Regular Polygons

Regular polygons with equal sides and angles Polygons are two dimensional geometric objects composed of points and line segments connected together to close and form a single shape and regular polygon have all equal angles and all equal side lengths.

Polygons that are not regular are considered to be irregular polygons with unequal sides, or angles or both. A third set of polygons are known as complex polygons. Irregular polygons

Regular polygons with convex angles have particular properties associated with their angles, area, perimeter, and more that are valuable for key concepts in algebra and geometry.

Any \(n\)-sided regular polygon can be divided into \((n-2)\) triangles, as shown in the figures below.

Triangles in Polygons

Since the sum of all the interior angles of a triangle is \(180^\circ\), the sum of all the interior angles of an \(n\)-sided polygon would be equal to the sum of all the interior angles of \((n -2) \) triangles, which is \( (n-2)180^\circ.\) This leads to two important theorems.

In a regular polygon, the sum of the measures of its interior angles is \((n-2)180^{\circ}.\) It follows that the measure of one angle is

\[{\frac{(n-2)180}{n}}^\circ.\]

The sum of the measures of the exterior angles of a regular polygon is \(360^\circ\). It follows that the measure of one exterior angle is

\[{\frac{360}{n}}^\circ.\]

These theorems can be helpful for relating the number of sides of a regular polygon to information about its angles.

What is the sum of the interior angles of a dodecagon?

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A dodecagon is a polygon with 12 sides. The sum of its interior angles will be

\[180 \times (12 - 2)^\circ = 180 \times 10^\circ =1800^\circ.\ _\square\]

The sum of the interior angles of a polygon is \(1260^\circ\). How many sides does it have?

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Let the polygon have \(n\) sides. Then, \(1260^\circ = 180 \times (n-2)^\circ\), which gives us

\[ 7 = n-2 \Rightarrow n = 9. \ _\square\]

The area of a regular polygon can be found using different methods, depending on the variables that are given. Commonly, one is given the side length \(s \), the apothem \(a\) (the distance from center to side--it is also the radius of the tangential incircle, often given as \(r\)), or the radius \(R\) (the distance from center to vertex--it is also the radius of the circumcircle).

The area of a regular polygon (\(n\)-gon) is

\[ n a^2 \tan \left( \frac{180^\circ } { n } \right ) = \frac{ ns^2 } { 4} \cot \left( \frac{180^\circ } { n } \right ) = \frac{ nR^2}{2} \sin \left( \frac{360^\circ } { n } \right ) = \frac{ n a s }{ 2 }. \ _\square \]

The proof follows from using the variable to calculate the area of an isosceles triangle, and then multiplying for the \(n\) triangles. Here is the proof or derivation of the above formula of the area of a regular polygon.

The figure below shows one of the \(n\) isosceles triangles that form a regular polygon.

The side length is labeled \(s\), the radius is labeled \(R\), and half central angle is labeled \( \theta \). The apothem is the distance from the center of the regular polygon to the midpoint of the side, which meets at right angle and is labeled \(a\).

Since \(\theta\) is just half the value of the full angle which is equal to \(\frac{360^\circ}{n}\), where \(n\) is the number of sides, it follows that \( \theta=\frac{180^\circ}{n}.\) Thus, we obtain \( \frac{s}{2a} = \tan\frac{180^\circ}{n}~\text{ and }~\frac{a}{R} = \cos \frac{ 180^\circ } { n} .\) \(_\square\)

It can be useful to know the formulas for some common regular polygons, especially triangles, squares, and hexagons.

Areas of Regular Polygons

1. Equilateral triangle\[A=\frac{\sqrt{3}}{4}s^2\]
2. Square\[A=s^2\]
3. Regular pentagon\[A=\frac{1}{4}\sqrt{5\left(5+2\sqrt{5}\right)}~s^2\]
4. Regular hexagon\[A=\frac{3\sqrt{3}}{2}s^2\]
5. Regular octagon\[A=2\left(1+\sqrt{2}\right)s^2\]

The following examples are based on the application of the above formulas:

What is the area of a regular hexagon whose side length is \(5 \text{ cm}\)?

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Using the area formula given the side length with \(n=6\), we have

\[\begin{align} A_{p}&=\frac{5(6^{2})}{4}\cdot \cot\frac{180^\circ}{6}\\ &=45\cdot \cot 30^\circ\\ &\approx 77.9 \ \big(\text{cm}^{2}\big). \ _\square \end{align}\]

The apothem of a regular hexagon measures 6. Find the area of the hexagon.

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Let \(C\) be the center of the regular hexagon, and \(AB\) one of its sides. Draw \(CA,CB,\) and the apothem \(CD\) \((\)which, you need to remember, is perpendicular to \(AB\) at point \(D).\) Then, since \(CA \cong CB\), \(\triangle ABC\) is isosceles, and in particular, for a regular hexagon, \(\triangle ABC\) is equilateral.

The length of \(CD\) \((\)which, in this case, is also an altitude of equilateral \(\triangle ABC)\) is \(\frac{\sqrt{3}}{2}\) times the length of one side \((\)here \(AB).\) Thus, \[CD=\frac{\sqrt{3}}{2}{AB} \implies AB=\frac{2}{\sqrt{3}}{CD}=\frac{2\sqrt{3}}{3}(6)=4\sqrt{3}.\] Now that we have found the length of one side, we proceed with finding the area. There are (at least) 3 ways for this:

First method: Use the perimeter-apothem formula.
It follows that the perimeter of the hexagon is \(P=6s=6\big(4\sqrt{3}\big)=24\sqrt{3}\). Therefore, \[A=\frac{1}{2}aP=\frac{1}{2}CD \cdot P=\frac{1}{2}(6)\big(24\sqrt{3}\big)=72\sqrt{3}.\ _\square\]

Second method: Use the area formula for a regular hexagon.
Thus the area of the hexagon is \[A=\frac{3s^2}{2}\sqrt{3}=\frac{3\big(4\sqrt{3}\big)^2}{2}\sqrt{3}=72\sqrt{3}\] as before. \(_\square\)

Third method: Use the general area formula for regular polygons.
Here, we will only show that this is equivalent to using the area formula for regular hexagons. Taking \(n=6\), we obtain \[A=\frac{ns^2}{4}\cot\frac{180^\circ}{n}=\frac{6s^2}{4}\cot\frac{180^\circ}{6}=\frac{3s^2}{2}\cot 30^\circ=\frac{3s^2}{2}\sqrt{3}=72\sqrt{3}.\ _\square\]

The area of a regular polygon can be determined in many ways, depending on what is given. These are discussed below, but the key takeaway is to understand how these formulas are all related and how they can be derived.

Area when the apothem \(a\) is given:

The area of the triangle is half the apothem times the side length, which is \[ A_{t}=\frac{1}{2}2a\tan \frac{180^\circ}{n} \cdot a=a^{2}\tan \frac{180^\circ}{n} .\] Since an \(n\)-sided polygon is made up of \(n\) congruent isosceles triangles, the total area is \[A_{p}=n a^{2} \tan \frac{180^\circ}{n}.\]

Area when the side length \(s\) is given:

From the trigonometric formula, we get \( a = \frac{s}{2 \tan \theta} \).
Substituting this into the area, we get \[A_{p}= n \left(\frac{s}{2 \tan \theta}\right)^2 \tan \frac{180^\circ}{n} = \frac{ns^{2}}{4}\cdot \cot \frac{180^\circ}{n}.\]

Area when the radius \(r\) is given:

From the trigonometric formula, we get \( a = r \cos \frac{ 180^\circ } { n}\). Substituting this into the area, we get \[\begin{align} A_{p} & =n \left( r \cos \frac{ 180^\circ } { n} \right)^2 \tan \frac{180^\circ}{n} \\ & = n r^2 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n} \\ & = \frac{nr^2}{2} \sin\frac{360^\circ}{n}. \end{align}\]

Area when the apothem \(a\) and the side length \(s\) are given:

Using \( a \tan \frac{180^\circ}{n} = \frac{s}{2} \), we obtain \[ A_{p}=n a^{2} \tan \frac{180^\circ}{n} = \frac{ n a s }{ 2 }. \] This should be obvious, because the area of the isosceles triangle is \( \frac{1}{2} \times \text{ base } \times \text { height } = \frac{ as } { 2} \).

Finding the perimeter of a regular polygon follows directly from the definition of perimeter, given the side length and the number of sides of the polygon:

The perimeter of a regular polygon with \(n\) sides with side length \(s\) is \(P=ns.\)

However, one might be interested in determining the perimeter of a regular polygon which is inscribed in or circumscribed about a circle.

What is the perimeter of a square inscribed in a circle of radius 1?

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First, we divide the square into small triangles by drawing the radii to the vertices of the square:

Then, by right triangle trigonometry, half of the side length is \(\sin\left(45^\circ\right) = \frac{1}{\sqrt{2}}.\)

Thus, the perimeter is \(2 \cdot 4 \cdot \frac{1}{\sqrt{2}} = 4\sqrt{2}.\) \(_\square\)

Using the same method as in the example above, this result can be generalized to regular polygons with \(n\) sides.

The perimeter of a regular polygon with \(n\) sides that is inscribed in a circle of radius \(r\) is \(2nr\sin\left(\frac{\pi}{n}\right).\)

Using similar methods, one can determine the perimeter of a regular polygon circumscribed about a circle of radius 1.

What is the perimeter of a regular hexagon circumscribed about a circle of radius 1?

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First, we divide the hexagon into small triangles by drawing the radii to the midpoints of the hexagon. These will form right angles via the property that tangent segments to a circle form a right angle with the radius.

Then, by right triangle trigonometry, half of the side length is \(\tan \left(30^\circ\right) = \frac{1}{\sqrt{3}}.\)

Thus, the perimeter is \(2 \cdot 6 \cdot \frac{1}{\sqrt{3}} = 4\sqrt{3}.\) \(_\square\)

Once again, this result generalizes directly to all regular polygons.

The perimeter of a regular polygon with \(n\) sides that is circumscribed about a circle of radius \(r\) is \(2nr\tan\left(\frac{\pi}{n}\right).\)

The number of diagonals of a regular polygon is \(\binom{n}{2}-n=\frac{n(n-3)}{2}.\)

What regular polygon has the same number of diagonals as the number of its sides?

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Let \(n\) be the number of sides. The number of diagonals is given by \(\frac{n(n-3)}{2}\). But since the number of sides equals the number of diagonals, we have \[n=\frac{n(n-3)}{2}, \] which becomes \[1=\frac{n-3}{2}\] since \(n\) is nonzero. Then \(2=n-3\), and thus \(n=5\). Therefore, the polygon desired is a regular pentagon. \( _\square \)

Try the following problem:

• The exterior angle of a regular hexagon is \( \frac{360^\circ}6 = 60^\circ\).

• The interior angle of a regular hexagon is the \(180^\circ - (\text{exterior angle}) = 120^\circ\).

• It consists of 6 equilateral triangles of side length \(R\), where \(R\) is the circumradius of the regular hexagon.

• The area of the regular hexagon is the sum of areas of these 6 equilateral triangles: \[ 6\times \frac12 R^2 \cdot \sin 60^\circ = \frac{3\sqrt3}2 R^2 .\]

Here are examples and problems that relate specifically to the regular hexagon.

There are two circles: one that is inscribed inside a regular hexagon with circumradius 1, and the other that is circumscribed outside the regular hexagon. What is the ratio between the areas of the two circles (larger circle to smaller circle)?

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Let \(r\) and \(R\) denote the radii of the inscribed circle and the circumscribed circle, respectively. Let \(O\) denote the center of both these circles. Dropping the altitude from \(O\) to the side length (of 1) shows that the \(r\) satisfies the equation \(r = \cos 30^\circ \) and \(R \) is simply the circumradius of the hexagon, so \(R = 1\). Taking the ratio of their areas, we have \[ \frac{ \pi R^2}{\pi r^2} = \sec^2 30^\circ = \frac43 = 4 :3. \ _\square \]

• Regular Polygons - Decomposition into Triangles
• Constructing Regular Polygons