Area

This article is about the geometric quantity. For other uses, seeArea (disambiguation).

Areais themeasureof aregion's size on asurface.The area of a plane region orplane arearefers to the area of ashapeorplanar lamina,whilesurface arearefers to the area of anopen surfaceor theboundaryof athree-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 ofpaintnecessary to cover the surface with a single coat.It is the two-dimensional analogue of thelengthof acurve(a one-dimensional concept) or thevolumeof a solid (a three-dimensional concept). Two different regions may have the same area (as insquaring the circle); bysynecdoche,"area" sometimes is used to refer to the region, as in a "polygonal area".

Area
The areas of this square and thisdiskare the same.
Common symbols	AorS
SI unit	Square metre[m2]
InSI base units	1m2
Dimension	${\mathsf {L}}^{2}$

The area of a shape can be measured by comparing the shape tosquaresof a fixed size.In theInternational System of Units(SI), the standard unit of area is thesquare metre(written as m²), which is the area of a square whose sides are onemetrelong.A shape with an area of three square metres would have the same area as three such squares. Inmathematics,theunit squareis defined to have area one, and the area of any other shape or surface is adimensionlessreal number.

There are several well-knownformulasfor the areas of simple shapes such astriangles,rectangles,andcircles.Using these formulas, the area of anypolygoncan be found bydividing the polygon into triangles.For shapes with curved boundary,calculusis usually required to compute the area. Indeed, the problem of determining the area of plane figures was a major motivation for thehistorical development of calculus.

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

Area plays an important role in modern mathematics. In addition to its obvious importance ingeometryand calculus, area is related to the definition ofdeterminantsinlinear algebra,and is a basic property of surfaces indifferential geometry.Inanalysis,the area of a subset of the plane is defined usingLebesgue measure,though not every subset is measurable if one supposes the axiom of choice.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

See also:Jordan measure

An approach to defining what is meant by "area" is throughaxioms."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 allSinM,a(S) ≥ 0.
• IfSandTare inMthen so areS∪TandS∩T,and alsoa(S∪T) =a(S) +a(T) −a(S∩T).
• IfSandTare inMwithS⊆TthenT−Sis inManda(T−S) =a(T) −a(S).
• If a setSis inMandSis congruent toTthenTis also inManda(S) =a(T).
• Every rectangleRis inM.If the rectangle has lengthhand breadthkthena(R) =hk.
• LetQbe a set enclosed between two step regionsSandT.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 numbercsuch thata(S) ≤ c ≤a(T)for all such step regionsSandT,thena(Q) =c.

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

Units

A square metrequadratmade of PVC pipe

Everyunit of lengthhas a corresponding unit of area, namely the area of a square with the given side length. Thus areas can be measured insquare 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 thesquaresof the corresponding length units.

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

Conversions

Although there are 10 mm in 1 cm, there are 100 mm²in 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 =1,000,000square metres
• 1 square metre =10,000square centimetres = 1,000,000 square millimetres
• 1 square centimetre =100square millimetres.

Non-metric units

In non-metric units, the conversion between two square units is thesquareof the conversion between the corresponding length units.

1foot= 12inches,

the relationship between square feet and square inches is

1 square foot = 144 square inches,

where 144 = 12²= 12 × 12. Similarly:

• 1 square yard =9square 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 =0.09290304square metres
• 1 square yard =0.83612736square metres
• 1 square mile =2.589988110336square kilometres

Other units including historical

See also:Category:Units of area

There are several other common units for area. Thearewas the original unit of area in themetric system,with:

• 1 are = 100 square metres

Though the are has fallen out of use, thehectareis 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 thetetrad,thehectad,and themyriad.

Theacreis 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 ofbarns,such that:

• 1 barn = 10⁻²⁸square meters.

The barn is commonly used in describing the cross-sectional area of interaction innuclear physics.

InSouth Asia(mainly Indians), although the countries use SI units as official, many South Asians still use traditional units. Each administrative division has its own area unit, some of them have same names, but with different values. There's no official consensus about the traditional units values. Thus, the conversions between the SI units and the traditional units may have different results, depending on what reference that has been used.

Some traditional South Asian units that have fixed value:

• 1 Killa = 1 acre
• 1 Ghumaon = 1 acre
• 1 Kanal = 0.125 acre (1 acre = 8 kanal)
• 1 Decimal = 48.4 square yards
• 1 Chatak = 180 square feet

History

Circle area

Main article:Area of a circle § History

In the 5th century BCE,Hippocrates of Chioswas the first to show that the area of a disk (the region enclosed by a circle) is proportional to the square of its diameter, as part of hisquadratureof thelune of Hippocrates,but did not identify theconstant of proportionality.Eudoxus of Cnidus,also in the 5th century BCE, also found that the area of a disk is proportional to its radius squared.

Subsequently, Book I ofEuclid'sElementsdealt with equality of areas between two-dimensional figures. The mathematicianArchimedesused the tools ofEuclidean geometryto show that the area inside a circle is equal to that of aright trianglewhose base has the length of the circle's circumference and whose height equals the circle's radius, in his bookMeasurement of a Circle.(The circumference is 2πr,and the area of a triangle is half the base times the height, yielding the areaπr²for the disk.) Archimedes approximated the value of π (and hence the area of a unit-radius circle) withhis doubling method,in which he inscribed a regular triangle in a circle and noted its area, then doubled the number of sides to give a regularhexagon,then repeatedly doubled the number of sides as the polygon's area got closer and closer to that of the circle (and did the same withcircumscribed polygons).

Triangle area

This section is an excerpt fromArea of a triangle § History.[edit]

Heron of Alexandriafound what is known asHeron's formulafor the area of a triangle in terms of its sides, and a proof can be found in his book,Metrica,written around 60 CE. It has been suggested thatArchimedesknew the formula over two centuries earlier,and sinceMetricais a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.In 300 BCE Greek mathematicianEuclidproved that the area of a triangle is half that of a parallelogram with the same base and height in his bookElements of Geometry.

In 499Aryabhata,a greatmathematician-astronomerfrom the classical age ofIndian mathematicsandIndian astronomy,expressed the area of a triangle as one-half the base times the height in theAryabhatiya.

A formula equivalent to Heron's was discovered by the Chinese independently of the Greeks. It was published in 1247 inShushu Jiuzhang( "Mathematical Treatise in Nine Sections"), written byQin Jiushao.

Quadrilateral area

In the 7th century CE,Brahmaguptadeveloped a formula, now known asBrahmagupta's formula,for the area of acyclic quadrilateral(aquadrilateralinscribedin a circle) in terms of its sides. In 1842, the German mathematiciansCarl Anton BretschneiderandKarl Georg Christian von Staudtindependently found a formula, known asBretschneider's formula,for the area of any quadrilateral.

General polygon area

The development ofCartesian coordinatesbyRené Descartesin the 17th century allowed the development of thesurveyor's formulafor the area of any polygon with knownvertexlocations byGaussin the 19th century.

Areas determined using calculus

The development ofintegral calculusin the late 17th century provided tools that could subsequently be used for computing more complicated areas, such as the area of anellipseand thesurface areasof various curved three-dimensional objects.

Area formulas

Polygon formulas

Main article:Polygon § Area

For a non-self-intersecting (simple) polygon, theCartesian coordinates$(x_{i},y_{i})$(i=0, 1,...,n-1) of whosenverticesare known, the area is given by thesurveyor's formula:

$A={\frac {1}{2}}{\Biggl \vert }\sum _{i=0}^{n-1}(x_{i}y_{i+1}-x_{i+1}y_{i}){\Biggr \vert }$

where wheni=n-1, theni+1 is expressed asmodulusnand so refers to 0.

Rectangles

The area of this rectangle islw.

The most basic area formula is the formula for the area of arectangle.Given a rectangle with lengthland widthw,the formula for the area is:

A=lw(rectangle).

That is, the area of the rectangle is the length multiplied by the width. As a special case, asl=win the case of a square, the area of a square with side lengthsis given by the formula:

A=s²(square).

The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as adefinitionoraxiom.On the other hand, ifgeometryis developed beforearithmetic,this formula can be used to definemultiplicationofreal numbers.

Dissection, parallelograms, and triangles

Main articles:Triangle areaandParallelogram § Area formula

A parallelogram can be cut up and re-arranged to form a rectangle.

Most other simple formulas for area follow from the method ofdissection. This involves cutting a shape into pieces, whose areas mustsumto the area of the original shape. For an example, anyparallelogramcan be subdivided into atrapezoidand aright 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:

A=bh(parallelogram).

A parallelogram split into two equal triangles

However, the same parallelogram can also be cut along adiagonalinto twocongruenttriangles, as shown in the figure to the right. It follows that the area of eachtriangleis half the area of the parallelogram:

$A={\frac {1}{2}}bh$(triangle).

Similar arguments can be used to find area formulas for thetrapezoidas well as more complicatedpolygons.

Area of curved shapes

Circles

A circle can be divided intosectorswhich rearrange to form an approximateparallelogram.

Main article:Area of a circle

The formula for the area of acircle(more properly called the area enclosed by a circle or the area of adisk) is based on a similar method. Given a circle of radiusr,it is possible to partition the circle intosectors,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 isr,and the width is half thecircumferenceof the circle, orπr.Thus, the total area of the circle isπr²:

A= πr²(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. Thelimitof the areas of the approximate parallelograms is exactlyπr²,which is the area of the circle.

This argument is actually a simple application of the ideas ofcalculus.In ancient times, themethod of exhaustionwas used in a similar way to find the area of the circle, and this method is now recognized as a precursor tointegral calculus.Using modern methods, the area of a circle can be computed using adefinite integral:

$A\;=\;2\int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\,dx\;=\;\pi r^{2}.$

Ellipses

Main article:Ellipse § Area

The formula for the area enclosed by anellipseis related to the formula of a circle; for an ellipse withsemi-majorandsemi-minoraxesxandythe formula is:

$A=\pi xy.$

Non-planar surface area

Main article:Surface area

Archimedesshowed that the surface area of asphereis exactly four times the area of a flatdiskof the same radius, and the volume enclosed by the sphere is exactly 2/3 of the volume of acylinderof the same height and radius.

Most basic formulas forsurface areacan be obtained by cutting surfaces and flattening them out (see:developable surfaces). For example, if the side surface of acylinder(or anyprism) is cut lengthwise, the surface can be flattened out into a rectangle. Similarly, if a cut is made along the side of acone,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 asphereis more difficult to derive: because a sphere has nonzeroGaussian curvature,it cannot be flattened out. The formula for the surface area of a sphere was first obtained byArchimedesin his workOn the Sphere and Cylinder.The formula is:

A= 4πr²(sphere),

whereris the radius of the sphere. As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar tocalculus.

General formulas

Areas of 2-dimensional figures

Triangle area$A={\tfrac {b\cdot h}{2}}$

• Atriangle:${\tfrac {1}{2}}Bh$(whereBis any side, andhis the distance from the line on whichBlies to the other vertex of the triangle). This formula can be used if the heighthis known. If the lengths of the three sides are known thenHeron's formulacan be used:${\sqrt {s(s-a)(s-b)(s-c)}}$wherea,b,care the sides of the triangle, and$s={\tfrac {1}{2}}(a+b+c)$is half of its perimeter.If an angle and its two included sides are given, the area is${\tfrac {1}{2}}ab\sin(C)$whereCis the given angle andaandbare 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${\tfrac {1}{2}}(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3})$.This formula is also known as theshoelace formulaand 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 usecalculusto find the area.
• Asimple polygonconstructed on a grid of equal-distanced points (i.e., points withintegercoordinates) such that all the polygon's vertices are grid points:$i+{\frac {b}{2}}-1$,whereiis the number of grid points inside the polygon andbis the number of boundary points. This result is known asPick's theorem.

Area in calculus

Integration can be thought of as measuring the area under a curve, defined byf(x), between two points (hereaandb).

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 valuesaandb(b is defined as the larger of the two values) on the horizontal axis, is given by the integral fromatobof the function that represents the curve:

$A=\int _{a}^{b}f(x)\,dx.$

• The area between thegraphsof two functions isequalto theintegralof onefunction,f(x),minusthe integral of the other function,g(x):

$A=\int _{a}^{b}(f(x)-g(x))\,dx,$where$f(x)$is the curve with the greater y-value.

• An area bounded by a function$r=r(\theta )$expressed inpolar coordinatesis:

$A={1 \over 2}\int r^{2}\,d\theta.$

• The area enclosed by aparametric curve${\vec {u}}(t)=(x(t),y(t))$with endpoints${\vec {u}}(t_{0})={\vec {u}}(t_{1})$is given by theline integrals:

$\oint _{t_{0}}^{t_{1}}x{\dot {y}}\,dt=-\oint _{t_{0}}^{t_{1}}y{\dot {x}}\,dt={1 \over 2}\oint _{t_{0}}^{t_{1}}(x{\dot {y}}-y{\dot {x}})\,dt$

or thez-component of

${1 \over 2}\oint _{t_{0}}^{t_{1}}{\vec {u}}\times {\dot {\vec {u}}}\,dt.$

(For details, seeGreen's theorem § Area calculation.) This is the principle of theplanimetermechanical device.

Bounded area between two quadratic functions

To find the bounded area between twoquadratic functions,we first subtract one from the other, writing the difference as $$f(x)-g(x)=ax^{2}+bx+c=a(x-\alpha )(x-\beta )$$ wheref(x) is the quadratic upper bound andg(x) is the quadratic lower bound. By the area integral formulas above andVieta's formula,we can obtain that $$A={\frac {(b^{2}-4ac)^{3/2}}{6a^{2}}}={\frac {a}{6}}(\beta -\alpha )^{3},\qquad a\neq 0.$$ The above remains valid if one of the bounding functions is linear instead of quadratic.

Surface area of 3-dimensional figures

• Cone:$\pi r\left(r+{\sqrt {r^{2}+h^{2}}}\right)$,whereris the radius of the circular base, andhis the height. That can also be rewritten as$\pi r^{2}+\pi rl$or$\pi r(r+l)\,\!$whereris the radius andlis the slant height of the cone.$\pi r^{2}$is the base area while$\pi rl$is the lateral surface area of the cone.
• Cube:$6s^{2}$,wheresis the length of an edge.
• Cylinder:$2\pi r(r+h)$,whereris the radius of a base andhis the height. The$2\pi r$can also be rewritten as$\pi d$,wheredis the diameter.
• Prism:$2B+Ph$,whereBis the area of a base,Pis the perimeter of a base, andhis the height of the prism.
• pyramid:$B+{\frac {PL}{2}}$,whereBis the area of the base,Pis the perimeter of the base, andLis the length of the slant.
• Rectangular prism:$2(\ell w+\ell h+wh)$,where$\ell$is the length,wis the width, andhis the height.

General formula for surface area

The general formula for the surface area of the graph of a continuously differentiable function$z=f(x,y),$where$(x,y)\in D\subset \mathbb {R} ^{2}$and$D$is a region in the xy-plane with the smooth boundary:

$A=\iint _{D}{\sqrt {\left({\frac {\partial f}{\partial x}}\right)^{2}+\left({\frac {\partial f}{\partial y}}\right)^{2}+1}}\,dx\,dy.$

An even more general formula for the area of the graph of aparametric surfacein the vector form$\mathbf {r} =\mathbf {r} (u,v),$where$\mathbf {r}$is a continuously differentiable vector function of$(u,v)\in D\subset \mathbb {R} ^{2}$is:

$A=\iint _{D}\left|{\frac {\partial \mathbf {r} }{\partial u}}\times {\frac {\partial \mathbf {r} }{\partial v}}\right|\,du\,dv.$

List of formulas

Additional common formulas for area:

Shape	Formula	Variables
Square	$A=s^{2}$
Rectangle	$A=ab$
Triangle	$A={\frac {1}{2}}bh\,\!$
Triangle	$A={\frac {1}{2}}ab\sin(\gamma )\,\!$
Triangle (Heron's formula)	$A={\sqrt {s(s-a)(s-b)(s-c)}}\,\!$	$s={\tfrac {1}{2}}(a+b+c)$
Isosceles triangle	$A={\frac {c}{4}}{\sqrt {4a^{2}-c^{2}}}$
Regulartriangle (equilateral triangle)	$A={\frac {\sqrt {3}}{4}}a^{2}\,\!$
Rhombus/Kite	$A={\frac {1}{2}}de$
Parallelogram	$A=ah_{a}\,\!$
Trapezoid	$A={\frac {(a+c)h}{2}}\,\!$
Regularhexagon	$A={\frac {3}{2}}{\sqrt {3}}a^{2}\,\!$
Regularoctagon	$A=2(1+{\sqrt {2}})a^{2}\,\!$
Regular polygon ($n$sides)	$A=n{\frac {ar}{2}}={\frac {pr}{2}}$ $\quad ={\tfrac {1}{4}}na^{2}\cot({\tfrac {\pi }{n}})$ $\quad =nr^{2}\tan({\tfrac {\pi }{n}})$ $\quad ={\tfrac {1}{4n}}p^{2}\cot({\tfrac {\pi }{n}})$ $\quad ={\tfrac {1}{2}}nR^{2}\sin({\tfrac {2\pi }{n}})\,\!$	$p=na\$(perimeter) $r={\tfrac {a}{2}}\cot({\tfrac {\pi }{n}}),$ ${\tfrac {a}{2}}=r\tan({\tfrac {\pi }{n}})=R\sin({\tfrac {\pi }{n}})$ $r:$incircleradius $R:$circumcircleradius
Circle	$A=\pi r^{2}={\frac {\pi d^{2}}{4}}$ ($d=2r:$diameter)
Circular sector	$A={\frac {\theta }{2}}r^{2}={\frac {L\cdot r}{2}}\,\!$
Ellipse	$A=\pi ab\,\!$
Integral	$A=\int _{a}^{b}f(x)\mathrm {d} x,\ f(x)\geq 0$
	Surface area
Sphere	$A=4\pi r^{2}=\pi d^{2}$
Cuboid	$A=2(ab+ac+bc)$
Cylinder (incl. bottom and top)	$A=2\pi r(r+h)$
Cone (incl. bottom)	$A=\pi r(r+{\sqrt {r^{2}+h^{2}}})$
Torus	$A=4\pi ^{2}\cdot R\cdot r$
Surface of revolution	$A=2\pi \int _{a}^{b}\!f(x){\sqrt {1+\left[f'(x)\right]^{2}}}\mathrm {d} x$ (rotation around the x-axis)

The above calculations show how to find the areas of many commonshapes.

The areas of irregular (and thus arbitrary) polygons can be calculated using the "Surveyor's formula"(shoelace formula).

Relation of area to perimeter

Theisoperimetric inequalitystates that, for a closed curve of lengthL(so the region it encloses hasperimeterL) and for areaAof the region that it encloses,

$4\pi A\leq L^{2},$

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

At the other extreme, a figure with given perimeterLcould have an arbitrarily small area, as illustrated by arhombusthat is "tipped over" arbitrarily far so that two of itsanglesare arbitrarily close to 0° and the other two are arbitrarily close to 180°.

For a circle, the ratio of the area to thecircumference(the term for the perimeter of a circle) equals half theradiusr.This can be seen from the area formulaπr²and the circumference formula 2πr.

The area of aregular polygonis half its perimeter times theapothem(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 afractaldrawn 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 thefractal dimensionof the fractal.

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 themediansof the triangle (which connect the sides' midpoints with the opposite vertices), and these areconcurrentat the triangle'scentroid;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 itsincircle). 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 anychordsthrough 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 aminimal surface.Familiar examples includesoap bubbles.

The question of thefilling areaof theRiemannian circleremains open.

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

Acyclic 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 theisoperimetric inequalityfor triangles states that the triangle of greatest area among all those with a given perimeter isequilateral.

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.

The ratio of the area of the incircle to the area of an equilateral triangle,${\frac {\pi }{3{\sqrt {3}}}}$,is larger than that of any non-equilateral triangle.

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

See also

• Brahmagupta quadrilateral,a cyclic quadrilateral with integer sides, integer diagonals, and integer area.
• Equiareal map
• Heronian triangle,a triangle with integer sides and integer area.
• List of triangle inequalities
• One-seventh area triangle,an inner triangle with one-seventh the area of the reference triangle.

• Routh's theorem,a generalization of the one-seventh area triangle.

• Orders of magnitude—A list of areas by size.
• Derivation of the formula of a pentagon
• Planimeter,an instrument for measuring small areas, e.g. on maps.
• Area of a convex quadrilateral
• Robbins pentagon,a cyclic pentagon whose side lengths and area are all rational numbers.