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 The picture shows a regular decagon with side length a and radius R of the
circumscribed circle. • The triangle E_{10}E_1M has two equally long legs with length R and a base with length a • The circle around E_1 with radius a intersects ]M\,E_{10}[ in a point P (not designated in the picture). • Now the triangle {E_{10}E_1P}\; is an
isosceles triangle with vertex E_1 and with base angles m\angle E_1 E_{10} P = m\angle E_{10} P E_1 = 72^\circ \;. • Therefore m\angle P E_1 E_{10} = 180^\circ -2\cdot 72^\circ = 36^\circ \;. So \; m\angle M E_1 P = 72^\circ- 36^\circ = 36^\circ\; and hence \; E_1 M P\; is also an isosceles triangle with vertex P. The length of its legs is a, so the length of [P\,E_{10}] is R-a. • The isosceles triangles E_{10} E_1 M\; and P E_{10} E_1\; have equal angles of 36° at the vertex, and so they are
similar, hence: \;\frac{a}{R}=\frac{R-a}{a} • Multiplication with the denominators R,a >0 leads to the quadratic equation: \;a^2=R^2-aR\; • This equation for the side length a\, has one positive solution: \;a=\frac{R}{2}(-1+\sqrt{5}) So the regular decagon can be constructed with
ruler and compass. ;Further conclusions: \;R=\frac{2a}{\sqrt{5}-1}=\frac{a}{2}(\sqrt{5}+1)\; and the base height of \Delta\,E_{10} E_1 M\, (i.e. the length of [M\,D]) is h = \sqrt{R^2-(a/2)^2}=\frac{a}{2}\sqrt{5+2\sqrt{5}}\; and the triangle has the area: 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: : 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: :A = 10 \tan\left(\frac{\pi}{10}\right) r^2 = 2r^2\sqrt{5\left(5-2\sqrt5\right)} \simeq 3.249196962\,r^2 In terms of the
circumradius R, the area is: : 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 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, :d=2a\left(\cos\tfrac{3\pi}{10}+\cos\tfrac{\pi}{10}\right), and it can be written
algebraically as :d=a\sqrt{5+2\sqrt{5}}.
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 ==