Reference: Angle Measures and Areas of Regular Polygons

Angle Measures

A reference of exact and approximate values of sine, cosine, and tangent of angle measures.

x°	x rad	sin(x)	cos(x)	tan(x)
0	0	0	1	0
15	\( \frac{\pi}{12} \)	\( \frac{\sqrt{2-\sqrt{3}}}{2} \) ≈ 0.25882	\( \frac{\sqrt{2+\sqrt{3}}}{2} \) ≈ 0.96592	\( 2-\sqrt{\text{3}} \) ≈ 0.26795
18	\( \frac{\pi}{10} \)	\( \frac{\sqrt{5}-1}{4} \) ≈ 0.309017	\( \frac{\sqrt{10+2\sqrt{5}}}{4} \) ≈ 0.95106	\( \frac{\sqrt{25-10\sqrt{5}}}{5} \) ≈ 0.32492
30	\( \frac{\pi}{6} \)	\( \frac{1}{2} \)	\( \frac{\sqrt{3}}{2} \) ≈ 0.86603	\( \frac{1}{\sqrt{3}} \) = \( \frac{\sqrt{3}}{3} \) ≈ 0.57735
36	\( \frac{\pi}{5} \)	\( \frac{\sqrt{10-2\sqrt{5}}}{4} \) ≈ 0.58779	\( \frac{\sqrt{5}+1}{4} \) ≈ 0.809017	\( \sqrt{5-2\sqrt{5}} \) ≈ 0.72654
45	\( \frac{\pi}{4} \)	\( \frac{\sqrt{2}}{2} \) ≈ 0.707107	\( \frac{\sqrt{2}}{2} \) ≈ 0.707107	1
54	\( \frac{3\pi}{10} \)	\( \frac{\sqrt{5}+1}{4} \) ≈ 0.809017	\( \frac{\sqrt{10-2\sqrt{5}}}{4} \) ≈ 0.58779	\( \frac{\sqrt{25+10\sqrt{5}}}{5} \) ≈ 1.37638
60	\( \frac{\pi}{3} \)	\( \frac{\sqrt{3}}{2} \) ≈ 0.86603	\( \frac{1}{2} \)	\( \sqrt{3} \) ≈ 1.73205
72	\( \frac{2\pi}{5} \)	\( \frac{\sqrt{10+2\sqrt{5}}}{4} \) ≈ 0.95106	\( \frac{\sqrt{5}-1}{4} \) ≈ 0.309017	\( \sqrt{5+2\sqrt{5}} \) ≈ 3.07768
75	\( \frac{5\pi}{12} \)	\( \frac{\sqrt{2+\sqrt{3}}}{2} \) ≈ 0.96592	\( \frac{\sqrt{2-\sqrt{3}}}{2} \) ≈ 0.25882	\( 2+\sqrt{3} \) ≈ 3.73205
90	\( \frac{\pi}{2} \)	1	0	undefined
108	\( \frac{3\pi}{5} \)	\( \frac{\sqrt{10+2\sqrt{5}}}{4} \) ≈ 0.95105	\( \frac{-\sqrt{5}+1}{4} \) ≈ -0.309017	\( -\sqrt{5+2\sqrt{5}} \) ≈ -3.07768
cos(108) and tan(108) are negative values since 108 is 2nd quadrant.

Areas of Regular Polygons

The following table is a reference of areas of regular polygons when the length of the side is known. A regular polygon is one whose sides are equal in length and the internal angles are equal.

Polygon	n	Area	Appx. area
Equilateral triangle	3	\( \frac{\sqrt{3}}{4}{{s}^{2}} \)	\( 0.433s^{2} \)
Square	4	\( s^{2} \)	-
Pentagon	5	\( \frac{\sqrt{25+10\sqrt{5}}}{4}{{s}^{2}} \)	\( 1.72s^{2} \)
Hexagon	6	\( \frac{3\sqrt{3}}{2}s^2 \)	\(2.598s^{2} \)
Heptagon	7	-	\(3.634s^{2} \)
Octagon	8	\( 2(\sqrt{2}+1){{s}^{2}} \)	\(4.828s^{2} \)
Nonagon	9	-	\(6.182s^{2} \)
Decagon	10	\( \frac{5\sqrt{2\sqrt{5}+5}}{2}{{s}^{2}} \)	\(7.694s^{2} \)
n-sided polygon	n	\( \frac{1}{4}n(\cot\frac{\pi}{n})s^{2} \)	-

Area of Inscribed Polygon

If a circle is cirmumscribed on a polygon and has a radius of r, then the area of the polygon with n number of sides is \( n(\sin\frac{\pi}{n})(\cos\frac{\pi}{n})r^{2} = (\frac{1}{2}n\sin\frac{2\pi}{n})r^{2} \)

We will use the image below to derive this formula.

A regular polygon can be divided into n congruent triangles. One triangle is shown in Figure 1a. The central angle measure will be \( \frac{2\pi}{n} \). If we draw an altitude, this divides the central angle into 2, or \( \frac{\pi}{n} \). We can find the length of the side in terms of the radius of the circle and use the polygon area formula \( \frac{n}{4}(\cot\frac{\pi}{n})s^{2} \).

Using the sine ratio, we have \( \sin\frac{\pi}{n} = \frac{s/2}{r} = \frac{s}{2r} \). Solving for s, we get \( s = 2(\sin\frac{\pi}{n})r \). Substituting this into the polygon area formula:

(i) \( A= \frac{n}{4}(\cot\frac{\pi}{n})(2(\sin\frac{\pi}{n})r)^{2} \)

(ii) \( A= \frac{n}{4}(\cot\frac{\pi}{n})\cdot 4(\sin^{2}\frac{\pi}{n})r^{2} \)

(iii) \( A= n(\sin\frac{\pi}{n})(\cos\frac{\pi}{n})r^{2} \)

(iv) \( A= \frac{1}{2}n(\sin\frac{2\pi}{n})r^{2} \)

Also note that the area of a triangle is also given by \( A = \frac{1}{2}ab\sin C \). In this case, a = b = r and the angle C is \( \frac{2\pi}{n} \). Therefore, the area of the triangle in Figure 1A is \( \frac{1}{2}r^{2}\sin\frac{2\pi}{n} \). There are n such triangles, so the formula that results is the same as (iv).

Limit of the Polygon Area

The more sides of a regular we have, the more it looks like a circle. If we take the limit of the polygon area formula at infinity, we should get the area of a circle.

However, we need to use the formula where the radius of the circle is constant. This formula is the one we just derived above: \( A= \frac{1}{2}n(\sin\frac{2\pi}{n})r^{2} \). If we use the formula \( \frac{n}{4}(\cot\frac{\pi}{n})s^{2} \), the area increases without bound as n increases because the length of the sides of the polygon remains constant at s and the polygon gets bigger and bigger. This can be seen here in the image below.

The polygon area formula in terms of the radius of the circle is \( A= \frac{1}{2}n(\sin\frac{2\pi}{n})r^{2} \).

The limit at infinity becomes: \( \frac{r^{2}}{2} \lim\limits_{n\to\infty} n\sin\frac{2\pi}{n} \, = \) \( \frac{r^{2}}{2} (\infty\cdot 0) \). This gives us an indeterminate form. Therefore, we have to change this limit so we get a 0/0 indeterminate form to apply L'Hôpital’s rule. We can accomplish this by putting the inverse of n in the denominator.

(i) \( \frac{r^{2}}{2} \lim\limits_{n\to\infty} \frac{\sin\frac{2\pi}{n}}{\frac{1}{n}} \)

The above gives us the indeterminate form 0/0. We can use L'Hôpital’s rule for the limit by taking the derivative of both top and the bottom.

(ii) \( \frac{r^{2}}{2} \lim\limits_{n\to\infty} \frac{\cos\frac{2\pi}{n}\cdot (-\frac{2\pi}{n^{2}})}{-\frac{1}{n^{2}}} \)

(iii) \( \frac{r^{2}}{2} \cdot 2\pi \lim\limits_{n\to\infty} \frac{\cos\frac{2\pi}{n}\cdot (-\frac{1}{n^{2}})}{-\frac{1}{n^{2}}} \)

(iv) \( \frac{r^{2}}{2} \cdot 2\pi \lim\limits_{n\to\infty} \cos\frac{2\pi}{n} \)

(v) \( \frac{r^{2}}{2} \cdot 2\pi \cos 0 = \pi r^{2} \)

We get the area of a circle by taking the limit at infinity.

Polygon Iterations

If an n-side polygon is iterated by connecting the midpoints of the sides of the polygon with length L₀, then the sum of the areas of all the polygons converges and is given by \( {{A}_{\text{T}}}=\frac{n\cos({\pi}/{n})}{4{\sin^{3}({\pi}/{n})}}L_{0}^{2} \), where A_T is the total area of all the polygons. The series of the areas is a geometric series.

Square Iteration

Let’s iterate a square at the midpoints. Image below shows a few of the squares.

The length of the side of the first square inside the original square is \( \frac{\sqrt{2}}{2}L_{0} \). Each subsequent square has a side that is \( \frac{\sqrt{2}}{2} \) times the previous square side length. The progression of the side lengths is:

\( L_{0} \), \( \frac{\sqrt{2}}{2}L_{0} \), \( \frac{2}{4}L_{0} \), \( \frac{2\sqrt{2}}{8}L_{0} \), \( \frac{4}{16}L_{0} \) ... In reduced form:

\( L_{0} \), \( \frac{\sqrt{2}}{2}L_{0} \), \( \frac{1}{2}L_{0} \), \( \frac{\sqrt{2}}{4}L_{0} \), \( \frac{1}{4}L_{0} \) ...

If we take the areas and add them, we would get:

\( A_{T} = L_{0}^{2} + \frac{2}{4}L_{0}^{2} + \frac{1}{4}L_{0}^{2} + \frac{2}{16}L_{0}^{2} + \frac{1}{16}L_{0}^{2} + ... \)

\( A_{T} = L_{0}^{2}(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...) \)

This is a geometric series with a common ratio of 1/2. The sum of a geometric series is \( S = \frac{1}{1-r} \), where r is the common ratio. Therefore the sum of the areas of all squares is:

\( A_{T} = L_{0}^{2}\left(\frac{1}{1-\frac{1}{2}}\right) = 2L_{0}^{2} \)

Now, we use the iteration formula above to see if we get the same result for \( n = 4 \):

\( {{A}_{\text{T}}} = \frac{4\cos({\pi}/{4})}{4{\sin^{3}({\pi}/{4})}}L_{0}^{2} = \frac{4\frac{\sqrt{2}}{2}}{4(\frac{\sqrt{2}}{2})^{3}}L_{0}^{2} = \frac{2\sqrt{2}}{\sqrt{2}}L_{0}^{2} = 2L_{0}^{2} \)

The formula checks out for a square.

Equilateral Triangle Iteration

By connecting the midpoints of an equilateral triangle, we get a triangle that has one-fourth the area of the original.

Hence, the area of all equilateral triangles would be \( A_{T} = \frac{\sqrt{3}}{4}L_{0}^{2}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) \, = \) \(\frac{\sqrt{3}}{4}L_{0}^{2}\left(\frac{1}{1-\frac{1}{4}}\right) \, = \) \( \frac{\sqrt{3}}{3}L_{0}^{2} \)

Using the formula with \( n = 3 \): \( {{A}_{\text{T}}} = \frac{3\cos ({\pi}/{3})}{4{\sin^{3}({\pi}/{3})}}L_{0}^{2} = \frac{3\cdot \frac{1}{2}}{4(\frac{\sqrt{3}}{2})^{3}}L_{0}^{2} \, = \) \( \frac{3}{8(\frac{3\sqrt{3}}{8})}L_{0}^{2} \, = \) \( \frac{\sqrt{3}}{3}L_{0}^{2} \)

The formula checks for an equilateral triangle.

Proof of Polygon Area Formula

The area of an n-sided polygon is given by \( \frac{n}{4}(\cot\frac{\pi}{n})s^{2} \). Let’s prove this formula. Consider the image of a polygon below. The side length is s. The height of one triangle (of the n triangles) drawn with the side and two radii is h.

From the triangle, we have the following relationship between s and h: \( \tan\frac{\theta}{2} = \frac{s}{2h} \). Therefore, \( h = \frac{s}{2}\cot\frac{\theta}{2} \).

Therefore the area of one triangle is \( \frac{1}{2}\cdot s \cdot \frac{s}{2}\cot\frac{\theta}{2} \, = \) \( \frac{1}{4}s^{2}\cot\frac{\theta}{2}\).

There are n such triangles in the polygon, and the central angle θ is equal to \( \frac{2\pi}{n} \). Therefore, the formula above becomes \( \frac{1}{4}n(\cot\frac{\pi}{n})s^{2}\).

Iteration Formula

Now consider an iterated polygon formed by connecting the midpoints of all sides. One side is shown in red above. The angle ϕ has a measure of \( \pi - \frac{2\pi}{n} \).

The side length of the new polygon, \( L_{1} \), is then calculated by the Law of Cosines:

(i) \( L_{1}^{2} = (\frac{s}{2})^{2} + (\frac{s}{2})^{2} - 2(\frac{s}{2})(\frac{s}{2})\cos(\pi - \frac{2\pi}{n}) \)

(ii) \( L_{1}^{2} = \frac{s^{2}}{2} - \frac{s^{2}}{2}\cos(\pi - \frac{2\pi}{n}) \)

(iii) \( L_{1}^{2} = \frac{s^{2}}{2}(1 - \cos(\pi - \frac{2\pi}{n})) \)

Using the sum of angles identity: \( \cos(\pi - \frac{2\pi}{n}) \, = \) \( \cos\pi \cos \frac{2\pi}{n} + \sin \pi \sin \frac{2\pi}{n} \, = \) \( -\cos\frac{2\pi}{n} \), (iii) becomes:

(iv) \( L_{1}^{2} = \frac{s^{2}}{2}(1 + \cos(\frac{2\pi}{n})) \)

(v) \( L_{1}^{2} = \frac{s^{2}}{2}(1 + \cos(\frac{2\pi}{n})) \)

(vi) \( L_{1} = s\sqrt{\frac{1 + \cos(\frac{2\pi}{n})}{2}} \)

(v) \( L_{1} = s\cdot \cos\frac{\pi}{n} \) (Using the half-angle identity)

Therefore, the side of the iterated polygon is equal to \((\cos\frac{\pi}{n})s \). The area of this polygon, in terms of the original polygon side length of s, would be \( \frac{n}{4}(\cot\frac{\pi}{n})\left( \cos^{2}\frac{\pi}{n}\right)s^{2} \).

The next iterated polygon would have a side of \( (\cos\frac{\pi}{n})(\cos\frac{\pi}{n})s = \cos^{2}\frac{\pi}{n}s \). The area of this polygon would be \( \frac{n}{4}(\cot\frac{\pi}{n})\left( \cos^{4}\frac{\pi}{n}\right)s^{2} \).

Each subsequent polygon has an area that is multiplied by \( \cos^{2}\frac{\pi}{n} \). The sum of all these areas is:

(vi) \( A_{T} = \frac{n}{4}\cot\frac{\pi}{n}(1 + \cos^{2}\frac{\pi}{n} + \cos^{4}\frac{\pi}{n} + \cos^{6}\frac{\pi}{n} + ...) \)

This is a geometric series with a common ratio of \( \cos^{2}\frac{\pi}{n} \) that has the sum:

(vii) \( A_{T} = \frac{n}{4}\cot\frac{\pi}{n}\left( \frac{1}{1-\cos^{2}\frac{\pi}{n}} \right) \)

(viii) \( A_{T} = \frac{n}{4}\cot\frac{\pi}{n}\cdot \frac{1}{\sin^{2}\frac{\pi}{n}} \)

Equation (viii) is the final area formula for the sum of all interated polygons and can be written in different ways depending on how to use the trig identities.