Areas of Regular Polygons

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Learning goals

Distinguish a regular polygon's center, radius, apothem, and central angle
Explain why congruent center triangles fill a convex regular polygon
Calculate area or recover an apothem using perimeter and perpendicular height
Apply special right triangles to regular triangles, squares, and hexagons
Use right-triangle trigonometry to find polygon dimensions and check their compatibility

Two distances from the center

A regular polygon has congruent sides and congruent interior angles. Throughout this lesson it is simple, convex, and nondegenerate, with an integer side count $n\ge3$ and positive lengths. Its center $O$ is the common center of its rotational symmetries. Chapter 8 established that a rotation by one vertex step matches its boundary to itself.

A radius is a segment from $O$ to a vertex; its length is $r$ . An apothem is a perpendicular segment from $O$ to a side; its length is $a$ . Rotational symmetry makes all the vertex distances equal and all the side distances equal.

For a regular hexagon of side 2 units, a center triangle has two equal radii and a 2-unit base. Its perpendicular height is $\sqrt3$ units, as the special-triangle calculation below will explain. Each such triangle has area $\sqrt3$ square units; the six together have area $6\sqrt3$ square units.

The apothem meets a side at a right angle; a radius ends at a vertex.

In any center triangle, the equal radii make it isosceles. Its altitude from $O$ bisects the side and the angle at $O$ : the two right halves are congruent by hypotenuse-leg. Thus the foot $M$ is a side midpoint, and the apothem is shorter than the radius, which is a hypotenuse.

The angles and triangles fit exactly

A central angle is the angle at $O$ between radii to adjacent vertices. The eight central angles of a regular octagon are equal and fill one full turn, so each measures 45 degrees. This uses angles around a point, rather than a circle theorem.

Why the center triangles cover the polygon#

For a regular square of side 10 units, the one-vertex-step rotation carries each directed side to the next, turning it through the square’s 90-degree exterior turn. It turns each radius through the same angle. The four radius sectors fill one turn, with a triangle in each sector extending from $O$ to a square side. Neighboring triangles share a whole radius, and they surround $O$ with no gap. Their only outside edges are the square’s four sides, so $O$ is inside the square. Every ray from the center meets the boundary in one point. Depending on which side it meets, the part of that ray inside the square lies in the triangle on that side. The four triangles have disjoint interiors and cover the square. Each has base 10 and height 5, so each has area 25 square units. Their sum is 100 square units.

For any convex regular $n$ -gon, Chapter 8’s one-vertex-step rotation carries each directed side to the next. Its angle is the regular polygon’s exterior turn, $360^\circ/n$ , as proved in Polygon Angles and Tessellations. The radii turn through that same angle, so their $n$ consecutive sectors fill one turn. Since $n\ge3$ , each sector is less than 180 degrees. Its two radii and the polygon side joining their endpoints form a nondegenerate triangle contained in that sector.

Neighboring triangles share a whole radius. All radial edges pair up, leaving just the polygon sides on the outer boundary. Each side-base stays a positive distance from $O$ , so the triangles cover a small region all around $O$ within the full turn of sectors. Thus $O$ lies strictly inside the region bounded by the polygon sides.

Now convexity keeps each center-to-boundary segment inside. A ray from $O$ meets exactly one boundary side, except at a vertex shared by neighboring sides. The part of the ray in the polygon lies in the triangle on that side. This covers every point. Triangles on different sides have disjoint interiors because their directions at $O$ lie in separate consecutive angle sectors; neighboring triangles share only a radius. Each center triangle has two sides of length $r$ and one of length $s$ , so SSS makes all $n$ triangles congruent. Nothing here depended on four sides.

Their equal angles around $O$ sum to a full turn, giving

\text{central angle}=\frac{360^\circ}{n}.

The apothem bisects this angle, giving

\text{half-central angle}=\frac{180^\circ}{n}.

Four center triangles cover the square without overlapping interiors.

Worked example 1 Read a center triangle

Problem. A convex regular octagon has center $O$ , and $B,C$ are adjacent vertices. Segment $OM$ is perpendicular to $BC$ at $M$ . Find the measures of angle $BOC$ and angle $BOM$ , and identify the radius and apothem in triangle $OMB$ .

The schematic marks the perpendicular, while the text supplies regularity.

Solution. Eight congruent center triangles fill one turn:

\begin{aligned}\angle BOC&=360^\circ\div8\\\angle BOC&=45^\circ.\end{aligned}

Since $OM$ bisects the isosceles triangle’s vertex angle,

\begin{aligned}\angle BOM&=45^\circ\div2\\\angle BOM&=22.5^\circ.\end{aligned}

The radius is $OB$ , the hypotenuse of right triangle $OMB$ . The apothem is $OM$ , perpendicular to the side. The base part $MB$ is half of $BC$ .

Check your understanding

A convex regular dodecagon has 12 sides. What are its central angle and half-central angle, respectively?

A 60 degrees and 30 degrees B 30 degrees and 15 degrees C 150 degrees and 75 degrees D 15 degrees and 7.5 degrees

Add the center-triangle areas

Each triangle contributes half its side-base times the common apothem. For a regular hexagon with side 7 units and apothem $7\sqrt3/2$ units, one center triangle has area $49\sqrt3/4$ square units. Six triangles give $147\sqrt3/2$ square units. Alternatively, the perimeter is 42 units, and half of 42 times $7\sqrt3/2$ gives the same total.

The triangle area formula uses a perpendicular height. The radius is slanted relative to a side-base and cannot replace that height.

The perimeter-apothem area formula#

For a square of side 16 units and apothem 8 units, summing the four triangle areas gives

A=4\left(\frac12\cdot16\cdot8\right).

Grouping the four bases first gives

A=\frac12\cdot64\cdot8.

For a convex regular polygon of positive side length $s$ , there are $n$ center triangles, each with base $s$ and perpendicular height $a$ . By area addition,

A=n\left(\frac12 sa\right).

Its perimeter is

P=ns.

Substitution gives

\boxed{A=\frac12 Pa}.

The square’s side count did not matter: the argument uses the nonoverlapping center decomposition and equal perpendicular heights. Regularity guarantees these heights agree. For an arbitrary polygon, one center-to-side distance need not be the height of every triangle, so a perimeter and one distance alone do not justify this formula.

Worked example 2 Use the full perimeter

Problem. A convex regular hexagon has perimeter 54 cm and apothem $\tfrac{9\sqrt3}{2}$ cm. Find its exact area.

Each side is one sixth of the given perimeter.

Solution. The given perpendicular distance is the height of every center triangle, so

\begin{aligned} A&=\frac12(54)\frac{9\sqrt3}{2}\\ A&=\frac{243\sqrt3}{2}\text{ cm}^2. \end{aligned}

The perimeter already includes all six bases. Multiplying this answer by six again would count each triangle six times.

The same formula can recover an unknown apothem. Since $P$ is positive, multiply by 2 and divide by $P$ :

a=\frac{2A}{P}.

For example, a square of area 169 square cm and perimeter 52 cm has apothem 6.5 cm. This matches half its 13 cm side.

Check your understanding

A square has perimeter 36 cm and apothem 4.5 cm. Which area and justification are correct?

A 162 square cm, by using perimeter times apothem B 40.5 square cm, by adding only two center triangles C 81 square cm, by adding center triangles with disjoint interiors that fill the square D 20.25 square cm, by squaring the apothem

Familiar special cases

Special right triangles let us find exact apothems without a calculator. For example, a regular hexagon of side 6 units has central angle 60 degrees. Each center triangle is equilateral: its two base angles are also 60 degrees. Its radius is therefore 6, its half-side is 3, and its apothem is $3\sqrt3$ . The total area is $54\sqrt3$ square units.

For any positive side length $s$ , the same special right-triangle ratios give these cases:

Regular polygon	Right half of a center triangle	Apothem	Area
Equilateral triangle	Half-central angle 60 degrees	$s\sqrt3/6$	$s^2\sqrt3/4$
Square	Half-central angle 45 degrees	$s/2$	$s^2$
Regular hexagon	Half-central angle 30 degrees	$s\sqrt3/2$	$3s^2\sqrt3/2$

For the triangle row, the half-side $s/2$ is opposite 60 degrees, so the apothem is $(s/2)/\sqrt3$ . For the square row the two legs are equal. For the hexagon row, the apothem is $\sqrt3$ times the half-side. Substituting each apothem and the appropriate perimeter into the proved formula gives the areas in the last column.

Worked example 3 Locate the shorter leg

Problem. An equilateral triangle has side length 14 cm. Find its apothem and exact area.

This is half a center triangle, not half the entire equilateral triangle.

Solution. The 7 cm half-side is opposite the 60-degree angle. The apothem is the shorter leg of this 30-60-90 triangle:

\begin{aligned} a&=\frac7{\sqrt3}\\ a&=\frac{7\sqrt3}{3}\text{ cm}. \end{aligned}

The perimeter is 42 cm. Therefore

\begin{aligned} A&=\frac12(42)\frac{7\sqrt3}{3}\\ A&=49\sqrt3\text{ cm}^2. \end{aligned}

The triangle’s full altitude is $7\sqrt3$ cm. Its apothem is the distance from the center, so those two heights have different starting points.

Check your understanding

A regular hexagon has side length 8 cm. Which is its exact area?

96\sqrt3\text{ cm}^2

48\sqrt3\text{ cm}^2

192\text{ cm}^2

64\sqrt3\text{ cm}^2

Find the apothem with trigonometry

For other side counts, use the same right half-triangle with degree-mode trigonometry. In a regular pentagon of side 2 units, the half-central angle is 36 degrees. Tangent compares the half-side 1 with the adjacent apothem, giving an apothem of $1/\tan36^\circ$ units. Keeping that expression unrounded avoids carrying a rounded height into the area.

The enlarged half-triangle names the radius, apothem, and half-side separately.

Here the angle $\theta$ is

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

In right triangle $OMB$ , the opposite leg is $s/2$ , the adjacent leg is $a$ , and the hypotenuse is $r$ . From the right-triangle ratios,

\tan\theta=\frac{s/2}{a}.

Solving gives

\boxed{a=\frac{s}{2\tan\theta}}.

If the radius is known instead, cosine gives

\boxed{a=r\cos\theta}.

Sine finds the half-side:

\frac{s}{2}=r\sin\theta.

These are right-triangle ratios, not a new general-triangle area formula. Since $n\ge3$ , the half-central angle is positive and at most 60 degrees, so the denominators above are nonzero.

Worked example 4 Start from a side length

Problem. A convex regular pentagon has side length 8 cm. Find its apothem and area to the nearest tenth, using a calculator in degree mode.

The half-side is opposite the 36-degree angle.

Solution. The pentagon’s half-central angle is 36 degrees. Tangent gives

\begin{aligned} a&=\frac4{\tan36^\circ}\\ a&\approx5.5\text{ cm}. \end{aligned}

Its perimeter is 40 cm. Use the unrounded expression for the apothem:

\begin{aligned} A&=\frac12(40)\frac4{\tan36^\circ}\\ A&\approx110.1\text{ cm}^2. \end{aligned}

The displayed 5.5 cm is rounded, so substituting it as an exact height would lose precision.

Check your understanding

A convex regular heptagon has side length 6 cm. Which expression gives its apothem in centimeters? All angles use degrees.

6\tan(180^\circ/7)

3\sin(180^\circ/7)

6/\cos(360^\circ/7)

3/\tan(180^\circ/7)

Start from a radius, or check the dimensions

Given a radius, find the apothem before using it as a height. In a square of radius $4\sqrt2$ units, the half-central angle is 45 degrees, so cosine gives apothem 4 units. Sine gives half-side 4 units. The perimeter is 32 units, and the area is 64 square units.

Worked example 5 Use a radius in two right-triangle ratios

Problem. A convex regular octagon has radius 9 cm. Find its apothem, side length, and area to the nearest tenth. Check that the dimensions are compatible with the given radius. Use degree mode and round only the final results.

The given radius is the hypotenuse; it is not perpendicular to the side.

Solution. The half-central angle is 22.5 degrees. First find the two legs:

\begin{aligned} a&=9\cos22.5^\circ\\ a&\approx8.3\text{ cm}. \end{aligned}

\begin{aligned} s&=18\sin22.5^\circ\\ s&\approx6.9\text{ cm}. \end{aligned}

Use eight sides for the perimeter and retain both exact trig expressions:

\begin{aligned} P&=8(18\sin22.5^\circ)\\ A&=\frac12 P(9\cos22.5^\circ)\\ A&\approx229.1\text{ cm}^2. \end{aligned}

Using the unrounded sine and cosine expressions, the Pythagorean identity from Chapter 7 gives, in square centimeters,

a^2+(s/2)^2=81.

This matches the square of the given 9 cm radius. Testing the rounded 8.3 and 6.9 instead would produce a small rounding discrepancy.

Dimensions in a regular polygon must fit the same half-triangle. Pythagoras gives

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

Tangent also links the side and apothem to the side count. For example, side $4\sqrt3$ units and apothem 2 units give tangent $\sqrt3$ for the half-central angle. The acute angle is 60 degrees, so the side count is 3. These are compatible equilateral-triangle dimensions. For given $s$ and $a$ , find the acute angle with inverse tangent, then check that $180^\circ/\theta$ is an integer at least 3. An exact noninteger side count cannot describe a regular polygon.

Check your understanding

A convex regular nonagon has radius 10 cm. Which expression is its apothem in centimeters? Use degrees.

10\sin20^\circ

10/\cos20^\circ

10\cos20^\circ

10\tan40^\circ

Areas of Regular Polygons

Learning goals

Two distances from the center

The angles and triangles fit exactly

Add the center-triangle areas

Familiar special cases

Find the apothem with trigonometry

Start from a radius, or check the dimensions

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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