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Apothem of a hexagon Graphs of side, s; apothem, a; and area, A of regular polygons of n sides and circumradius 1, with the base, b of a rectangle with the same area. The green line shows the case n = 6. The apothem (sometimes abbreviated as apo [1]) of a regular polygon is a line
The apothem is half the cotangent of /, and the area of each of the 14 small triangles is one-fourth of the apothem. The area of a regular heptagon inscribed in a circle of radius R is 7 R 2 2 sin 2 π 7 , {\displaystyle {\tfrac {7R^{2}}{2}}\sin {\tfrac {2\pi }{7}},} while the area of the circle itself is π R 2 ; {\displaystyle \pi R^{2 ...
All vertices of a regular polygon lie on a common circle (the circumscribed circle); i.e., they are concyclic points. That is, a regular polygon is a cyclic polygon . Together with the property of equal-length sides, this implies that every regular polygon also has an inscribed circle or incircle that is tangent to every side at the midpoint.
Contracting an edge of a polygon-circle graph results in another polygon-circle graph. A geometric representation of the new graph may be formed by replacing the polygons corresponding to the two endpoints of the contracted edge by their convex hull. Alternatively, in the alternating sequence representing the original graph, combining the ...
In geometry, a bigon, [1] digon, or a 2-gon, is a polygon with two sides and two vertices.Its construction is degenerate in a Euclidean plane because either the two sides would coincide or one or both would have to be curved; however, it can be easily visualised in elliptic space.
The common length of the sides equals the radius of the circumscribed circle or circumcircle, which equals times the apothem (radius of the inscribed circle). All internal angles are 120 degrees . A regular hexagon has six rotational symmetries ( rotational symmetry of order six ) and six reflection symmetries ( six lines of symmetry ), making ...
As proved by Archimedes, in his Measurement of a Circle, the area enclosed by a circle is equal to that of a triangle whose base has the length of the circle's circumference and whose height equals the circle's radius, [11] which comes to π multiplied by the radius squared: =.
Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides is ns, which is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals nhs/2.