Why is the side of a regular hexagon equal to its radius?

The radii cut the hexagon into six congruent triangles whose angles at the centre fill 360°, so each is 60°. Each triangle is isosceles with two radii, so its base angles are 60° too; with all angles 60° the triangle is equilateral and its third side, the hexagon side, equals the radius.

How do you construct a regular hexagon with a compass?

Draw a circle, keep the compass opening equal to the radius, and step it round the circle from any point: the six marks are the vertices of a regular hexagon, because the side of a regular hexagon equals the radius.

What are the angles of a regular hexagon?

Each interior angle is 120° (two base angles of 60° meet at every vertex), the angles at the centre are 60°, and the exterior angles are 60°.

Regular Hexagon: Side Equals Radius

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Drag $A$ to turn and resize the hexagon, or $O$ to move it; the readout under the canvas shows the side, the radius and the angle at the centre. Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Join the centre O to the six vertices. The six triangles OAB, OBC, … have two radii and one side each, so they are congruent (SSS), and their six equal angles at O fill a full turn: each is 60°. The triangle OAB is isosceles, so its base angles are (180° − 60°)/2 = 60° as well. All three angles equal: all three sides equal, AB = OA. Drag A to turn and resize the hexagon, or O to move it.Regular hexagon

DisplayLabelsMeasures

Drag A to turn and resize the hexagon, or O to move it

AB = 2.35,  OA = 2.35,  ∠AOB = 60°

Step 1 / 6

Step explanations

1A regular hexagon in its circle

Given
ABCDEF is a regular hexagon: its six sides are equal (two ticks each) and its vertices lie on a circle with centre O. To prove: AB = OA, the side of a regular hexagon equals the radius of its circle.

Drag A: the hexagon turns and resizes. Learn more about the statement

Getting Started with the Visualizer

The Statement

Six Congruent Triangles

The Angles at the Centre

The Base Angles

An Equilateral Triangle

Why It Holds

Dragging the Figure

Consequences

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag

A

to turn and resize the hexagon, or

O

to move it, and watch the side and the radius in the readout stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

Proof?Regular hexagonDisplayLabelsMeasuresDrag A to turn and resize the hexagon, or O to move it

AB = 2.35,  OA = 2.35,  ∠AOB = 60°

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1A regular hexagon in its circle

1234567

1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — A turns and resizes the hexagon, O moves the circle; the hollow points B to F follow, 60° apart

4Readout — the line under the figure: the side AB, the radius OA and the angle at the centre, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 6 of 6

Proof?Regular hexagonDisplayLabelsMeasuresDrag A to turn and resize the hexagon, or O to move it

AB = 2.35,  OA = 2.35,  ∠AOB = 60°

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1A regular hexagon in its circle

2Six congruent triangles

SSS
Join O to the six vertices. The six segments are radii, so they are equal (one tick each). Each of the triangles OAB, OBC, OCD, ODE, OEF, OFA has two radii and one side of the hexagon, so all six have the same three side lengths, and by SSS they are congruent.

3Each angle at O is 60°

Angles around a point
The congruent triangles have equal angles at O, θ each, and the six angles θ fill the full turn round O: 6θ = 360°, so θ = 60°.

4The base angles of OAB are 60°

Isosceles base angles
In the triangle OAB the sides OA and OB are radii, so it is isosceles and its base angles at A and B are equal, φ each. The angle sum gives θ + 2φ = 180°, so φ = (180° − 60°)/2 = 60°.

5OAB is equilateral

Equal angles, equal sides
All three angles of OAB are 60°. In a triangle, equal angles face equal sides: the angle at O faces AB and the angle at A faces OB, so AB = OB; and OB = OA. So OA = OB = AB: the triangle is equilateral.

6AB = OA: side equals radius

Combining
So AB = OA, and since all six triangles are congruent, every side of the regular hexagon equals the radius. The hexagon is made of six equilateral triangles meeting at the centre. ∎

12345

1The six radii — equal, one tick each, cutting the hexagon into six triangles

2The angles at the centre — six angles θ round O, 60° each

3The triangle OAB — shaded, with base angles φ at A and B: all three angles 60°

4Justification tags — Given, SSS, Angles around a point, Isosceles base angles, Equal angles equal sides, Combining

5Replay — at the last step Play becomes Replay and Next is off

The Statement

A regular hexagon has six equal sides and six equal angles, and its vertices lie on a circle, the circumscribed circle, with centre

O

. The theorem says that its side is exactly the radius of that circle:

AB = OA

This is the fact behind the oldest ruler-and-compass construction of a regular polygon: keep the compass at the radius and step it six times round the circle, and you come back to the start. It is also the first estimate of

\pi

: the hexagon's perimeter

6r

is less than the circle's

2\pi r

, so

\pi > 3

.

The proof needs only earlier pages: SSS, the angles around a point, the isosceles base angles and the angle sum. In the player

O

and

A

are free, and

B

F

are

A

turned about

O

60°

at a time; the readout shows the side, the radius and the angle at the centre.

Opening scene, frozen

The regular hexagon ABCDEF in its circle, six equal sides.

Six Congruent Triangles

Join

O

to each of the six vertices. The six segments are radii, so they are all equal. They cut the hexagon into six triangles

OAB

OBC

OCD

ODE

OEF

OFA

, and each has two radii and one side of the hexagon. The sides are equal because the hexagon is regular, so the six triangles have the same three side lengths, and by SSS they are congruent.

Six congruent triangles, frozen

Six equal radii: six triangles congruent by SSS.

The Angles at the Centre

Congruent triangles have equal angles, so the six angles at

O

are equal,

\theta

each. Together they fill the full turn round

O

, which is

360°

by the angles around a point page:

6\theta = 360°, \qquad \theta = 60°

The angles at the centre, frozen

Six equal angles θ round O: θ = 60°.

The Base Angles

In the triangle

OAB

the sides

OA

and

OB

are radii, so the triangle is isosceles with apex

O

and its base angles at

A

and

B

are equal,

\varphi

each. The angle sum of the triangle gives

\theta + 2\varphi = 180°, \qquad \varphi = \frac{180° - 60°}{2} = 60°

The base angles, frozen

OAB shaded: base angles φ = (180° − 60°)/2 = 60°.

An Equilateral Triangle

All three angles of

OAB

are

60°

. In a triangle equal angles face equal sides: if two sides facing equal angles were unequal, the larger side would face the larger angle. The angle at

O

faces

AB

and the angle at

A

faces

OB

, so

AB = OB

; and

OB = OA

as radii. So

OA = OB = AB

and the triangle is equilateral.

An equilateral triangle, frozen

All three angles 60°: OA = OB = AB.

Why It Holds

The side

AB

equals the radius

OA

, and since all six triangles are congruent, the same holds for every side:

AB = BC = CD = DE = EF = FA = OA

A regular hexagon is six equilateral triangles meeting at its centre. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, SSS, angles around a point, isosceles base angles, equal angles facing equal sides, and combining.

The number six is what makes it work: only for six sides is the central angle

60°

, the one angle that makes the isosceles triangle on each side equilateral. For a square the central angle is

90°

and the side is

r\sqrt2

; for a regular

n

-gon it is

2r\sin(180°/n)

The conclusion, frozen

The side of the hexagon equals the radius.

Dragging the Figure

Drag

A

: the whole hexagon turns with it and grows or shrinks, since

B

F

are always

A

turned about

O

60°

steps, and in the readout the side stays equal to the radius while the angle at the centre stays at

60°

. Drag

O

to move the circle;

A

stays where it is, so the hexagon turns and resizes about the new centre.

The player keeps the radius between about one and a half and two and a half readout units, so that the six angle marks at the centre stay readable and the hexagon stays in the frame; a drag outside that range is refused.

After dragging A, frozen

A moved: the hexagon turned and grew, its side still equal to the radius.

Consequences

• The compass construction. With the compass opened to the radius, six steps round a circle mark the vertices of a regular hexagon; every second mark gives an equilateral triangle, and bisecting the arcs gives the regular 12-gon.
• A first bound for π. The hexagon's perimeter is

6r

and it lies inside the circle, so

6r < 2\pi r

and

\pi > 3

. Archimedes doubled the sides to 96 and found

3\tfrac{10}{71} < \pi < 3\tfrac17

.
• Area. Six equilateral triangles of side

r

give the area

\tfrac{3\sqrt3}{2} r^2

, by the Pythagorean theorem in half of one triangle.
• Tiling and honeycombs. The interior angle is

120°

(also from the polygon angle sum,

720°/6

), so three hexagons meet round a point: the honeycomb tiling.
• Chords of 60°. A chord that subtends

60°

at the centre equals the radius; it is the entry Ptolemy started his table of chords from, together with the equal chords page's fact that equal arcs have equal chords.