Editing Decagon (section)

==Regular decagon==
A ''[[regular polygon|regular]] decagon'' has all sides of equal length and each internal angle will always be equal to 144°.<ref name="sidebotham"/> Its [[Schläfli symbol]] is {10} <ref>{{citation|title=Polyhedron Models|first=Magnus J.|last=Wenninger|publisher=Cambridge University Press|year=1974|page=9|isbn=9780521098595|url=https://books.google.com/books?id=N8lX2T-4njIC&pg=PA9}}.</ref> and can also be constructed as a [[Truncation (geometry)|truncated]] [[pentagon]], t{5}, a quasiregular decagon alternating two types of edges.

{{multiple image
 | align = right | perrow = 2 | total_width = 270
 | image1 = Timurid Qur'an decagon.jpg
 | image2 = Ioanniskepplerih00kepl 0078.jpg
 | image3 = SSS Penrose tiling, 7iter.svg
 | footer = Decagons often appear in tilings with (partial) 5-fold symmetry. The images show an [[Islamic geometric patterns|Islamic geometric pattern]] (15th century), an illustration in Kepler's [[Harmonices Mundi]] (1619) and a [[Penrose tiling]].
}}

=== Side length ===
[[File:01-Zehneck-Seitenlänge.svg|300px|right]]
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 [[Similarity (geometry)|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 ''[[Straightedge and compass construction|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:<ref>{{citation|title=The elements of plane and spherical trigonometry|publisher=Society for Promoting Christian Knowledge|year=1850|page=59|url=https://books.google.com/books?id=AW7qzTr_f1sC&pg=PA59}}. Note that this source uses ''a'' as the edge length and gives the argument of the cotangent as an angle in degrees rather than in radians.</ref>
:<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 figure|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 [[algebraic expression|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 polygon|constructible]] using [[compass and straightedge]], or by an edge-[[bisection]] of a regular [[pentagon]].<ref name="ludlow">{{citation|title=Geometric Construction of the Regular Decagon and Pentagon Inscribed in a Circle|first=Henry H.|last=Ludlow|publisher=The Open Court Publishing Co.|year=1904|url=https://books.google.com/books?id=vLMlw7uL8kgC}}.</ref>
<div class="skin-invert-image">{{multiple image
| align = left
| image1 = Regular Decagon Inscribed in a Circle.gif
| width1 = 260
| alt1 = 
| caption1 = Construction of decagon
| image2 = Regular Pentagon Inscribed in a Circle.gif
| width2 = 260
| alt2 = 
| caption2 = Construction of pentagon
| footer = 
}}</div>
{{clear}}

An alternative (but similar) method is as follows:
#Construct a pentagon in a circle by one of the methods shown in [[Pentagon#Construction of a regular pentagon|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,  [[Image (mathematics)#Image_of_a_subset|the image]] of a regular pentagon under a [[point reflection]] with respect of [[Regular polygon#Symmetry|its center]] is a [[Concentric objects|concentric]] ''[[Congruence (geometry)|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.