Side length

The picture shows a regular decagon with side length ${\displaystyle a}$ and radius ${\displaystyle R}$ of the circumscribed circle.

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

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

Further conclusions

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

Area

The area of a regular decagon of side length a is given by:^[3]

${\displaystyle A={\frac {5}{2}}a^{2}\cot \left({\frac {\pi }{10}}\right)={\frac {5}{2}}a^{2}{\sqrt {5+2{\sqrt {5}}}}\simeq 7.694208843\,a^{2}}$

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

${\displaystyle A=10\tan \left({\frac {\pi }{10}}\right)r^{2}=2r^{2}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\simeq 3.249196962\,r^{2}}$

In terms of the circumradius R, the area is:

${\displaystyle A=5\sin \left({\frac {\pi }{5}}\right)R^{2}={\frac {5}{2}}R^{2}{\sqrt {\frac {5-{\sqrt {5}}}{2}}}\simeq 2.938926261\,R^{2}}$

An alternative formula is ${\displaystyle A=2.5da}$ 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,

${\displaystyle d=2a\left(\cos {\tfrac {3\pi }{10}}+\cos {\tfrac {\pi }{10}}\right),}$

and it can be written algebraically as

${\displaystyle d=a{\sqrt {5+2{\sqrt {5}}}}.}$

Sides

A regular decagon has 10 sides and is equilateral. It has 35 diagonals

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.^[4]

Construction of decagon

Construction of pentagon

An alternative (but similar) method is as follows:

1. Construct a pentagon in a circle by one of the methods shown in constructing a pentagon.
2. 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.
3. The five corners of the pentagon constitute alternate corners of the decagon. Join these points to the adjacent new points to form the decagon.

Nonconvex regular decagon

This tiling by golden triangles, a regular pentagon, contains a stellation of regular decagon, the Schäfli symbol of which is {10/3}.

The length ratio of two inequal edges of a golden triangle is the golden ratio, denoted by ${\displaystyle \Phi }$, or its multiplicative inverse:

$${\displaystyle \Phi -1={\frac {1}{\Phi }}=2\,\cos 72\,^{\circ }={\frac {1}{\,2\,\cos 36\,^{\circ }}}={\frac {\,{\sqrt {5}}-1\,}{2}}{\text{.}}}$$

So we can get the properties of a regular decagonal star, through a tiling by golden triangles that fills this star polygon.

Petrie polygons

The regular skew decagon is the Petrie polygon for many higher-dimensional polytopes, shown in these orthogonal projections in various Coxeter planes:^[9] The number of sides in the Petrie polygon is equal to the Coxeter number, h, for each symmetry family.

A9	D6		B5
9-simplex	411	131	5-orthoplex	5-cube