Geometry

The apothem of a regular polygon

How far is the middle of a regular polygon from its sides, and how does that distance tie together the side, the radius and the number of sides?

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The apothem of a regular polygon

Why this exists

A regular polygon has a center: one point the same distance from every corner and the same distance from every side, as regular polygons and their circles shows. The distance out to a corner is the polygon’s radius. The distance out to a side is just as useful, and it gets its own name, the apothem.

You meet it whenever a regular polygon is gripped, measured or stacked by its flat sides. A spanner for a hexagonal bolt head closes on two opposite flat sides, so the size stamped on the spanner is the distance across the flats, which is twice the apothem. A hexagonal pencil lies on a desk on one flat side, and its height off the desk is again twice the apothem. And when a regular polygon is cut into its triangles at the center, the apothem is the height of every one of those triangles, which is what makes the area of a regular polygon a short calculation.

The apothem, the radius and the side are three lengths of one right triangle. Know the number of sides and any one of them, and the other two follow. This lesson is about that triangle.

Definition

Let a regular polygon have n sides, each of length s, and center O. Its radius R is the distance from O to each vertex.

The apothem of the polygon is the segment from the center O to a side that meets the side at a right angle. Like the word radius, the word apothem names both the segment and its length, and the length is written a. A polygon with n sides has n apothems, one to each side, and they are all the same length.

In Figure 1, \overline{OM} is an apothem of the regular pentagon ABCDE: M is on \overline{AB} and \overline{OM} \perp \overline{AB}.

Figure 1The apothem OM, the radius OB and half a side make a right triangle, with the angle \theta = \dfrac{180^\circ}{n} at the center.

The apothem is the radius of the polygon’s inscribed circle, the circle with center O that touches every side. Only regular polygons are said to have an apothem; Boundaries says why.

Intuition

Stand at the center. The corners are the furthest the polygon reaches from you, and they are all at the radius R. The middles of the sides are the nearest the edge comes, and they are all at the apothem a. Everything on the edge lies between those two distances.

The triangle in Figure 1 holds all of it. The apothem OM goes straight out to the middle of side \overline{AB}, the radius OB goes out to the corner B, and the half side MB joins them. The right angle sits at M. The angle at O is half of the central angle \angle AOB, so it is half of \dfrac{360^\circ}{n}:

\theta = \frac{180^\circ}{n}

That angle depends only on n. So every regular polygon with n sides, of whatever size, has the same shape of right triangle at its center, and only its size changes. With more sides, \theta is smaller, the triangle is thinner, and the apothem comes closer and closer to the radius: the polygon is getting rounder.

Results

Where the apothem lands. The apothem to a side meets it at its midpoint, cuts the central angle over that side into two halves of \theta = \dfrac{180^\circ}{n}, and lies along a line of symmetry of the polygon.

The apothem triangle. The apothem a, half a side \tfrac{s}{2} and the radius R form a right triangle with hypotenuse R and the angle \theta = \dfrac{180^\circ}{n} at the center. So

a^2 + \left(\frac{s}{2}\right)^2 = R^2

and, with the trigonometric ratios of the angle \theta,

\begin{aligned} a &= R\cos\theta \\ \frac{s}{2} &= R\sin\theta \\ \frac{s}{2} &= a\tan\theta \end{aligned}

Three polygons with exact values. For n = 3, 4 and 6 the angle \theta is 60^\circ, 45^\circ and 30^\circ, and the apothem triangle is one of the special right triangles:

Polygon	From the side	From the radius
triangle	\displaystyle a = \tfrac{\sqrt{3}}{6}s	a = \tfrac{1}{2}R
square	a = \tfrac{1}{2}s	\displaystyle a = \tfrac{\sqrt{2}}{2}R
hexagon	\displaystyle a = \tfrac{\sqrt{3}}{2}s	\displaystyle a = \tfrac{\sqrt{3}}{2}R

The triangle’s value \displaystyle \tfrac{\sqrt{3}}{6}s is \displaystyle \dfrac{s}{2\sqrt{3}} with the root moved to the top.

Nearest and furthest. Every point on the edge of a regular polygon is at least a and at most R from the center. The distance is exactly a only at the midpoints of the sides and exactly R only at the vertices. In particular a < R.

Across the polygon. Stand the polygon on one side.

If n is even, the side opposite is parallel to it, and the distance between them, the width across the flats, is 2a. The distance between opposite vertices, the width across the corners, is 2R.
If n is odd, there is a vertex opposite the side instead, straight above the midpoint of the side, and the height of the polygon is a + R.

Why it is true

Where the apothem lands. Let \overline{AB} be a side and M the foot of the perpendicular from O. OA = OB, since both are radii, and OM is shared, so the right triangles OMA and OMB are congruent by HL. Hence AM = MB, so M is the midpoint, and \angle AOM = \angle MOB, so each is half the central angle, \dfrac{180^\circ}{n}. The line OM is then the perpendicular bisector of \overline{AB}. The same argument is in the proof of the inscribed circle in regular polygons and their circles.

All apothems are equal. The n triangles at the center are congruent, so their heights from O are equal.

The apothem triangle. Triangle OMB has its right angle at M, legs OM = a and MB = \tfrac{s}{2}, and hypotenuse OB = R. The first formula is the Pythagorean theorem. For the angle \theta at O, the side next to it is a and the side opposite it is \tfrac{s}{2}, so \cos\theta = \dfrac{a}{R}, \sin\theta = \dfrac{s/2}{R} and \tan\theta = \dfrac{s/2}{a}, which rearrange to the three formulas.

Nearest and furthest. Take a point P on side \overline{AB}, which lies between M and one of A or B, say B. Triangle OMP has a right angle at M (or P is M), so

OP^2 = a^2 + MP^2

with 0 \le MP \le MB. So OP^2 is at least a^2, with equality only when P = M, and at most a^2 + MB^2 = R^2, with equality only when P = B. That is the shortest distance from a point to a line at work: the perpendicular is the shortest way from O to the side.

Across the polygon. Measure directions at O from the apothem \overrightarrow{OM}. Going round, the vertices sit at \theta, 3\theta, 5\theta, and so on, the odd multiples of \theta, because each vertex is one central angle 2\theta round from the last and the first, B, is \theta from OM. The other apothems sit halfway between neighboring vertices: at 2\theta, 4\theta, and so on, the even multiples. The direction straight opposite \overrightarrow{OM} is 180^\circ, and

180^\circ = n \cdot \frac{180^\circ}{n} = n\theta

If n is even, n\theta is an even multiple of \theta, so an apothem \overline{ON} points straight away from \overline{OM}, and M, O, N lie on one line. Both sides are perpendicular to that line, so they are parallel, and the distance between them is MN = a + a = 2a. Opposite vertices are joined through O the same way, so the width across the corners is R + R = 2R.

If n is odd, n\theta is an odd multiple of \theta, so a vertex V sits straight opposite. The line MOV is perpendicular to the side, so V is MV = a + R above it. No point of the polygon is higher, because every point of the polygon is within R of O.

Worked examples

Example 1

A regular hexagon from its side

A regular hexagon has side 10, as in Figure 2. Find its apothem, its radius, and its widths across the flats and across the corners.

Figure 2A regular hexagon with side 10. The two apothems OM and ON lie on one line.

Solution

The central angle is \dfrac{360^\circ}{6} = 60^\circ, so triangle OAB has OA = OB and 60^\circ between them. Its base angles are equal and share 120^\circ, so all three angles are 60^\circ and the triangle is equilateral. So the radius is R = OB = 10.

The apothem OM lands on the midpoint of \overline{AB}, so MB = 5. In the right triangle OMB, by the Pythagorean theorem,

\begin{aligned} a^2 &= 10^2 - 5^2 = 75 \\ a &= \sqrt{75} = 5\sqrt{3} \approx 8.66 \end{aligned}

A hexagon has an even number of sides, so the top side is parallel to the bottom one and the two apothems OM and ON lie on one line. The width across the flats is \displaystyle 2a = 10\sqrt{3} \approx 17.32, and the width across the corners is 2R = 20.

Check. The table gives \displaystyle a = \dfrac{\sqrt{3}}{2}s = \dfrac{\sqrt{3}}{2} \cdot 10 = 5\sqrt{3}.

Example 2

A regular octagon, two ways

A regular octagon has side 6. Find its apothem exactly, and its radius to two decimal places.

Solution

With trigonometry. Here \theta = \dfrac{180^\circ}{8} = 22.5^\circ, and the apothem triangle has the half side 3 opposite \theta and the apothem a next to it. So

a = \frac{3}{\tan 22.5^\circ} \approx \frac{3}{0.41421} \approx 7.243

That is a decimal. An exact value comes from a picture.

With a square. Each interior angle of a regular octagon is \dfrac{6 \cdot 180^\circ}{8} = 135^\circ. Extend the top, bottom, left and right sides, as in Figure 3. Going round the octagon the direction of the sides turns 45^\circ at each vertex, so every second side has turned 90^\circ: the four extended sides meet at right angles and make a square. Each corner of the square cuts off a triangle whose slanted side is a side of the octagon, and whose angles beside it are 180^\circ - 135^\circ = 45^\circ, each forming a linear pair with an interior angle. So each corner triangle is a right isosceles triangle with hypotenuse 6, and its legs are \displaystyle \dfrac{6}{\sqrt{2}} = 3\sqrt{2}.

Figure 3A regular octagon is a square with its corners cut off.

The side of the square is then a leg, a side of the octagon and another leg:

3\sqrt{2} + 6 + 3\sqrt{2} = 6 + 6\sqrt{2}

The square’s top and bottom are two parallel sides of the octagon, and the distance between them is 2a. So

a = \frac{6 + 6\sqrt{2}}{2} = 3 + 3\sqrt{2} \approx 7.243

the same as before. For the radius, use the apothem triangle:

\begin{aligned} R^2 &= (3 + 3\sqrt{2})^2 + 3^2 \\ &= 36 + 18\sqrt{2} \end{aligned}

so \displaystyle R = \sqrt{36 + 18\sqrt{2}} \approx 7.84.

Check. R = \dfrac{3}{\sin 22.5^\circ} \approx \dfrac{3}{0.38268} \approx 7.84.

Example 3

A pentagon standing on a side

A regular pentagon has radius 10 and stands on side \overline{AB}, as in Figure 4. Find its apothem, its side and its height, each to two decimal places.

Figure 4A regular pentagon of radius 10 standing on a side. The line from M through O reaches the top vertex D.

Solution

The angle at the center of the apothem triangle is \theta = \dfrac{180^\circ}{5} = 36^\circ, and the hypotenuse is OB = 10. The apothem is next to \theta and the half side is opposite it:

\begin{aligned} a &= 10\cos 36^\circ \approx 8.09 \\ \frac{s}{2} &= 10\sin 36^\circ \approx 5.878 \end{aligned}

so the side is s \approx 11.76. (Doubling the unrounded half side avoids a rounding slip.)

5 is odd, so the line from M through O meets the vertex D straight above, and OD = 10 is a radius. The height of the pentagon is

MD = a + R \approx 8.09 + 10 = 18.09

It is not 2a: a pentagon has no side parallel to \overline{AB}.

Example 4

The spanner for a hexagonal bolt

A spanner that fits a hexagonal bolt head grips it across the flats, and the spanner is marked 24 mm. Find the side of the bolt head and its width across the corners.

Solution

The width across the flats is 2a = 24, so a = 12. For a hexagon \theta = 30^\circ, so the apothem triangle is a 30^\circ, 60^\circ, 90^\circ triangle with the half side opposite 30^\circ:

\frac{s}{2} = a \tan 30^\circ = \frac{12}{\sqrt{3}} = 4\sqrt{3}

So the side is \displaystyle s = 8\sqrt{3} \approx 13.86 mm. A regular hexagon’s radius equals its side, so \displaystyle R = 8\sqrt{3}, and the width across the corners is \displaystyle 2R = 16\sqrt{3} \approx 27.71 mm.

Check. a^2 + \left(\tfrac{s}{2}\right)^2 = 144 + 48 = 192 and R^2 = 64 \cdot 3 = 192.

A bolt needs more room to turn than its spanner size suggests: the corners sweep a circle about 3.7 mm wider than the 24 mm across the flats.

Example 5

Working back to the number of sides

A regular polygon has radius 20 cm, and its apothem measures 19.32 cm to the nearest hundredth. How many sides does it have, and how long is each side, to two decimal places?

Solution

In the apothem triangle \cos\theta = \dfrac{a}{R}, so

\cos\theta \approx \frac{19.32}{20} = 0.966

and, with the inverse cosine, \theta \approx 14.98^\circ. Since \theta = \dfrac{180^\circ}{n},

n = \frac{180^\circ}{\theta} \approx \frac{180}{14.98} \approx 12.01

A polygon has a whole number of sides, and the 0.01 comes from the apothem being rounded, so n = 12. As a test, n = 12 gives 20\cos 15^\circ \approx 19.3185, which rounds to 19.32, while 11 and 13 sides would give about 19.19 and 19.42.

With \theta = 15^\circ exactly, \dfrac{s}{2} = 20\sin 15^\circ, so the side is s = 40\sin 15^\circ \approx 10.35 cm.

Example 6

Trapping the circumference between two hexagons

A circle has radius r. One regular hexagon is drawn inside it with its vertices on the circle, and another around it with its sides touching the circle, as in Figure 5. Find the perimeters of both hexagons, and use them to show that \displaystyle 3 < \pi < 2\sqrt{3}.

Figure 5One hexagon inside the circle and one around it. The radius r is the radius of the inner one and the apothem of the outer one.

Solution

The inner hexagon. Its radius is r, and a regular hexagon’s side equals its radius, so its perimeter is 6r.

The outer hexagon. Each side touches the circle, so it is tangent to it and perpendicular to the radius at the touching point. That radius is the outer hexagon’s apothem: a = r. With \theta = 30^\circ,

\begin{aligned} \frac{s}{2} &= r\tan 30^\circ = \frac{r}{\sqrt{3}} \\ s &= \frac{2r}{\sqrt{3}} \end{aligned}

and its perimeter is \displaystyle 6 \cdot \dfrac{2r}{\sqrt{3}} = \dfrac{12r}{\sqrt{3}} = 4\sqrt{3}\,r \approx 6.93r.

The squeeze. The Circumference of the circle is 2\pi r. Each side of the inner hexagon is a straight chord and is shorter than the arc it cuts off, so 6r < 2\pi r. The circle sits inside the outer hexagon, and a closed curve that bends only one way and lies inside another such curve is the shorter of the two; taking that fact about lengths as given, \displaystyle 2\pi r < 4\sqrt{3}\,r. Dividing through by 2r,

3 < \pi < 2\sqrt{3} \approx 3.46

More sides squeeze harder, because the apothem of the outer polygon and the radius of the inner one are both r, and the gap between apothem and radius shrinks as n grows. Archimedes ran this same squeeze with polygons of 96 sides.

Boundaries

Regular polygons only. The apothem needs one point that is the same distance from every side. A rectangle measuring 6 by 8 has a center, but the distances from it to the sides are 3 and 4, so there is no single apothem. Every triangle does have a point the same distance from all three sides, its Incenter, and that distance is called the inradius; the word apothem is kept for regular polygons, where the same center also serves the vertices.

The three lengths are not free. For a given n, one of a, s and R fixes the other two. And two lengths chosen at random usually belong to no regular polygon at all: radius 13 and side 10 give \sin\theta = \dfrac{5}{13}, so \theta \approx 22.6^\circ and n \approx 7.96, which is not a whole number.

Across the flats is 2a only for an even number of sides. With an odd number there are no parallel sides, and the distance from a side to the far side of the polygon is a + R.

The apothem goes to a side, not to a corner. It is the shortest distance from the center to the edge; the radius is the longest. The two are equal for no polygon, though they come closer as n grows.

Where this goes next

Each triangle at the center has base s and height a, so its area is \tfrac{1}{2}as, and the n of them together give the area of a regular polygon as half the apothem times the Perimeter. Every number in that formula comes out of the apothem triangle.

Letting n grow, the apothem closes in on the radius and the polygon on the circle, which is how the Circumference and the area of a circle are first measured, as the last example began to do. For triangles that are not equilateral the single center splits in two: the Incenter, the same distance from every side, plays the apothem’s part.

Exercises

Without trigonometry

A regular hexagon has side 8. Find its apothem and its radius.
A square has side 14. Find its apothem and its radius.
An equilateral triangle has side 18. Find its apothem and its radius.
A regular hexagon has apothem 9. Find its side and its radius.
An equilateral triangle has apothem 5. Find its radius and its side.

With trigonometry

Give lengths to two decimal places.

A regular pentagon has side 8. Find its apothem and its radius.
A regular octagon has radius 10. Find its apothem and its side.
A regular decagon has apothem 15. Find its side and its radius.
A regular 12-gon has side 4. Using \displaystyle \tan 15^\circ = 2 - \sqrt{3}, find its apothem exactly.
A regular nonagon has radius 6. Find its apothem and its side.

Widths and heights

A stop sign is a regular octagon measuring 75 cm across the flats. Find the length of each side.
A hexagonal pencil measures 7 mm across the flats. Find its width across the corners.
A regular pentagon with side 10 stands on one side. Find its height.
An equilateral triangle has side 20. Find its apothem and its radius, and check that their sum is the height of the triangle.
A regular hexagonal floor tile measures 30 cm across the corners. Find its width across the flats.

Working back to the polygon

A regular polygon has radius 10 and apothem 9.51, to the nearest hundredth. How many sides does it have?
A regular polygon has side 2 and apothem 2.41, to the nearest hundredth. How many sides does it have?
The apothem of a regular polygon is half its side. Which polygon is it?
The apothem of a regular polygon is half its radius. Which polygon is it?
For which regular polygons is the apothem longer than the side?

Reasoning

Explain why the apothem of a regular polygon is always shorter than its radius.
One regular hexagon has twice the side of another. Explain why its apothem is also twice as long.
Explain why a rectangle measuring 6 by 8 has no apothem.
Explain why no two sides of a regular heptagon are parallel.
Regular polygons with more and more sides are all drawn with radius 1. What happens to the apothem, and why can it never reach 1?

Answers

\displaystyle a = 4\sqrt{3} \approx 6.93, R = 8. The triangles at the center are equilateral, so R = s = 8; then a^2 = 8^2 - 4^2 = 48.
a = 7, \displaystyle R = 7\sqrt{2} \approx 9.90. The apothem of a square is half a side, and R^2 = 7^2 + 7^2 = 98.
\displaystyle a = 3\sqrt{3} \approx 5.20, \displaystyle R = 6\sqrt{3} \approx 10.39. From the table, \displaystyle a = \tfrac{18}{2\sqrt{3}} = \tfrac{9}{\sqrt{3}} = 3\sqrt{3}, and R = 2a.
\displaystyle s = R = 6\sqrt{3} \approx 10.39. With \theta = 30^\circ, \displaystyle \tfrac{s}{2} = 9\tan 30^\circ = 3\sqrt{3}; a hexagon’s radius equals its side.
R = 10, \displaystyle s = 10\sqrt{3} \approx 17.32. For a triangle R = 2a; then \left(\tfrac{s}{2}\right)^2 = 10^2 - 5^2 = 75, so \displaystyle \tfrac{s}{2} = 5\sqrt{3}.
a \approx 5.51, R \approx 6.81. \theta = 36^\circ and \tfrac{s}{2} = 4: a = \tfrac{4}{\tan 36^\circ} and R = \tfrac{4}{\sin 36^\circ}.
a \approx 9.24, s \approx 7.65. \theta = 22.5^\circ: a = 10\cos 22.5^\circ and s = 20\sin 22.5^\circ.
s \approx 9.75, R \approx 15.77. \theta = 18^\circ: s = 2 \cdot 15\tan 18^\circ and R = \tfrac{15}{\cos 18^\circ}.
\displaystyle a = 4 + 2\sqrt{3} \approx 7.46. \theta = 15^\circ and \tfrac{s}{2} = 2, so \displaystyle a = \tfrac{2}{2 - \sqrt{3}}. Multiplying top and bottom by \displaystyle 2 + \sqrt{3} gives \displaystyle a = 2(2 + \sqrt{3}), since the denominator is 4 - 3 = 1.
a \approx 5.64, s \approx 4.10. \theta = 20^\circ: a = 6\cos 20^\circ and s = 12\sin 20^\circ.
\displaystyle 75(\sqrt{2} - 1) \approx 31.07 cm. Here a = 37.5 and \theta = 22.5^\circ, so s = 2 \cdot 37.5\tan 22.5^\circ = 75\tan 22.5^\circ. Or, from Example 2, the width across the flats is \displaystyle s(1 + \sqrt{2}), so \displaystyle s = \tfrac{75}{1 + \sqrt{2}}.
\displaystyle \tfrac{14\sqrt{3}}{3} \approx 8.08 mm. a = 3.5 and \displaystyle R = \tfrac{a}{\cos 30^\circ} = \tfrac{7}{\sqrt{3}}; across the corners is 2R.
About 15.39. The height is a + R, with a = \tfrac{5}{\tan 36^\circ} \approx 6.88 and R = \tfrac{5}{\sin 36^\circ} \approx 8.51.
\displaystyle a = \tfrac{10\sqrt{3}}{3} \approx 5.77, \displaystyle R = \tfrac{20\sqrt{3}}{3} \approx 11.55, and \displaystyle a + R = 10\sqrt{3} \approx 17.32. The height splits the triangle into two right triangles with hypotenuse 20 and leg 10, so it is \displaystyle \sqrt{400 - 100} = 10\sqrt{3}, the same.
\displaystyle 15\sqrt{3} \approx 25.98 cm. R = 15, and for a hexagon \displaystyle a = \tfrac{\sqrt{3}}{2}R, so \displaystyle 2a = 15\sqrt{3}.
10 sides. \cos\theta \approx 0.951 gives \theta \approx 18.01^\circ and n \approx \tfrac{180}{18.01} \approx 9.99, so n = 10.
8 sides. \tan\theta = \tfrac{1}{2.41} gives \theta \approx 22.54^\circ and n \approx 7.99, so n = 8; indeed the octagon with side 2 has \displaystyle a = 1 + \sqrt{2} \approx 2.41.
A square. \tan\theta = \tfrac{s/2}{a} = 1, so \theta = 45^\circ and n = \tfrac{180}{45} = 4.
An equilateral triangle. \cos\theta = \tfrac{a}{R} = \tfrac{1}{2}, so \theta = 60^\circ and n = 3.
Those with 7 or more sides. a > s means \tfrac{s}{2\tan\theta} > s, that is \tan\theta < \tfrac{1}{2}, so \theta < 26.57^\circ and n > \tfrac{180}{26.57} \approx 6.78. A hexagon has a \approx 0.87s; a heptagon has a \approx 1.04s.
The apothem and the radius to an end of the same side are a leg and the hypotenuse of one right triangle, the apothem triangle. The hypotenuse is opposite the right angle, the largest angle, so it is the longest side: a < R.
Both hexagons have apothem triangles with angles 30^\circ, 60^\circ and 90^\circ, so the triangles are similar. The half sides are in the ratio 2 to 1, so every pair of matching sides is, and the apothems too. The same holds for any n: a = \tfrac{s}{2\tan\theta} is a fixed multiple of s.
An apothem needs a point the same distance from all four sides. Points the same distance from the two long sides lie on the line midway between them, and points the same distance from the two short sides on the line midway between those. The only point on both is the center, which is 4 from the long sides and 3 from the short ones. So no point works. The apothem belongs to regular polygons, and a rectangle that is not a square is not regular.
Two parallel sides are perpendicular to one line through O, so their apothems point in opposite directions along it. But for n = 7 the direction opposite an apothem is 7\theta, an odd multiple of \theta, where a vertex sits, not an apothem. So no side has a parallel partner.
The apothem is a = \cos\dfrac{180^\circ}{n}. As n grows the angle shrinks toward 0^\circ, so a grows and comes as close to 1 as you like. It never reaches 1, because a is a leg and R = 1 the hypotenuse of a right triangle, and a leg is always shorter.