Perimeter Area - Revision history

Khanh at 04:09, 16 December 2021

2021-12-16T04:09:21Z

Khanh: /* Area formulas */

2021-12-16T03:51:31Z

Area formulas

Khanh: /* Area bisectors */

2021-12-16T03:34:40Z

Area bisectors

Khanh at 03:33, 16 December 2021

2021-12-16T03:33:13Z

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Created page with "right|thumb|alt=Three shapes on a square grid|The combined area of these three shapes is approximately 15.57 squares. '''Area''' is the quantity that expres..."

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[[File:Area.svg|right|thumb|alt=Three shapes on a square grid|The combined area of these three shapes is approximately 15.57 squares.]]

'''Area''' is the quantity that expresses the extent of a two-dimensional region, shape, or planar lamina, in the plane. Surface area is its analog on the two-dimensional surface of a three-dimensional object. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept).

The area of a shape can be measured by comparing the shape to squares of a fixed size. In the International System of Units (SI), the standard unit of area is the square metre (written as m2), which is the area of a square whose sides are one metre long. A shape with an area of three square metres would have the same area as three such squares. In mathematics, the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number.
[[File:Squaring the circle.svg|right|thumb|This square and this disk both have the same area (see: squaring the circle).]]

There are several well-known formulas for the areas of simple shapes such as triangles, rectangles, and circles. Using these formulas, the area of any polygon can be found by dividing the polygon into triangles. For shapes with curved boundary, calculus is usually required to compute the area. Indeed, the problem of determining the area of plane figures was a major motivation for the historical development of calculus.

For a solid shape such as a sphere, cone, or cylinder, the area of its boundary surface is called the surface area. Formulas for the surface areas of simple shapes were computed by the ancient Greeks, but computing the surface area of a more complicated shape usually requires multivariable calculus.

Area plays an important role in modern mathematics. In addition to its obvious importance in geometry and calculus, area is related to the definition of determinants in linear algebra, and is a basic property of surfaces in differential geometry. In analysis, the area of a subset of the plane is defined using Lebesgue measure, though not every subset is measurable. In general, area in higher mathematics is seen as a special case of volume for two-dimensional regions.

Area can be defined through the use of axioms, defining it as a function of a collection of certain plane figures to the set of real numbers. It can be proved that such a function exists.

==Formal definition==
An approach to defining what is meant by "area" is through axioms. "Area" can be defined as a function from a collection M of a special kinds of plane figures (termed measurable sets) to the set of real numbers, which satisfies the following properties:
* For all ''S'' in ''M'', ''a''(''S'') ≥ 0.
* If ''S'' and ''T'' are in ''M'' then so are ''S'' ∪ ''T'' and ''S'' ∩ ''T'', and also ''a''(''S''∪''T'') = ''a''(''S'') + ''a''(''T'') − ''a''(''S''∩''T'').
* If ''S'' and ''T'' are in ''M'' with ''S'' ⊆ ''T'' then ''T'' − ''S'' is in ''M'' and ''a''(''T''−''S'') = ''a''(''T'') − ''a''(''S'').
* If a set ''S'' is in ''M'' and ''S'' is congruent to ''T'' then ''T'' is also in ''M'' and ''a''(''S'') = ''a''(''T'').
* Every rectangle ''R'' is in ''M''. If the rectangle has length ''h'' and breadth ''k'' then ''a''(''R'') = ''hk''.
* Let ''Q'' be a set enclosed between two step regions ''S'' and ''T''. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. ''S'' ⊆ ''Q'' ⊆ ''T''. If there is a unique number ''c'' such that ''a''(''S'') ≤ c ≤ ''a''(''T'') for all such step regions ''S'' and ''T'', then ''a''(''Q'') = ''c''.

It can be proved that such an area function actually exists.

==Units==
[[File:SquareMeterQuadrat.JPG|thumb|right|alt=A square made of PVC pipe on grass|A square metre quadrat made of PVC pipe.]]
Every unit of length has a corresponding unit of area, namely the area of a square with the given side length. Thus areas can be measured in square metres (m2), square centimetres (cm2), square millimetres (mm2), square kilometres (km2), square feet (ft2), square yards (yd2), square miles (mi2), and so forth. Algebraically, these units can be thought of as the squares of the corresponding length units.

The SI unit of area is the square metre, which is considered an SI derived unit.

===Conversions===
[[File:Area conversion - square mm in a square cm.png|thumb|right|alt=A diagram showing the conversion factor between different areas|Although there are 10 mm in 1 cm, there are 100 mm2 in 1 cm2.]]
Calculation of the area of a square whose length and width are 1 metre would be:

1 metre × 1 metre = 1 m2

and so, a rectangle with different sides (say length of 3 metres and width of 2 metres) would have an area in square units that can be calculated as:

3 metres × 2 metres = 6 m2. This is equivalent to 6 million square millimetres. Other useful conversions are:
* 1 square kilometre = 1,000,000 square metres
* 1 square metre = 10,000 square centimetres = 1,000,000 square millimetres
* 1 square centimetre = 100 square millimetres.

==== Non-metric units ====
In non-metric units, the conversion between two square units is the square of the conversion between the corresponding length units.
:1 foot = 12 inches,
the relationship between square feet and square inches is
:1 square foot = 144 square inches,
where 144 = 122 = 12 × 12. Similarly:
* 1 square yard = 9 square feet
* 1 square mile = 3,097,600 square yards = 27,878,400 square feet
In addition, conversion factors include:
* 1 square inch = 6.4516 square centimetres
* 1 square foot = 0.092 903 04 square metres
* 1 square yard = 0.836 127 36 square metres
* 1 square mile = 2.589 988 110 336 square kilometres

===Other units including historical===
There are several other common units for area. The are was the original unit of area in the metric system, with:
* 1 are = 100 square metres
Though the are has fallen out of use, the hectare is still commonly used to measure land:
* 1 hectare = 100 ares = 10,000 square metres = 0.01 square kilometres
Other uncommon metric units of area include the tetrad, the hectad, and the myriad.

The acre is also commonly used to measure land areas, where
* 1 acre = 4,840 square yards = 43,560 square feet.
An acre is approximately 40% of a hectare.

On the atomic scale, area is measured in units of barns, such that:
* 1 barn = 10−28 square meters.
The barn is commonly used in describing the cross-sectional area of interaction in nuclear physics.

In India,
* 20 dhurki = 1 dhur
* 20 dhur = 1 khatha
* 20 khata = 1 bigha
* 32 khata = 1 acre

==Area formulas==

===Polygon formulas===

For a non-self-intersecting (simple) polygon, the Cartesian coordinates <math>(x_i, y_i)</math> (''i''=0, 1, ..., ''n''-1) of whose ''n'' vertices are known, the area is given by the surveyor's formula:

:<math>A = \frac{1}{2} \Biggl\vert \sum_{i = 0}^{n - 1}( x_i y_{i + 1} - x_{i + 1} y_i) \Biggr\vert</math>

where when ''i''=''n''-1, then ''i''+1 is expressed as modulus ''n'' and so refers to 0.

====Rectangles====
[[File:RectangleLengthWidth.svg|thumb|right|upright|alt=A rectangle with length and width labelled|The area of this rectangle is {{mvar|lw}}.]]
The most basic area formula is the formula for the area of a rectangle. Given a rectangle with length {{mvar|l}} and width {{mvar|w}}, the formula for the area is:

:{{bigmath|''A'' {{=}} ''lw''}}  (rectangle).
That is, the area of the rectangle is the length multiplied by the width. As a special case, as {{math|''l'' {{=}} ''w''}} in the case of a square, the area of a square with side length {{mvar|s}} is given by the formula:
:{{bigmath|''A'' {{=}} ''s''2}}  (square).

The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as a definition or axiom. On the other hand, if geometry is developed before arithmetic, this formula can be used to define multiplication of real numbers.

====Dissection, parallelograms, and triangles====

Most other simple formulas for area follow from the method of dissection.
This involves cutting a shape into pieces, whose areas must sum to the area of the original shape.
[[File:ParallelogramArea.svg|thumb|left|upright|A diagram showing how a parallelogram can be re-arranged into the shape of a rectangle.]]
For an example, any [[parallelogram]] can be subdivided into a [[trapezoid]] and a [[right triangle]], as shown in figure to the left. If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle. It follows that the area of the parallelogram is the same as the area of the rectangle:<ref name=AF/>
:{{bigmath|''A'' {{=}} ''bh''}}  (parallelogram).
[[File:TriangleArea.svg|thumb|right|upright|A parallelogram split into two equal triangles.]]
However, the same parallelogram can also be cut along a [[diagonal]] into two [[congruence (geometry)|congruent]] triangles, as shown in the figure to the right. It follows that the area of each [[triangle]] is half the area of the parallelogram:<ref name=AF/>
:<math>A = \frac{1}{2}bh</math>  (triangle).
Similar arguments can be used to find area formulas for the [[trapezoid]]<ref>{{citation|title=Problem Solving Through Recreational Mathematics|first1=Bonnie|last1=Averbach|first2=Orin|last2=Chein|publisher=Dover|year=2012|isbn=978-0-486-13174-0|page=306|url=https://books.google.com/books?id=Dz_CAgAAQBAJ&pg=PA306|url-status=live|archive-url=https://web.archive.org/web/20160513101526/https://books.google.com/books?id=Dz_CAgAAQBAJ&pg=PA306|archive-date=2016-05-13}}</ref> as well as more complicated [[polygon]]s.

===Area of curved shapes===

====Circles====
[[File:CircleArea.svg|thumb|right|alt=A circle divided into many sectors can be re-arranged roughly to form a parallelogram|A circle can be divided into [[Circular sector|sectors]] which rearrange to form an approximate [[parallelogram]].]]
{{main article|Area of a circle}}
The formula for the area of a [[circle]] (more properly called the area enclosed by a circle or the area of a [[disk (mathematics)|disk]]) is based on a similar method. Given a circle of radius {{math|''r''}}, it is possible to partition the circle into [[Circular sector|sectors]], as shown in the figure to the right. Each sector is approximately triangular in shape, and the sectors can be rearranged to form an approximate parallelogram. The height of this parallelogram is {{math|''r''}}, and the width is half the [[circumference]] of the circle, or {{math|π''r''}}. Thus, the total area of the circle is {{math|π''r''2}}:<ref name=AF/>
:{{bigmath|''A'' {{=}} π''r''2}}  (circle).
Though the dissection used in this formula is only approximate, the error becomes smaller and smaller as the circle is partitioned into more and more sectors. The [[limit (mathematics)|limit]] of the areas of the approximate parallelograms is exactly {{math|π''r''2}}, which is the area of the circle.<ref name=Surveyor/>

This argument is actually a simple application of the ideas of [[calculus]]. In ancient times, the [[method of exhaustion]] was used in a similar way to find the area of the circle, and this method is now recognized as a precursor to [[integral calculus]]. Using modern methods, the area of a circle can be computed using a [[definite integral]]:
:<math>A \;=\;2\int_{-r}^r \sqrt{r^2 - x^2}\,dx \;=\; \pi r^2.</math>

====Ellipses====
{{main article|Ellipse#Area}}
The formula for the area enclosed by an [[ellipse]] is related to the formula of a circle; for an ellipse with [[semi-major axis|semi-major]] and [[semi-minor axis|semi-minor]] axes {{math|''x''}} and {{math|''y''}} the formula is:<ref name=AF/>
:<math>A = \pi xy .</math>

====Surface area====
{{main article|Surface area}}
[[File:Archimedes sphere and cylinder.svg|right|thumb|alt=A blue sphere inside a cylinder of the same height and radius|[[Archimedes]] showed that the surface area of a [[sphere]] is exactly four times the area of a flat [[disk (mathematics)|disk]] of the same radius, and the volume enclosed by the sphere is exactly 2/3 of the volume of a [[cylinder (geometry)|cylinder]] of the same height and radius.]]
Most basic formulas for [[surface area]] can be obtained by cutting surfaces and flattening them out. For example, if the side surface of a [[cylinder (geometry)|cylinder]] (or any [[prism (geometry)|prism]]) is cut lengthwise, the surface can be flattened out into a rectangle. Similarly, if a cut is made along the side of a [[cone (geometry)|cone]], the side surface can be flattened out into a sector of a circle, and the resulting area computed.

The formula for the surface area of a [[sphere]] is more difficult to derive: because a sphere has nonzero [[Gaussian curvature]], it cannot be flattened out. The formula for the surface area of a sphere was first obtained by [[Archimedes]] in his work ''[[On the Sphere and Cylinder]]''. The formula is:<ref name=MathWorldSurfaceArea/>
:{{math|''A'' {{=}} 4''πr''2}}  (sphere),
where {{math|''r''}} is the radius of the sphere. As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar to [[calculus]].

===General formulas===

====Areas of 2-dimensional figures====
[[File:Triangle_GeometryArea.svg|thumb|Triangle area <math>A=\tfrac{b\cdot h}{2}</math>]]
* A [[triangle]]: <math>\tfrac12Bh</math> (where ''B'' is any side, and ''h'' is the distance from the line on which ''B'' lies to the other vertex of the triangle). This formula can be used if the height ''h'' is known. If the lengths of the three sides are known then ''[[Heron's formula]]'' can be used: <math>\sqrt{s(s-a)(s-b)(s-c)}</math> where ''a'', ''b'', ''c'' are the sides of the triangle, and <math>s = \tfrac12(a + b + c)</math> is half of its perimeter.<ref name=AF/> If an angle and its two included sides are given, the area is <math>\tfrac12 a b \sin(C)</math> where {{math|''C''}} is the given angle and {{math|''a''}} and {{math|''b''}} are its included sides.<ref name=AF/> If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of <math>\tfrac12(x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3)</math>. This formula is also known as the [[shoelace formula]] and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points ''(x1,y1)'', ''(x2,y2)'', and ''(x3,y3)''. The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use [[calculus]] to find the area.
* A [[simple polygon]] constructed on a grid of equal-distanced points (i.e., points with [[integer]] coordinates) such that all the polygon's vertices are grid points: <math>i + \frac{b}{2} - 1</math>, where ''i'' is the number of grid points inside the polygon and ''b'' is the number of boundary points. This result is known as [[Pick's theorem]].<ref name="Pick">{{cite journal|last=Trainin|first=J.|date=November 2007|title=An elementary proof of Pick's theorem|journal=[[Mathematical Gazette]]|volume=91|issue=522|pages=536–540|doi=10.1017/S0025557200182270|s2cid=124831432}}</ref>

====Area in calculus====
[[File:Integral as region under curve.svg|thumb|alt=A diagram showing the area between a given curve and the x-axis|Integration can be thought of as measuring the area under a curve, defined by ''f''(''x''), between two points (here ''a'' and ''b'').]]
[[File:Areabetweentwographs.svg|thumb|alt=A diagram showing the area between two functions|The area between two graphs can be evaluated by calculating the difference between the integrals of the two functions]]
* The area between a positive-valued curve and the horizontal axis, measured between two values ''a'' and ''b'' (b is defined as the larger of the two values) on the horizontal axis, is given by the integral from ''a'' to ''b'' of the function that represents the curve:<ref name=MathWorld/>
:<math> A = \int_a^{b} f(x) \, dx.</math>
* The area between the [[graph of a function|graphs]] of two functions is [[equality (mathematics)|equal]] to the [[integral]] of one [[function (mathematics)|function]], ''f''(''x''), [[subtraction|minus]] the integral of the other function, ''g''(''x''):
:<math> A = \int_a^{b} ( f(x) - g(x) ) \, dx, </math> where <math> f(x) </math> is the curve with the greater y-value.
* An area bounded by a function <math>r = r(\theta)</math> expressed in [[polar coordinates]] is:<ref name=MathWorld/>
:<math>A = {1 \over 2} \int r^2 \, d\theta. </math>
* The area enclosed by a [[parametric curve]] <math>\vec u(t) = (x(t), y(t)) </math> with endpoints <math> \vec u(t_0) = \vec u(t_1) </math> is given by the [[line integral]]s:
::<math> \oint_{t_0}^{t_1} x \dot y \, dt = - \oint_{t_0}^{t_1} y \dot x \, dt = {1 \over 2} \oint_{t_0}^{t_1} (x \dot y - y \dot x) \, dt </math>
: or the ''z''-component of
::<math>{1 \over 2} \oint_{t_0}^{t_1} \vec u \times \dot{\vec u} \, dt.</math>
:(For details, see {{slink|Green's theorem|Area calculation}}.) This is the principle of the [[planimeter]] mechanical device.

====Bounded area between two quadratic functions====
To find the bounded area between two [[quadratic function]]s, we subtract one from the other to write the difference as
:<math>f(x)-g(x)=ax^2+bx+c=a(x-\alpha)(x-\beta)</math>
where ''f''(''x'') is the quadratic upper bound and ''g''(''x'') is the quadratic lower bound. Define the [[discriminant]] of ''f''(''x'')-''g''(''x'') as
:<math>\Delta=b^2-4ac.</math>
By simplifying the integral formula between the graphs of two functions (as given in the section above) and using [[Vieta's formulas|Vieta's formula]], we can obtain<ref>{{cite book|title=Matematika|url=https://books.google.com/books?id=NFkVfrZBqpUC&pg=PA51|publisher=PT Grafindo Media Pratama|isbn=978-979-758-477-1|pages=51–|url-status=live|archive-url=https://web.archive.org/web/20170320100900/https://books.google.com/books?id=NFkVfrZBqpUC&pg=PA51|archive-date=2017-03-20}}</ref><ref>{{cite book|title=Get Success UN +SPMB Matematika|url=https://books.google.com/books?id=uwqvITs8OaUC&pg=PA157|publisher=PT Grafindo Media Pratama|isbn=978-602-00-0090-9|pages=157–|url-status=live|archive-url=https://web.archive.org/web/20161223115304/https://books.google.com/books?id=uwqvITs8OaUC&pg=PA157|archive-date=2016-12-23}}</ref>
:<math>A=\frac{\Delta\sqrt{\Delta}}{6a^2}=\frac{a}{6}(\beta-\alpha)^3,\qquad a\neq0.</math>
The above remains valid if one of the bounding functions is linear instead of quadratic.

====Surface area of 3-dimensional figures====
* [[Cone]]:<ref name=MathWorldCone>{{cite web|url=http://mathworld.wolfram.com/Cone.html|title=Cone|publisher=[[Wolfram MathWorld]]|author-link=Eric W. Weisstein|author=Weisstein, Eric W.|access-date=6 July 2012|url-status=live|archive-url=https://web.archive.org/web/20120621230050/http://mathworld.wolfram.com/Cone.html|archive-date=21 June 2012}}</ref> <math>\pi r\left(r + \sqrt{r^2 + h^2}\right)</math>, where ''r'' is the radius of the circular base, and ''h'' is the height. That can also be rewritten as <math>\pi r^2 + \pi r l </math><ref name=MathWorldCone/> or <math>\pi r (r + l) \,\!</math> where ''r'' is the radius and ''l'' is the slant height of the cone. <math>\pi r^2 </math> is the base area while <math>\pi r l </math> is the lateral surface area of the cone.<ref name=MathWorldCone/>
* [[cube]]: <math>6s^2</math>, where ''s'' is the length of an edge.<ref name=MathWorldSurfaceArea/>
* [[cylinder]]: <math>2\pi r(r + h)</math>, where ''r'' is the radius of a base and ''h'' is the height. The ''2<math>\pi</math>r'' can also be rewritten as ''<math>\pi</math> d'', where ''d'' is the diameter.
* [[Prism (geometry)|prism]]: 2B + Ph, where ''B'' is the area of a base, ''P'' is the perimeter of a base, and ''h'' is the height of the prism.
* [[pyramid (geometry)|pyramid]]: <math>B + \frac{PL}{2}</math>, where ''B'' is the area of the base, ''P'' is the perimeter of the base, and ''L'' is the length of the slant.
* [[rectangular prism]]: <math>2 (\ell w + \ell h + w h)</math>, where <math>\ell</math> is the length, ''w'' is the width, and ''h'' is the height.

====General formula for surface area====
The general formula for the surface area of the graph of a continuously differentiable function <math>z=f(x,y),</math> where <math>(x,y)\in D\subset\mathbb{R}^2</math> and <math>D</math> is a region in the xy-plane with the smooth boundary:
: <math> A=\iint_D\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}\,dx\,dy. </math>
An even more general formula for the area of the graph of a [[parametric surface]] in the vector form <math>\mathbf{r}=\mathbf{r}(u,v),</math> where <math>\mathbf{r}</math> is a continuously differentiable vector function of <math>(u,v)\in D\subset\mathbb{R}^2</math> is:<ref name="doCarmo"/>
: <math> A=\iint_D \left|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right|\,du\,dv. </math>

===List of formulas===

{| class="wikitable"
|+ Additional common formulas for area:
! Shape
! Formula
! Variables
|-
|[[Rectangle]]
|<math>A=ab</math>
|[[File:Rechteck-ab.svg|120px]]
|-
|[[Triangle]]
|<math>A=\frac12bh \,\!</math>
|[[File:Dreieck-allg-bh.svg|100px]]
|-
|[[Triangle]]
|<math>A=\frac12 a b \sin(\gamma)\,\!</math>
|[[File:Dreieck-allg-w.svg|100px]]
|-
|[[Triangle]] 
([[Heron's formula]])
|<math>A=\sqrt{s(s-a)(s-b)(s-c)}\,\!</math>
|[[File:Dreieck-allg.svg|100px]] <math> s =\tfrac 1 2 (a+b+c)</math>
|-
|[[Isosceles triangle]]
|<math>A=\frac{b}{4}\sqrt{4a^2-c^2}</math>
|[[File:Dreieck-gsch.svg|80px]]
|-
|Regular [[triangle]] 
([[equilateral triangle]])
||<math>A=\frac{\sqrt{3}}{4}a^2\,\!</math>
||[[File:Dreieck-gseit.svg|100px]]
|-
|[[Rhombus]]/[[Kite (geometry)|Kite]]
|<math>A=\frac12de</math>
|[[File:Raute-de.svg|160px]]
|-
|[[Parallelogram]]
|<math>A=ah_a\,\!</math>
|[[File:Parallelog-aha.svg|160px]]
|-
|[[Trapezoid]]
|<math>A=\frac{(a+c)h}{2} \,\!</math>
|[[File:Trapez-abcdh.svg|150px]]
|-
|Regular [[hexagon]]
|<math>A=\frac{3}{2} \sqrt{3}a^2\,\!</math>
|[[File:Hexagon-a.svg|100px]]
|-
|Regular [[octagon]]
|<math>A=2(1+\sqrt{2})a^2\,\!</math>
|[[File:Oktagon-a.svg|120px]]
|-
|[[Regular polygon]] 
(<math>n</math> sides)
|<math>A=n\frac{ar}{2}=\frac{pr}{2}</math> 
<math>\quad =\tfrac 1 4 na^2\cot(\tfrac \pi n)</math> 
<math>\quad = nr^2 \tan(\tfrac \pi n)</math> 
<math>\quad =\tfrac{1}{4n}p^2\cot(\tfrac \pi n)</math> 
<math>\quad =\tfrac{1}{2}nR^2 \sin(\tfrac{2\pi}{n}) \,\!</math>
|[[File:Oktagon-a-r-R.svg|150px|left]]
<math>p=na\ </math> ([[perimeter]]) 
<math>r=\tfrac a 2 \cot(\tfrac \pi n),</math> 
<math>\tfrac a 2= r\tan(\tfrac \pi n)=R\sin(\tfrac \pi n)</math> 
<math>r:</math> [[incircle]] radius 
<math>R:</math> [[circumcircle]] radius
|-
|[[Circle]]
|<math>A=\pi r^2=\frac{\pi d^2}{4}</math> 
(<math> d=2r: </math> [[diameter]])
|[[fILE:Kreis-r-tab.svg|100px]]
|-
|[[Circular sector]]
|<math>A=\frac{\theta}{2}r^2=\frac{L \cdot r}{2}\,\!</math>
|[[File:Circle arc.svg|120px]]
|-
|[[Ellipse]]
|<math>A=\pi ab \,\!</math>
|[[File:Ellipse-ab-tab.svg|120px]]
|-
|[[Integral]]
| <math>A=\int_a^b f(x)\mathrm{d}x ,\ f(x)\ge 0</math>
| [[File:Vase-f-fx-tab.svg|hochkant=0.2]]
|-
|
|'''[[Surface area]]'''
|
|-
| [[sphere (geometry)|Sphere]] 
| <math>A = 4\pi r^2 = \pi d^2</math>
| [[File:Kugel-1-tab.svg|100px]]
|-
| [[Cuboid]]
| <math>A = 2(ab+ac+bc)</math>
| [[File:Quader-1-tab.svg|150px]]
|-
| [[Cylinder (geometry)|cylinder]] 
(incl. bottom and top)
| <math>A = 2 \pi r (r + h)</math>
| [[File:Zylinder-1-tab.svg|120px]]
|-
| [[Cone]] 
(incl. bottom)
| <math>A = \pi r (r + \sqrt{r^2+h^2})</math>
| [[File:Kegel-1-tab.svg|120px]]
|-
| [[Torus]]
| <math>A = 4\pi^2 \cdot R \cdot r</math>
| [[File:Torus-1-tab.svg|200px]]
|-
| [[Surface of revolution]]
| <math>A = 2\pi\int_a^b\! f(x)\sqrt{1+\left[f'(x)\right]^2}\mathrm{d}x</math> 
(rotation around the x-axis)
| [[File:Vase-1-tab.svg|220px]]
|-
|}

The above calculations show how to find the areas of many common [[shapes]].

The areas of irregular (and thus arbitrary) polygons can be calculated using the "[[Surveyor's formula]]" (shoelace formula).<ref name=Surveyor>{{cite journal|last1=Braden|first1=Bart|date=September 1986|title=The Surveyor's Area Formula|journal=The College Mathematics Journal|volume=17|issue=4|pages=326–337|doi=10.2307/2686282|url=http://www.maa.org/pubs/Calc_articles/ma063.pdf|access-date=15 July 2012|url-status=live|archive-url=https://web.archive.org/web/20120627180152/http://www.maa.org/pubs/Calc_articles/ma063.pdf|archive-date=27 June 2012|jstor=2686282}}</ref>

===Relation of area to perimeter===

The [[isoperimetric inequality]] states that, for a closed curve of length ''L'' (so the region it encloses has [[perimeter]] ''L'') and for area ''A'' of the region that it encloses,

:<math>4\pi A \le L^2,</math>

and equality holds if and only if the curve is a [[circle]]. Thus a circle has the largest area of any closed figure with a given perimeter.

At the other extreme, a figure with given perimeter ''L'' could have an arbitrarily small area, as illustrated by a [[rhombus]] that is "tipped over" arbitrarily far so that two of its [[angle]]s are arbitrarily close to 0° and the other two are arbitrarily close to 180°.

For a circle, the ratio of the area to the [[circumference]] (the term for the perimeter of a circle) equals half the [[radius]] ''r''. This can be seen from the area formula ''πr''2 and the circumference formula 2''πr''.

The area of a [[regular polygon]] is half its perimeter times the [[apothem]] (where the apothem is the distance from the center to the nearest point on any side).

===Fractals===

Doubling the edge lengths of a polygon multiplies its area by four, which is two (the ratio of the new to the old side length) raised to the power of two (the dimension of the space the polygon resides in). But if the one-dimensional lengths of a [[fractal]] drawn in two dimensions are all doubled, the spatial content of the fractal scales by a power of two that is not necessarily an integer. This power is called the [[fractal dimension]] of the fractal.
<ref name="Mandelbrot1983">
{{cite book
|last = Mandelbrot
|first = Benoît B.
|title = The fractal geometry of nature 
|url = https://books.google.com/books?id=JFX9mQEACAAJ
|access-date = 1 February 2012
|year = 1983
|publisher = Macmillan
|isbn = 978-0-7167-1186-5
|url-status = live
|archive-url = https://web.archive.org/web/20170320115652/https://books.google.com/books?id=JFX9mQEACAAJ
|archive-date = 20 March 2017
}}</ref>

==Area bisectors==
{{Main article|Bisection#Area bisectors and perimeter bisectors}}

There are an infinitude of lines that bisect the area of a triangle. Three of them are the [[Median (triangle)|medians]] of the triangle (which connect the sides' midpoints with the opposite vertices), and these are [[Concurrent lines|concurrent]] at the triangle's [[centroid]]; indeed, they are the only area bisectors that go through the centroid. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its [[incircle]]). There are either one, two, or three of these for any given triangle.

Any line through the midpoint of a parallelogram bisects the area.

All area bisectors of a circle or other ellipse go through the center, and any [[Chord (geometry)|chords]] through the center bisect the area. In the case of a circle they are the diameters of the circle.

==Optimization==
Given a wire contour, the surface of least area spanning ("filling") it is a [[minimal surface]]. Familiar examples include [[soap bubble]]s.

The question of the [[filling area conjecture|filling area]] of the [[Riemannian circle]] remains open.<ref>{{citation
|last = Gromov
|first = Mikhael
|issue = 1
|journal = Journal of Differential Geometry
|mr = 697984
|pages = 1–147
|title = Filling Riemannian manifolds
|url = http://projecteuclid.org/euclid.jdg/1214509283
|volume = 18
|year = 1983
|doi = 10.4310/jdg/1214509283
|citeseerx = 10.1.1.400.9154
|url-status = live
|archive-url = https://web.archive.org/web/20140408110006/http://projecteuclid.org/euclid.jdg/1214509283
|archive-date = 2014-04-08
}}</ref>

The circle has the largest area of any two-dimensional object having the same perimeter.

A [[cyclic polygon]] (one inscribed in a circle) has the largest area of any polygon with a given number of sides of the same lengths.

A version of the [[isoperimetric inequality]] for triangles states that the triangle of greatest area among all those with a given perimeter is [[equilateral]].<ref name=Chakerian/>

The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral.<ref>Dorrie, Heinrich (1965), ''100 Great Problems of Elementary Mathematics'', Dover Publ., pp. 379–380.</ref>

The ratio of the area of the incircle to the area of an equilateral triangle, <math>\frac{\pi}{3\sqrt{3}}</math>, is larger than that of any non-equilateral triangle.<ref>{{cite journal|author1=Minda, D.|author2=Phelps, S.|title=Triangles, ellipses, and cubic polynomials|journal=[[American Mathematical Monthly]]|volume=115|issue=8|date=October 2008|pages=679–689: Theorem 4.1|doi=10.1080/00029890.2008.11920581|url=https://www.researchgate.net/publication/228698127|jstor=27642581|s2cid=15049234|url-status=live|archive-url=https://web.archive.org/web/20161104141707/https://www.researchgate.net/publication/228698127_Triangles_ellipses_and_cubic_polynomials|archive-date=2016-11-04}}</ref>

The ratio of the area to the square of the perimeter of an equilateral triangle, <math>\frac{1}{12\sqrt{3}},</math> is larger than that for any other triangle.

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Perimeter Perimeter, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Area Area, Wikipedia] under a CC BY-SA license

@@ Line 95: / Line 95: @@
 The most basic area formula is the formula for the area of a rectangle. Given a rectangle with length {{mvar|l}} and width {{mvar|w}}, the formula for the area is:
-:{{bigmath|''A'' {{=}} ''lw''}} &nbsp;(rectangle).
+:''A'' = ''lw'' &nbsp;(rectangle).
-That is, the area of the rectangle is the length multiplied by the width.  As a special case, as {{math|''l'' {{=}} ''w''}} in the case of a square, the area of a square with side length {{mvar|s}} is given by the formula:
+That is, the area of the rectangle is the length multiplied by the width.  As a special case, as ''l'' = ''w'' in the case of a square, the area of a square with side length {{mvar|s}} is given by the formula:
-:{{bigmath|''A'' {{=}} ''s''<sup>2</sup>}} &nbsp;(square).
+:''A'' = ''s''<sup>2</sup> &nbsp;(square).
 The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as a definition or axiom. On the other hand, if geometry is developed before arithmetic, this formula can be used to define multiplication of real numbers.
@@ Line 106: / Line 106: @@
 This involves cutting a shape into pieces, whose areas must sum to the area of the original shape.
 [[File:ParallelogramArea.svg|thumb|left|upright|A diagram showing how a parallelogram can be re-arranged into the shape of a rectangle.]]
-For an example, any [[parallelogram]] can be subdivided into a [[trapezoid]] and a [[right triangle]], as shown in figure to the left.  If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle.  It follows that the area of the parallelogram is the same as the area of the rectangle:<ref name=AF/>
+For an example, any parallelogram can be subdivided into a trapezoid and a right triangle, as shown in figure to the left.  If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle.  It follows that the area of the parallelogram is the same as the area of the rectangle:
-:{{bigmath|''A'' {{=}} ''bh''}} &nbsp;(parallelogram).
+:''A'' = ''bh'' &nbsp;(parallelogram).
 [[File:TriangleArea.svg|thumb|right|upright|A parallelogram split into two equal triangles.]]
-However, the same parallelogram can also be cut along a [[diagonal]] into two [[congruence (geometry)|congruent]] triangles, as shown in the figure to the right.  It follows that the area of each [[triangle]] is half the area of the parallelogram:<ref name=AF/>
+However, the same parallelogram can also be cut along a diagonal into two congruent triangles, as shown in the figure to the right.  It follows that the area of each triangle is half the area of the parallelogram:
 :<math>A = \frac{1}{2}bh</math> &nbsp;(triangle).
-Similar arguments can be used to find area formulas for the [[trapezoid]]<ref>{{citation|title=Problem Solving Through Recreational Mathematics|first1=Bonnie|last1=Averbach|first2=Orin|last2=Chein|publisher=Dover|year=2012|isbn=978-0-486-13174-0|page=306|url=https://books.google.com/books?id=Dz_CAgAAQBAJ&pg=PA306|url-status=live|archive-url=https://web.archive.org/web/20160513101526/https://books.google.com/books?id=Dz_CAgAAQBAJ&pg=PA306|archive-date=2016-05-13}}</ref> as well as more complicated [[polygon]]s.
+Similar arguments can be used to find area formulas for the trapezoid as well as more complicated polygons.
 ===Area of curved shapes===
@@ Line 117: / Line 117: @@
 ====Circles====
 [[File:CircleArea.svg|thumb|right|alt=A circle divided into many sectors can be re-arranged roughly to form a parallelogram|A circle can be divided into sectors which rearrange to form an approximate parallelogram.]]
-{{main article|Area of a circle}}
 The formula for the area of a circle (more properly called the area enclosed by a circle or the area of a disk) is based on a similar method.  Given a circle of radius {{math|''r''}}, it is possible to partition the circle into sectors, as shown in the figure to the right.  Each sector is approximately triangular in shape, and the sectors can be rearranged to form an approximate parallelogram.  The height of this parallelogram is {{math|''r''}}, and the width is half the circumference of the circle, or {{math|π''r''}}.  Thus, the total area of the circle is {{math|π''r''<sup>2</sup>}}:
-:{{bigmath|''A'' {{=}} π''r''<sup>2</sup>}} &nbsp;(circle).
+:''A'' = π''r''<sup>2</sup> &nbsp;(circle).
 Though the dissection used in this formula is only approximate, the error becomes smaller and smaller as the circle is partitioned into more and more sectors.  The limit of the areas of the approximate parallelograms is exactly {{math|π''r''<sup>2</sup>}}, which is the area of the circle.
@@ Line 126: / Line 126: @@
 ====Ellipses====
-The formula for the area enclosed by an [[ellipse]] is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes {{math|''x''}} and {{math|''y''}} the formula is:<ref name=AF/>
+The formula for the area enclosed by an ellipse is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes {{math|''x''}} and {{math|''y''}} the formula is:
 :<math>A = \pi xy .</math>
@@ Line 135: / Line 135: @@
 The formula for the surface area of a sphere is more difficult to derive: because a sphere has nonzero Gaussian curvature, it cannot be flattened out.  The formula for the surface area of a sphere was first obtained by Archimedes in his work ''On the Sphere and Cylinder''.  The formula is:
-:{{math|''A'' {{=}} 4''πr''<sup>2</sup>}} &nbsp;(sphere),
+:''A'' = 4''πr''<sup>2</sup> &nbsp;(sphere),
 where {{math|''r''}} is the radius of the sphere.  As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar to calculus.
@@ Line 142: / Line 142: @@
 ====Areas of 2-dimensional figures====
 [[File:Triangle_GeometryArea.svg|thumb|Triangle area <math>A=\tfrac{b\cdot h}{2}</math>]]
-* A [[triangle]]: <math>\tfrac12Bh</math> (where ''B'' is any side, and ''h'' is the distance from the line on which ''B'' lies to the other vertex of the triangle). This formula can be used if the height ''h'' is known. If the lengths of the three sides are known then ''Heron's formula'' can be used: <math>\sqrt{s(s-a)(s-b)(s-c)}</math> where ''a'', ''b'', ''c'' are the sides of the triangle, and <math>s = \tfrac12(a + b + c)</math> is half of its perimeter. If an angle and its two included sides are given, the area is <math>\tfrac12 a b \sin(C)</math> where {{math|''C''}} is the given angle and {{math|''a''}} and {{math|''b''}} are its included sides.<ref name=AF/> If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of <math>\tfrac12(x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3)</math>. This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points ''(x<sub>1</sub>,y<sub>1</sub>)'', ''(x<sub>2</sub>,y<sub>2</sub>)'', and ''(x<sub>3</sub>,y<sub>3</sub>)''. The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use calculus to find the area.
+* A triangle: <math>\tfrac12Bh</math> (where ''B'' is any side, and ''h'' is the distance from the line on which ''B'' lies to the other vertex of the triangle). This formula can be used if the height ''h'' is known. If the lengths of the three sides are known then ''Heron's formula'' can be used: <math>\sqrt{s(s-a)(s-b)(s-c)}</math> where ''a'', ''b'', ''c'' are the sides of the triangle, and <math>s = \tfrac12(a + b + c)</math> is half of its perimeter. If an angle and its two included sides are given, the area is <math>\tfrac12 a b \sin(C)</math> where {{math|''C''}} is the given angle and {{math|''a''}} and {{math|''b''}} are its included sides. If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of <math>\tfrac12(x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3)</math>. This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points ''(x<sub>1</sub>,y<sub>1</sub>)'', ''(x<sub>2</sub>,y<sub>2</sub>)'', and ''(x<sub>3</sub>,y<sub>3</sub>)''. The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use calculus to find the area.
 * A simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points: <math>i + \frac{b}{2} - 1</math>, where ''i'' is the number of grid points inside the polygon and ''b'' is the number of boundary points. This result is known as Pick's theorem.
 ====Surface area of 3-dimensional figures====
-* Cone: <math>\pi r\left(r + \sqrt{r^2 + h^2}\right)</math>, where ''r'' is the radius of the circular base, and ''h'' is the height. That can also be rewritten as <math>\pi r^2 + \pi r l </math><ref name=MathWorldCone/> or <math>\pi r (r + l) \,\!</math> where ''r'' is the radius and ''l'' is the slant height of the cone. <math>\pi r^2 </math> is the base area while <math>\pi r l </math> is the lateral surface area of the cone.
+* Cone: <math>\pi r\left(r + \sqrt{r^2 + h^2}\right)</math>, where ''r'' is the radius of the circular base, and ''h'' is the height. That can also be rewritten as <math>\pi r^2 + \pi r l </math> or <math>\pi r (r + l) \,\!</math> where ''r'' is the radius and ''l'' is the slant height of the cone. <math>\pi r^2 </math> is the base area while <math>\pi r l </math> is the lateral surface area of the cone.
 * Cube: <math>6s^2</math>, where ''s'' is the length of an edge.
 * Cylinder: <math>2\pi r(r + h)</math>, where ''r'' is the radius of a base and ''h'' is the height. The ''2<math>\pi</math>r'' can also be rewritten as ''<math>\pi</math> d'', where ''d'' is the diameter.
@@ Line 156: / Line 156: @@
 The general formula for the surface area of the graph of a continuously differentiable function <math>z=f(x,y),</math> where <math>(x,y)\in D\subset\mathbb{R}^2</math> and <math>D</math> is a region in the xy-plane with the smooth boundary:
 : <math> A=\iint_D\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}\,dx\,dy. </math>
-An even more general formula for the area of the graph of a parametric surface in the vector form <math>\mathbf{r}=\mathbf{r}(u,v),</math> where <math>\mathbf{r}</math> is a continuously differentiable vector function of <math>(u,v)\in D\subset\mathbb{R}^2</math> is:<ref name="doCarmo"/>
+An even more general formula for the area of the graph of a parametric surface in the vector form <math>\mathbf{r}=\mathbf{r}(u,v),</math> where <math>\mathbf{r}</math> is a continuously differentiable vector function of <math>(u,v)\in D\subset\mathbb{R}^2</math> is:
 : <math> A=\iint_D \left|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right|\,du\,dv. </math>
@@ Line 167: / Line 167: @@
 ! Variables
 |-
-|[[Rectangle]]
+|Rectangle
 |<math>A=ab</math>
 |[[File:Rechteck-ab.svg|120px]]
@@ Line 221: / Line 221: @@
 <math>\quad =\tfrac{1}{2}nR^2 \sin(\tfrac{2\pi}{n}) \,\!</math>
 |[[File:Oktagon-a-r-R.svg|150px|left]]
-<math>p=na\ </math> ([[perimeter]])<br>
+<math>p=na\ </math> (perimeter)<br>
 <math>r=\tfrac a 2 \cot(\tfrac \pi n),</math><br>
 <math>\tfrac a 2= r\tan(\tfrac \pi n)=R\sin(\tfrac \pi n)</math><br>
-<math>r:</math> [[incircle]] radius<br>
+<math>r:</math> incircle radius<br>
-<math>R:</math> [[circumcircle]] radius
+<math>R:</math> circumcircle radius
 |-
 |Circle

← Older revision		Revision as of 03:34, 16 December 2021
Line 300:		Line 300:

	==Area bisectors==		==Area bisectors==
−	~~{{Main article\|Bisection#Area bisectors and perimeter bisectors}}~~

	There are an infinitude of lines that bisect the area of a triangle. Three of them are the medians of the triangle (which connect the sides' midpoints with the opposite vertices), and these are concurrent at the triangle's centroid; indeed, they are the only area bisectors that go through the centroid. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.		There are an infinitude of lines that bisect the area of a triangle. Three of them are the medians of the triangle (which connect the sides' midpoints with the opposite vertices), and these are concurrent at the triangle's centroid; indeed, they are the only area bisectors that go through the centroid. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.

@@ Line 1: / Line 1: @@
 = Area =
@@ Line 323: / Line 404: @@
 The ratio of the area to the square of the perimeter of an equilateral triangle, <math>\frac{1}{12\sqrt{3}},</math> is larger than that for any other triangle.
-== Licensing ==
+= Licensing =
 Content obtained and/or adapted from:
 * [https://en.wikipedia.org/wiki/Perimeter Perimeter, Wikipedia] under a CC BY-SA license
 * [https://en.wikipedia.org/wiki/Area Area, Wikipedia] under a CC BY-SA license