Regular polygon
A polygon with all sides equal and all angles equal, such as an equilateral triangle, a square or a regular pentagon
Convex
Having no dents: the segment between any two points of the solid stays inside the solid
Angle sum at a vertex
The sum of the face angles that meet at one corner; for a solid it must be less than 360°360°
Dual solid
The solid you get by placing a vertex at the center of each face; vertices and faces swap
Net
A flat pattern of all the faces that folds up into the solid

Formula Card

Vertex condition

k×α<360°k \times \alpha < 360°

$k$ = faces at a vertex, $\alpha$ = one angle of a face

Edges from faces

E=F×n2E = \frac{F \times n}{2}

$n$ = sides of one face; each edge is shared by 2 faces

Vertices from faces

V=F×nkV = \frac{F \times n}{k}

$k$ = faces meeting at each vertex

Euler's formula

V−E+F=2V - E + F = 2

True for all five Platonic solids

Surface areas (edge $a$)

3a2, 6a2, 23a2, 53a2\sqrt{3}a^2,\ 6a^2,\ 2\sqrt{3}a^2,\ 5\sqrt{3}a^2

Tetrahedron, cube, octahedron, icosahedron

Today's challengeOne problem, three tries

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