Editing Polygon (section)

====Simple polygons====
{{further|Shoelace formula}}

If the polygon is non-self-intersecting (that is, [[simple polygon|simple]]), the signed [[area (geometry)|area]] is
:<math>A = \frac{1}{2} \sum_{i = 0}^{n - 1}( x_i y_{i + 1} - x_{i + 1} y_i) \quad \text {where } x_{n}=x_{0} \text{ and } y_n=y_{0}, </math>
or, using [[determinant]]s
:<math>16 A^{2} = \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} \begin{vmatrix} Q_{i,j} & Q_{i,j+1} \\
Q_{i+1,j} & Q_{i+1,j+1} \end{vmatrix} , </math>
where <math> Q_{i,j} </math> is the squared distance between <math>(x_i, y_i)</math> and <math>(x_j, y_j).</math><ref>B.Sz. Nagy, L. Rédey: Eine Verallgemeinerung der Inhaltsformel von Heron. Publ.
Math. Debrecen 1, 42–50 (1949)</ref><ref>{{cite web
 |url      = http://www.seas.upenn.edu/~sys502/extra_materials/Polygon%20Area%20and%20Centroid.pdf
 |title      = Calculating The Area And Centroid Of A Polygon
 |last      = Bourke
 |first      = Paul
 |date      = July 1988
 |access-date      = 6 Feb 2013
 |archive-date      = 16 September 2012
 |archive-url      = https://web.archive.org/web/20120916104133/http://www.seas.upenn.edu/~sys502/extra_materials/Polygon%20Area%20and%20Centroid.pdf
 |url-status      = dead
 }}</ref>

The signed area depends on the ordering of the vertices and of the [[orientation (vector space)|orientation]] of the plane. Commonly, the positive orientation is defined by the (counterclockwise) rotation that maps the positive {{mvar|x}}-axis to the positive {{mvar|y}}-axis. If the vertices are ordered counterclockwise (that is, according to positive orientation), the signed area is positive; otherwise, it is negative. In either case, the area formula is correct in [[absolute value]]. This is commonly called the ''[[shoelace formula]]'' or ''surveyor's formula''.<ref>{{cite journal |author=Bart Braden |title=The Surveyor's Area Formula |journal=The College Mathematics Journal |volume=17 |issue=4 |year=1986 |pages=326–337 |url=http://www.maa.org/pubs/Calc_articles/ma063.pdf|archive-url=https://web.archive.org/web/20121107190918/http://www.maa.org/pubs/Calc_articles/ma063.pdf|archive-date=2012-11-07 |doi=10.2307/2686282|jstor=2686282 }}</ref>

The area ''A'' of a simple polygon can also be computed if the lengths of the sides, ''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., ''a<sub>n</sub>'' and the [[exterior angle]]s, ''θ''<sub>1</sub>, ''θ''<sub>2</sub>, ..., ''θ<sub>n</sub>'' are known, from:
:<math>\begin{align}A = \frac12 ( a_1[a_2 \sin(\theta_1) + a_3 \sin(\theta_1 + \theta_2) + \cdots + a_{n-1} \sin(\theta_1 + \theta_2 + \cdots + \theta_{n-2})] \\
{} + a_2[a_3 \sin(\theta_2) + a_4 \sin(\theta_2 + \theta_3) + \cdots + a_{n-1} \sin(\theta_2 + \cdots + \theta_{n-2})] \\
{} + \cdots + a_{n-2}[a_{n-1} \sin(\theta_{n-2})] ). \end{align}</math>
The formula was described by Lopshits in 1963.<ref name="lopshits">{{cite book |title=Computation of areas of oriented figures |author=A.M. Lopshits |publisher=D C Heath and Company: Boston, MA |others=translators: J Massalski and C Mills Jr. |year=1963}}</ref>

If the polygon can be drawn on an equally spaced grid such that all its vertices are grid points, [[Pick's theorem]] gives a simple formula for the polygon's area based on the numbers of interior and boundary grid points: the former number plus one-half the latter number, minus 1.

In every polygon with perimeter ''p'' and area ''A '', the [[isoperimetric inequality]] <math>p^2 > 4\pi A</math> holds.<ref>{{cite web| url = http://forumgeom.fau.edu/FG2002volume2/FG200215.pdf| title = Dergiades, Nikolaos, "An elementary proof of the isoperimetric inequality", ''Forum Mathematicorum'' 2, 2002, 129–130.}}</ref>

For any two simple polygons of equal area, the [[Bolyai–Gerwien theorem]] asserts that the first can be cut into polygonal pieces which can be reassembled to form the second polygon.

The lengths of the sides of a polygon do not in general determine its area.<ref>Robbins, "Polygons inscribed in a circle", ''American Mathematical Monthly'' 102, June–July 1995.</ref> However, if the polygon is simple and cyclic then the sides ''do'' determine the area.<ref>{{cite journal|last=Pak|first=Igor|author-link=Igor Pak|doi=10.1016/j.aam.2004.08.006|issue=4|journal=[[Advances in Applied Mathematics]]|mr=2128993|pages=690–696|title=The area of cyclic polygons: recent progress on Robbins' conjectures|volume=34|year=2005|arxiv=math/0408104|s2cid=6756387}}</ref> Of all ''n''-gons with given side lengths, the one with the largest area is cyclic. Of all ''n''-gons with a given perimeter, the one with the largest area is regular (and therefore cyclic).<ref>Chakerian, G. D. "A Distorted View of Geometry." Ch. 7 in ''Mathematical Plums'' (R. Honsberger, editor). Washington, DC: Mathematical Association of America, 1979: 147.</ref>