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The quotients formed by the area of these polygons divided by the square of the circle radius can be made arbitrarily close to π as the number of polygon sides becomes large, proving that the area inside the circle of radius r is πr 2, π being defined as the ratio of the circumference to the diameter (C/d).
A regular polygon with n sides can be constructed with ruler, compass, and angle trisector if and only if =, where r, s, k ≥ 0 and where the p i are distinct Pierpont primes greater than 3 (primes of the form +). [8]: Thm. 2 These polygons are exactly the regular polygons that can be constructed with Conic section, and the regular polygons ...
[2]: p. 1 They could also construct half of a given angle, a square whose area is twice that of another square, a square having the same area as a given polygon, and regular polygons of 3, 4, or 5 sides [2]: p. xi (or one with twice the number of sides of a given polygon [2]: pp. 49–50 ).
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
Some regular polygons are easy to construct with compass and straightedge; other regular polygons are not constructible at all. The ancient Greek mathematicians knew how to construct a regular polygon with 3, 4, or 5 sides, [11]: p. xi and they knew how to construct a regular polygon with double the number of sides of a given regular polygon.
Shoelace scheme for determining the area of a polygon with point coordinates (,),..., (,). The shoelace formula, also known as Gauss's area formula and the surveyor's formula, [1] is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. [2]
The Delian problem, for instance, was to construct a length x so that the cube of side x contained the same volume as the rectangular box a 2 b for given sides a and b. Menaechmus (c. 350 BC) considered the problem geometrically by intersecting the pair of plane conics ay = x 2 and xy = ab. [2]
A general approach that works for non-simple polygons as well would be to choose a line not parallel to any of the sides of the polygon and draw a line parallel to this one through each of the vertices of the polygon. This will divide the polygon into triangles and trapezoids, which in turn can be converted into triangles.
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