Decagon - WikiHQ

In geometry, a decagon (from the Greek δέκα déka and γωνία gonía, "ten angles") is a ten-sided polygon or 10-gon. The total sum of the interior angles of a simple decagon is 1440°.

Regular decagon

A regular decagon has all sides of equal length and each internal angle will always be equal to 144°. and can also be constructed as a truncated pentagon, t{5}, a quasiregular decagon alternating two types of edges.

Side length

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The picture shows a regular decagon with side length <math>a</math> and radius <math>R</math> of the circumscribed circle.

The triangle <math>E_{10}E_1M</math> has two equally long legs with length <math>R</math> and a base with length <math>a</math>
The circle around <math>E_1</math> with radius <math>a</math> intersects <math>]M\,E_{10}[</math> in a point <math>P</math> (not designated in the picture).
Now the triangle <math>{E_{10}E_1P}\;</math> is an isosceles triangle with vertex <math>E_1</math> and with base angles <math>m\angle E_1 E_{10} P = m\angle E_{10} P E_1 = 72^\circ \;</math>.
Therefore <math>m\angle P E_1 E_{10} = 180^\circ -2\cdot 72^\circ = 36^\circ \;</math>. So <math>\; m\angle M E_1 P = 72^\circ- 36^\circ = 36^\circ\;</math> and hence <math>\; E_1 M P\;</math> is also an isosceles triangle with vertex <math>P</math>. The length of its legs is <math>a</math>, so the length of <math>[P\,E_{10}]</math> is <math>R-a</math>.
The isosceles triangles <math>E_{10} E_1 M\;</math> and <math>P E_{10} E_1\;</math> have equal angles of 36° at the vertex, and so they are similar, hence: <math>\;\frac{a}{R}=\frac{R-a}{a}</math>
Multiplication with the denominators <math>R,a >0</math> leads to the quadratic equation: <math>\;a^2=R^2-aR\;</math>
This equation for the side length <math>a\,</math> has one positive solution: <math>\;a=\frac{R}{2}(-1+\sqrt{5})</math>

So the regular decagon can be constructed with ruler and compass.

;Further conclusions:

<math>\;R=\frac{2a}{\sqrt{5}-1}=\frac{a}{2}(\sqrt{5}+1)\;</math> and the base height of <math>\Delta\,E_{10} E_1 M\,</math> (i.e. the length of <math>[M\,D]</math>) is <math>h = \sqrt{R^2-(a/2)^2}=\frac{a}{2}\sqrt{5+2\sqrt{5\;</math> and the triangle has the area: <math>A_\Delta=\frac{a}{2}\cdot h = \frac{a^2}{4}\sqrt{5+2\sqrt{5</math>.

Area

The area of a regular decagon of side length a is given by:

:<math> A = \frac{5}{2} a^2\cot\left(\frac{\pi}{10} \right) =

\frac{5}{2} a^2\sqrt{5+2\sqrt{5

\simeq 7.694208843\,a^2

</math>

In terms of the apothem r (see also inscribed figure), the area is:

:<math>A = 10 \tan\left(\frac{\pi}{10}\right) r^2 =

2r^2\sqrt{5\left(5-2\sqrt5\right)}

\simeq 3.249196962\,r^2

</math>

In terms of the circumradius R, the area is:

:<math>

A = 5 \sin\left(\frac{\pi}{5}\right) R^2

= \frac{5}{2}R^2\sqrt{\frac{5-\sqrt{5{2

\simeq 2.938926261\,R^2

</math>

An alternative formula is <math>A=2.5da</math> where d is the distance between parallel sides, or the height when the decagon stands on one side as base, or the diameter of the decagon's inscribed circle.

By simple trigonometry,

:<math>d=2a\left(\cos\tfrac{3\pi}{10}+\cos\tfrac{\pi}{10}\right),</math>

and it can be written algebraically as

:<math>d=a\sqrt{5+2\sqrt{5.</math>

Construction

As 10 = 2 × 5, a power of two times a Fermat prime, it follows that a regular decagon is constructible using compass and straightedge, or by an edge-bisection of a regular pentagon.

An alternative (but similar) method is as follows:

Construct a pentagon in a circle by one of the methods shown in constructing a pentagon.
Extend a line from each vertex of the pentagon through the center of the circle to the opposite side of that same circle. Where each line cuts the circle is a vertex of the decagon. In other words, the image of a regular pentagon under a point reflection with respect of its center is a concentric congruent pentagon, and the two pentagons have in total the vertices of a concentric regular decagon.
The five corners of the pentagon constitute alternate corners of the decagon. Join these points to the adjacent new points to form the decagon.

The golden ratio in decagon

Both in the construction with given circumcircle as well as with given side length is the golden ratio dividing a line segment by exterior division the

determining construction element.

In the construction with given circumcircle the circular arc around G with radius produces the segment , whose division corresponds to the golden ratio.

:<math>\frac{\overline{AM{\overline{MH = \frac{\overline{AH{\overline{AM = \frac{1+ \sqrt{5{2} = \Phi \approx 1.618 \text{.}</math>

In the construction with given side length the circular arc around D with radius produces the segment , whose division corresponds to the golden ratio.

:<math>\frac{\overline{E_1 E_{10}{\overline{E_1 F = \frac{\overline{E_{10} F{\overline{E_1 E_{10} = \frac{R}{a} = \frac{1+ \sqrt{5{2} =\Phi \approx 1.618 \text{.}</math>