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The regular decagon has Dih 10 symmetry, order 20. There are 3 subgroup dihedral symmetries: Dih 5, Dih 2, and Dih 1, and 4 cyclic group symmetries: Z 10, Z 5, Z 2, and Z 1. These 8 symmetries can be seen in 10 distinct symmetries on the decagon, a larger number because the lines of reflections can either pass through vertices or edges.
A regular triangle, decagon, and pentadecagon can completely fill a plane vertex. However, due to the triangle's odd number of sides, the figures cannot alternate around the triangle, so the vertex cannot produce a semiregular tiling.
The regular hexadecagon has Dih 16 symmetry, order 32. There are 4 dihedral subgroups: Dih 8, Dih 4, Dih 2, and Dih 1, and 5 cyclic subgroups: Z 16, Z 8, Z 4, Z 2, and Z 1, the last implying no symmetry.
As 14 = 2 × 7, a regular tetradecagon cannot be constructed using a compass and straightedge. [1] However, it is constructible using neusis with use of the angle trisector, [2] or with a marked ruler, [3] as shown in the following two examples.
Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities; List of volume formulas – Quantity of three-dimensional space
Regular convex and star polygons with 3 to 12 vertices labelled with their Schläfli symbols. These properties apply to all regular polygons, whether convex or star: . A regular n-sided polygon has rotational symmetry of order n.
Three squares of sides R can be cut and rearranged into a dodecagon of circumradius R, yielding a proof without words that its area is 3R 2. A regular dodecagon is a figure with sides of the same length and internal angles of the same size. It has twelve lines of reflective symmetry and rotational symmetry of order 12.
Let the circle on AF as diameter cut OB in K, and let the circle whose centre is E and radius EK cut OA in N 3 and N 5; then if ordinates N 3 P 3, N 5 P 5 are drawn to the circle, the arcs AP 3, AP 5 will be 3/17 and 5/17 of the circumference." The point N 3 is very close to the center point of Thales' theorem over AF.