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Decagon

Shape with ten sides

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  • 5 chapters

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Through Geometry, Polygon and Internal and external angles

Regular decagon

A regular decagon has all sides of equal length and each internal angle will always be equal to 144°. Its Schläfli symbol is {10} 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 E10E1M has two equally long legs with length R and a base with length a
  • The circle around E1 with radius a intersects ]ME10[ in a point P (not designated in the picture).
  • Now the triangle E10E1P is an isosceles triangle with vertex E1 and with base angles m∠E1E10P=m∠E10PE1=72∘.
  • Therefore m∠PE1E10=180∘−2⋅72∘=36∘. So m∠ME1P=72∘−36∘=36∘ and hence E1MP is also an isosceles triangle with vertex P. The length of its legs is a, so the length of [PE10] is R−a.
  • The isosceles triangles E10E1M and PE10E1 have equal angles of 36° at the vertex, and so they are similar, hence: aR=R−aa
  • Multiplication with the denominators R,a>0 leads to the quadratic equation: a2=R2−aR
  • This equation for the side length a has one positive solution: a=R2(−1+5)

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

Further conclusions

R=2a5−1=a2(5+1) and the base height of ΔE10E1M (i.e. the length of [MD]) is h=R2−(a/2)2=a25+25 and the triangle has the area: AΔ=a2⋅h=a245+25.

Area

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

A=52a2cot⁡(π10)=52a25+25≃7.694208843a2

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

A=10tan⁡(π10)r2=2r25(5−25)≃3.249196962r2

In terms of the circumradius R, the area is:

A=5sin⁡(π5)R2=52R25−52≃2.938926261R2

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(cos⁡3π10+cos⁡π10),

and it can be written algebraically as

d=a5+25.

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:

  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.

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