Area - wikipedia Area This article is about the geometric quantity. For other uses, see Area ( disambiguation ). The combined area of these three shapes is approximately 15.57 squares. Area is the quantity that expresses the extent of a two - dimensional figure or 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 m ), 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. 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. Contents 1 Formal definition 2 Units 2.1 Conversions 2.1. 1 Non-metric units 2.2 Other units including historical 3 History 3.1 Circle area 3.2 Triangle area 3.3 Quadrilateral area 3.4 General polygon area 3.5 Areas determined using calculus 4 Area formulas 4.1 Polygon formulas 4.1. 1 Rectangles 4.1. 2 Dissection, parallelograms, and triangles 4.2 Area of curved shapes 4.2. 1 Circles 4.2. 2 Ellipses 4.2. 3 Surface area 4.3 General formulas 4.3. 1 Areas of 2 - dimensional figures 4.3. 2 Area in calculus 4.3. 3 Bounded area between two quadratic functions 4.3. 4 Surface area of 3 - dimensional figures 4.3. 5 General formula for surface area 4.4 List of formulas 4.5 Relation of area to perimeter 4.6 Fractals 5 Area bisectors 6 Optimization 7 See also 8 References 9 External links Formal definition See also : Jordan measure An approach to defining what is meant by `` area '' is through axioms. `` Area '' can be defined as a function from a collection M of special kind 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 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 ( m ), square centimetres ( cm ), square millimetres ( mm ), square kilometres ( km ), square feet ( ft ), square yards ( yd ), square miles ( mi ), 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 Although there are 10 mm in 1 cm, there are 100 mm in 1 cm. Calculation of the area of a square whose length and width are 1 metre would be : 1 metre x 1 metre = 1 m and therefore, another square with different sides can be calculated as : 3 metres x 2 metres = 6 m. This is, however, equivalent to 6 million millimetres square. Following this, 1 kilometre square = 1,000,000 metres square 1 metre square = 10,000 centimetres square = 1,000,000 millimetres square 1 centimetre square = 100 millimetres square 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 = 12 = 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 See also : Category : Units of area 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 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 History Circle area In the 5th century BCE, Hippocrates of Chios was 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 his quadrature of the lune of Hippocrates, but did not identify the constant 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 of Euclid 's Elements dealt with equality of areas between two - dimensional figures. The mathematician Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle 's circumference and whose height equals the circle 's radius, in his book Measurement of a Circle. ( The circumference is 2πr, and the area of a triangle is half the base times the height, yielding the area πr for the disk. ) Archimedes approximated the value of π ( and hence the area of a unit - radius circle ) with his 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 regular hexagon, 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 with circumscribed polygons ). Swiss scientist Johann Heinrich Lambert in 1761 proved that π, the ratio of a circle 's area to its squared radius, is irrational, meaning it is not equal to the quotient of any two whole numbers. In 1794 French mathematician Adrien - Marie Legendre proved that π is irrational ; this also proves that π is irrational. In 1882, German mathematician Ferdinand von Lindemann proved that π is transcendental ( not the solution of any polynomial equation with rational coefficients ), confirming a conjecture made by both Legendre and Euler. Triangle area Heron ( or Hero ) of Alexandria found what is known as Heron 's formula for 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 that Archimedes knew the formula over two centuries earlier, and since Metrica is 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 499 Aryabhata, a great mathematician - astronomer from the classical age of Indian mathematics and Indian astronomy, expressed the area of a triangle as one - half the base times the height in the Aryabhatiya ( section 2.6 ). A formula equivalent to Heron 's was discovered by the Chinese independently of the Greeks. It was published in 1247 in Shushu Jiuzhang ( `` Mathematical Treatise in Nine Sections '' ), written by Qin Jiushao. Quadrilateral area In the 7th century CE, Brahmagupta developed a formula, now known as Brahmagupta 's formula, for the area of a cyclic quadrilateral ( a quadrilateral inscribed in a circle ) in terms of its sides. In 1842 the German mathematicians Carl Anton Bretschneider and Karl Georg Christian von Staudt independently found a formula, known as Bretschneider 's formula, for the area of any quadrilateral. General polygon area The development of Cartesian coordinates by René Descartes in the 17th century allowed the development of the surveyor 's formula for the area of any polygon with known vertex locations by Gauss in the 19th century. Areas determined using calculus The development of integral calculus in the late 17th century provided tools that could subsequently be used for computing more complicated areas, such as the area of an ellipse and the surface areas of various curved three - dimensional objects. Area formulas Polygon formulas Main article : Polygon § Area and centroid For a non-self - intersecting ( simple ) polygon, the Cartesian coordinates ( x i, y i ) ( displaystyle ( x_ ( i ), y_ ( i ) ) ) ( i = 0, 1,..., n - 1 ) of whose n vertices are known, the area is given by the surveyor 's formula : A = 1 2 ∑ i = 0 n − 1 ( x i y i + 1 − x i + 1 y i ), ( displaystyle A = ( \frac ( 1 ) ( 2 ) ) sum _ ( i = 0 ) ^ ( n - 1 ) ( x_ ( i ) y_ ( i + 1 ) - x_ ( i + 1 ) y_ ( i ) ), ) where when i = n - 1, then i + 1 is expressed as modulus n and so refers to 0. Rectangles The area of this rectangle is lw. The most basic area formula is the formula for the area of a rectangle. Given a rectangle with length l and width w, 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, as l = w in the case of a square, the area of a square with side length s is given by the formula : A = s ( 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. Equal area figures. Dissection, parallelograms, and triangles Main articles : Triangle § Computing the area of a triangle, and Parallelogram § Area formula 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. 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 : A = bh ( parallelogram ). Two equal triangles. 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 : A = 1 2 b h ( displaystyle A = ( \frac ( 1 ) ( 2 ) ) bh ) ( triangle ). Similar arguments can be used to find area formulas for the trapezoid as well as more complicated polygons. Area of curved shapes Circles 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 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 and approximate parallelogram. The height of this parallelogram is r, and the width is half the circumference of the circle, or πr. Thus, the total area of the circle is πr : A = πr ( 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 πr, which is the area of the circle. 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 : A = 2 ∫ − r r r 2 − x 2 d x = π r 2. ( displaystyle 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 an ellipse is related to the formula of a circle ; for an ellipse with semi-major and semi-minor axes x and y the formula is : A = π x y. ( displaystyle A = pi xy. ) Surface area Main article : Surface area Archimedes showed that the surface area of a sphere is exactly four times the area of a flat disk of the same radius, and the volume enclosed by the sphere is exactly 2 / 3 of the volume of a 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 ( or any 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, 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 can not 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 : A = 4πr ( sphere ), where 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 A triangle : 1 2 B h ( displaystyle ( \tfrac ( 1 ) ( 2 ) ) Bh ) ( 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 : s ( s − a ) ( s − b ) ( s − c ) ( displaystyle ( sqrt ( s ( s-a ) ( s-b ) ( s-c ) ) ) ) where a, b, c are the sides of the triangle, and s = 1 2 ( a + b + c ) ( displaystyle 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 1 2 a b sin ⁡ ( C ) ( displaystyle ( \tfrac ( 1 ) ( 2 ) ) ab sin ( C ) ) where C is the given angle and a and 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 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 ) ( displaystyle ( \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 the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points ( x, y ), ( x, y ), and ( x, y ). 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 : i + b 2 − 1 ( displaystyle i+ ( \frac ( b ) ( 2 ) ) - 1 ), 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. Area in calculus Integration can be thought of as measuring the area under a curve, defined by f ( x ), between two points ( here a and b ). 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 : A = ∫ a b f ( x ) d x. ( displaystyle A = int _ ( a ) ^ ( b ) f ( x ) , dx. ) The area between the graphs of two functions is equal to the integral of one function, f ( x ), minus the integral of the other function, g ( x ) : A = ∫ a b ( f ( x ) − g ( x ) ) d x, ( displaystyle A = int _ ( a ) ^ ( b ) ( f ( x ) - g ( x ) ) , dx, ) where f ( x ) ( displaystyle f ( x ) ) is the curve with the greater y - value. An area bounded by a function r = r ( θ ) expressed in polar coordinates is : A = 1 2 ∫ r 2 d θ. ( displaystyle A = ( 1 over 2 ) int r ^ ( 2 ) , d \theta. ) The area enclosed by a parametric curve u → ( t ) = ( x ( t ), y ( t ) ) ( displaystyle ( \vec ( u ) ) ( t ) = ( x ( t ), y ( t ) ) ) with endpoints u → ( t 0 ) = u → ( t 1 ) ( displaystyle ( \vec ( u ) ) ( t_ ( 0 ) ) = ( \vec ( u ) ) ( t_ ( 1 ) ) ) is given by the line integrals : ∮ t 0 t 1 x y  ̇ d t = − ∮ t 0 t 1 y x  ̇ d t = 1 2 ∮ t 0 t 1 ( x y  ̇ − y x  ̇ ) d t ( displaystyle 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 ) ( see Green 's theorem ) or the z - component of 1 2 ∮ t 0 t 1 u → × u →  ̇ d t. ( displaystyle ( 1 over 2 ) oint _ ( t_ ( 0 ) ) ^ ( t_ ( 1 ) ) ( \vec ( u ) ) \times ( dot ( \vec ( u ) ) ) , dt. ) Bounded area between two quadratic functions To find the bounded area between two quadratic functions, we subtract one from the other to write the difference as f ( x ) − g ( x ) = a x 2 + b x + c = a ( x − α ) ( x − β ) ( displaystyle f ( x ) - g ( x ) = ax ^ ( 2 ) + bx + c = a ( x - alpha ) ( x - \beta ) ) 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 Δ = b 2 − 4 a c. ( displaystyle Delta = b ^ ( 2 ) - 4ac. ) By simplifying the integral formula between the graphs of two functions ( as given in the section above ) and using Vieta 's formula, we can obtain A = Δ Δ 6 a 2 = a 6 ( β − α ) 3, a ≠ 0. ( displaystyle A = ( \frac ( Delta ( sqrt ( Delta ) ) ) ( 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 : π r ( r + r 2 + h 2 ) ( displaystyle pi r left ( r+ ( sqrt ( r ^ ( 2 ) + h ^ ( 2 ) ) ) \right ) ), where r is the radius of the circular base, and h is the height. That can also be rewritten as π r 2 + π r l ( displaystyle pi r ^ ( 2 ) + pi rl ) or π r ( r + l ) ( displaystyle pi r ( r + l ) , ! ) where r is the radius and l is the slant height of the cone. π r 2 ( displaystyle pi r ^ ( 2 ) ) is the base area while π r l ( displaystyle pi rl ) is the lateral surface area of the cone. cube : 6 s 2 ( displaystyle 6s ^ ( 2 ) ), where s is the length of an edge. cylinder : 2 π r ( r + h ) ( displaystyle 2 pi r ( r + h ) ), where r is the radius of a base and h is the height. The 2 π ( displaystyle pi ) r can also be rewritten as π ( displaystyle pi ) d, where d is the diameter. 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 : B + P L 2 ( displaystyle B+ ( \frac ( PL ) ( 2 ) ) ), 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 : 2 ( l w + l h + w h ) ( displaystyle 2 ( ell w+ ell h + wh ) ), where l ( displaystyle ell ) 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 z = f ( x, y ), ( displaystyle z = f ( x, y ), ) where ( x, y ) ∈ D ⊂ R 2 ( displaystyle ( x, y ) in D subset mathbb ( R ) ^ ( 2 ) ) and D ( displaystyle D ) is a region in the xy - plane with the smooth boundary : A = ∫∫ D ( ∂ f ∂ x ) 2 + ( ∂ f ∂ y ) 2 + 1 d x d y. ( displaystyle 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 a parametric surface in the vector form r = r ( u, v ), ( displaystyle mathbf ( r ) = mathbf ( r ) ( u, v ), ) where r ( displaystyle mathbf ( r ) ) is a continuously differentiable vector function of ( u, v ) ∈ D ⊂ R 2 ( displaystyle ( u, v ) in D subset mathbb ( R ) ^ ( 2 ) ) is : A = ∫∫ D ∂ r ∂ u × ∂ r ∂ v d u d v. ( displaystyle 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 Regular triangle ( equilateral triangle ) 3 4 s 2 ( displaystyle ( \frac ( sqrt ( 3 ) ) ( 4 ) ) s ^ ( 2 ) , ! ) s ( displaystyle s ) is the length of one side of the triangle. Triangle s ( s − a ) ( s − b ) ( s − c ) ( displaystyle ( sqrt ( s ( s-a ) ( s-b ) ( s-c ) ) ) , ! ) s ( displaystyle s ) is half the perimeter, a ( displaystyle a ), b ( displaystyle b ) and c ( displaystyle c ) are the length of each side. Triangle 1 2 a b sin ⁡ ( C ) ( displaystyle ( \tfrac ( 1 ) ( 2 ) ) ab sin ( C ) , ! ) a ( displaystyle a ) and b ( displaystyle b ) are any two sides, and C ( displaystyle C ) is the angle between them. Triangle 1 2 b h ( displaystyle ( \tfrac ( 1 ) ( 2 ) ) bh , ! ) b ( displaystyle b ) and h ( displaystyle h ) are the base and altitude ( measured perpendicular to the base ), respectively. Isosceles triangle 1 2 b a 2 − b 2 4 = b 4 4 a 2 − b 2 ( displaystyle ( \frac ( 1 ) ( 2 ) ) b ( sqrt ( a ^ ( 2 ) - ( \frac ( b ^ ( 2 ) ) ( 4 ) ) ) ) = ( \frac ( b ) ( 4 ) ) ( sqrt ( 4a ^ ( 2 ) - b ^ ( 2 ) ) ) ) a ( displaystyle a ) is the length of one of the two equal sides and b ( displaystyle b ) is the length of a different side. Rhombus / Kite 1 2 a b ( displaystyle ( \tfrac ( 1 ) ( 2 ) ) ab ) a ( displaystyle a ) and b ( displaystyle b ) are the lengths of the two diagonals of the rhombus or kite. Parallelogram b h ( displaystyle bh , ! ) b ( displaystyle b ) is the length of the base and h ( displaystyle h ) is the perpendicular height. Trapezoid ( a + b ) h 2 ( displaystyle ( \frac ( ( a + b ) h ) ( 2 ) ) , ! ) a ( displaystyle a ) and b ( displaystyle b ) are the parallel sides and h ( displaystyle h ) the distance ( height ) between the parallels. Regular hexagon 3 2 3 s 2 ( displaystyle ( \frac ( 3 ) ( 2 ) ) ( sqrt ( 3 ) ) s ^ ( 2 ) , ! ) s ( displaystyle s ) is the length of one side of the hexagon. Regular octagon 2 ( 1 + 2 ) s 2 ( displaystyle 2 ( 1 + ( sqrt ( 2 ) ) ) s ^ ( 2 ) , ! ) s ( displaystyle s ) is the length of one side of the octagon. Regular polygon 1 4 n l 2 ⋅ cot ⁡ ( π / n ) ( displaystyle ( \frac ( 1 ) ( 4 ) ) nl ^ ( 2 ) cdot cot ( pi / n ) , ! ) l ( displaystyle l ) is the side length and n ( displaystyle n ) is the number of sides. Regular polygon 1 4 n p 2 ⋅ cot ⁡ ( π / n ) ( displaystyle ( \frac ( 1 ) ( 4n ) ) p ^ ( 2 ) cdot cot ( pi / n ) , ! ) p ( displaystyle p ) is the perimeter and n ( displaystyle n ) is the number of sides. Regular polygon 1 2 n R 2 ⋅ sin ⁡ ( 2 π / n ) = n r 2 tan ⁡ ( π / n ) ( displaystyle ( \frac ( 1 ) ( 2 ) ) nR ^ ( 2 ) cdot sin ( 2 pi / n ) = nr ^ ( 2 ) \tan ( pi / n ) , ! ) R ( displaystyle R ) is the radius of a circumscribed circle, r ( displaystyle r ) is the radius of an inscribed circle, and n ( displaystyle n ) is the number of sides. Regular polygon 1 2 a p = 1 2 n s a ( displaystyle ( \tfrac ( 1 ) ( 2 ) ) ap = ( \tfrac ( 1 ) ( 2 ) ) nsa , ! ) n ( displaystyle n ) is the number of sides, s ( displaystyle s ) is the side length, a ( displaystyle a ) is the apothem, or the radius of an inscribed circle in the polygon, and p ( displaystyle p ) is the perimeter of the polygon. Circle π r 2 or π d 2 4 ( displaystyle pi r ^ ( 2 ) ( \text ( or ) ) ( \frac ( pi d ^ ( 2 ) ) ( 4 ) ) , ! ) r ( displaystyle r ) is the radius and d ( displaystyle d ) the diameter. Circular sector θ 2 r 2 or L ⋅ r 2 ( displaystyle ( \frac ( \theta ) ( 2 ) ) r ^ ( 2 ) ( \text ( or ) ) ( \frac ( L cdot r ) ( 2 ) ) , ! ) r ( displaystyle r ) and θ ( displaystyle \theta ) are the radius and angle ( in radians ), respectively and L ( displaystyle L ) is the length of the perimeter. Ellipse π a b ( displaystyle pi ab , ! ) a ( displaystyle a ) and b ( displaystyle b ) are the semi-major and semi-minor axes, respectively. Total surface area of a cylinder 2 π r ( r + h ) ( displaystyle 2 pi r ( r + h ) , ! ) r ( displaystyle r ) and h ( displaystyle h ) are the radius and height, respectively. Lateral surface area of a cylinder 2 π r h ( displaystyle 2 pi rh , ! ) r ( displaystyle r ) and h ( displaystyle h ) are the radius and height, respectively. Total surface area of a sphere 4 π r 2 or π d 2 ( displaystyle 4 pi r ^ ( 2 ) ( \text ( or ) ) pi d ^ ( 2 ) , ! ) r ( displaystyle r ) and d ( displaystyle d ) are the radius and diameter, respectively. Total surface area of a pyramid B + P L 2 ( displaystyle B+ ( \frac ( PL ) ( 2 ) ) , ! ) B ( displaystyle B ) is the base area, P ( displaystyle P ) is the base perimeter and L ( displaystyle L ) is the slant height. Total surface area of a pyramid frustum B + P L 2 ( displaystyle B+ ( \frac ( PL ) ( 2 ) ) , ! ) B ( displaystyle B ) is the base area, P ( displaystyle P ) is the base perimeter and L ( displaystyle L ) is the slant height. Square to circular area conversion 4 π A ( displaystyle ( \frac ( 4 ) ( pi ) ) A , ! ) A ( displaystyle A ) is the area of the square in square units. Circular to square area conversion π 4 C ( displaystyle ( \frac ( pi ) ( 4 ) ) C , ! ) C ( displaystyle C ) is the area of the circle in circular units. The above calculations show how to find the areas of many common shapes. The areas of irregular polygons can be calculated using the `` Surveyor 's formula ''. 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, 4 π A ≤ L 2, ( displaystyle 4 pi A leq L ^ ( 2 ), ) 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 angles 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 and the circumference formula 2πr. 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. 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. 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 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 bubbles. The question of the filling area of the Riemannian circle remains open. 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. 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, π 3 3 ( displaystyle ( \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, 1 12 3, ( displaystyle ( \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. Equi - areal mapping Heronian triangle, a triangle with integer sides and integer area. List of triangle inequalities # Area 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 ( area ) -- A list of areas by size. Pentagon # Derivation of the area formula Planimeter, an instrument for measuring small areas, e.g. on maps. Quadrilateral # Area of a convex quadrilateral Robbins pentagon, a cyclic pentagon whose side lengths and area are all rational numbers. References ^ Jump up to : Weisstein, Eric W. `` Area ''. Wolfram MathWorld. Retrieved 3 July 2012. ^ Jump up to : `` Area Formulas ''. Math.com. Retrieved 2 July 2012. ^ Jump up to : Bureau International des Poids et Mesures Resolution 12 of the 11th meeting of the CGPM ( 1960 ), retrieved 15 July 2012 Jump up ^ Mark de Berg ; Marc van Kreveld ; Mark Overmars ; Otfried Schwarzkopf ( 2000 ). `` Chapter 3 : Polygon Triangulation ''. Computational Geometry ( 2nd revised ed. ). Springer - Verlag. pp. 45 -- 61. ISBN 3 - 540 - 65620 - 0. Jump up ^ Boyer, Carl B. ( 1959 ). A History of the Calculus and Its Conceptual Development. Dover. ISBN 0 - 486 - 60509 - 4. ^ Jump up to : Weisstein, Eric W. `` Surface Area ''. Wolfram MathWorld. Retrieved 3 July 2012. ^ Jump up to : do Carmo, Manfredo. Differential Geometry of Curves and Surfaces. Prentice - Hall, 1976. Page 98, ISBN 978 - 0 - 13 - 212589 - 5 Jump up ^ Walter Rudin, Real and Complex Analysis, McGraw - Hill, 1966, ISBN 0 - 07 - 100276 - 6. Jump up ^ Gerald Folland, Real Analysis : modern techniques and their applications, John Wiley & Sons, Inc., 1999, Page 20, ISBN 0 - 471 - 31716 - 0 Jump up ^ Moise, Edwin ( 1963 ). Elementary Geometry from an Advanced Standpoint. Addison - Wesley Pub. Co. Retrieved 15 July 2012. ^ Jump up to : Bureau international des poids et mesures ( 2006 ). `` The International System of Units ( SI ) '' ( PDF ). 8th ed. Retrieved 2008 - 02 - 13. Chapter 5. Jump up ^ Heath, Thomas L. ( 2003 ), A Manual of Greek Mathematics, Courier Dover Publications, pp. 121 -- 132, ISBN 0 - 486 - 43231 - 9 Jump up ^ Stewart, James ( 2003 ). Single variable calculus early transcendentals. ( 5th. ed. ). Toronto ON : Brook / Cole. p. 3. ISBN 0 - 534 - 39330 - 6. However, by indirect reasoning, Eudoxus ( fifth century B.C. ) used exhaustion to prove the familiar formula for the area of a circle : A = π r 2. ( displaystyle A = pi r ^ ( 2 ). ) ^ Jump up to : Arndt, Jörg ; Haene l, Christoph ( 2006 ). Pi Unleashed. Springer - Verlag. ISBN 978 - 3 - 540 - 66572 - 4. Retrieved 2013 - 06 - 05. English translation by Catriona and David Lischka. Jump up ^ Eves, Howard ( 1990 ), An Introduction to the History of Mathematics ( 6th ed. ), Saunders, p. 121, ISBN 0 - 03 - 029558 - 0 Jump up ^ Heath, Thomas L. ( 1921 ). A History of Greek Mathematics ( Vol II ). Oxford University Press. pp. 321 -- 323. Jump up ^ Weisstein, Eric W. `` Heron 's Formula ''. MathWorld. Jump up ^ Bourke, Paul ( July 1988 ). `` Calculating The Area And Centroid Of A Polygon '' ( PDF ). Retrieved 6 Feb 2013. Jump up ^ `` Area of Parallelogram / Rectangle ''. ProofWiki.org. Retrieved 29 May 2016. Jump up ^ `` Area of Square ''. ProofWiki.org. Retrieved 29 May 2016. Jump up ^ Averbach, Bonnie ; Chein, Orin ( 2012 ), Problem Solving Through Recreational Mathematics, Dover, p. 306, ISBN 978 - 0 - 486 - 13174 - 0 Jump up ^ Joshi, K.D. ( 2002 ), Calculus for Scientists and Engineers : An Analytical Approach, CRC Press, p. 43, ISBN 978 - 0 - 8493 - 1319 - 6 ^ Jump up to : Braden, Bart ( September 1986 ). `` The Surveyor 's Area Formula '' ( PDF ). The College Mathematics Journal. 17 ( 4 ) : 326 -- 337. doi : 10.2307 / 2686282. Retrieved 15 July 2012. ^ Jump up to : Trainin, J. ( November 2007 ). `` An elementary proof of Pick 's theorem ''. Mathematical Gazette. 91 ( 522 ) : 536 -- 540. Jump up ^ Matematika. PT Grafindo Media Pratama. pp. 51 --. ISBN 978 - 979 - 758 - 477 - 1. Jump up ^ Get Success UN + SPMB Matematika. PT Grafindo Media Pratama. pp. 157 --. ISBN 978 - 602 - 00 - 0090 - 9. ^ Jump up to : Weisstein, Eric W. `` Cone ''. Wolfram MathWorld. Retrieved 6 July 2012. Jump up ^ Mandelbrot, Benoît B. ( 1983 ). The fractal geometry of nature. Macmillan. ISBN 978 - 0 - 7167 - 1186 - 5. Retrieved 1 February 2012. Jump up ^ Gromov, Mikhael ( 1983 ), `` Filling Riemannian manifolds '', Journal of Differential Geometry, 18 ( 1 ) : 1 -- 147, CiteSeerX 10.1. 1.400. 9154, MR 697984 ^ Jump up to : Chakerian, G.D. ( 1979 ) `` A Distorted View of Geometry. '' Ch. 7 in Mathematical Plums. R. Honsberger ( ed. ). Washington, DC : Mathematical Association of America, p. 147. Jump up ^ Dorrie, Heinrich ( 1965 ), 100 Great Problems of Elementary Mathematics, Dover Publ., pp. 379 -- 380. Jump up ^ Minda, D. ; Phelps, S. ( October 2008 ). `` Triangles, ellipses, and cubic polynomials ''. American Mathematical Monthly. 115 : 679 -- 689 : Theorem 4.1. JSTOR 27642581. External links Find more aboutAreaat Wikipedia 's sister projects Definitions from Wiktionary Media from Commons Data from Wikidata Classical mechanics SI units Linear / translational quantities Angular / rotational quantities Dimensions Dimensions time : t s absement : A ms time : t s distance : d, position : r, s, x, displacement m area : A m angle : θ, angular displacement : θ rad solid angle : Ω rad, sr frequency : f s, Hz speed : v, velocity : v ms kinematic viscosity : ν, specific angular momentum : h m s frequency : f s, Hz angular speed : ω, angular velocity : ω rad s acceleration : a ms angular acceleration : α rad s jerk : j ms angular jerk : ζ rad s mass : m kg ML moment of inertia : I kg m MT momentum : p, impulse : J kg m s, N s action : S, actergy : א kg m s, J s ML T angular momentum : L, angular impulse : ΔL kg m s action : S, actergy : א kg m s, J s MT force : F, weight : F kg ms, N energy : E, work : W kg m s, J ML T torque : τ, moment : M kg m s, N m energy : E, work : W kg m s, J MT yank : Y kg ms, N s power : P kg m s, W ML T rotatum : P kg m s, N ms power : P kg m s, W GND : 4193807 - 0 Retrieved from `` https://en.wikipedia.org/w/index.php?title=Area&oldid=800088069 '' Categories : Area Hidden categories : Wikipedia articles with GND identifiers Talk Contents About Wikipedia Адыгэбзэ Afrikaans Alemannisch Aragonés ܐܪܡܝܐ অসমীয়া Asturianu Авар Azərbaycanca تۆرکجه বাংলা Bân - lâm - gú Башҡортса Беларуская Беларуская ( тарашкевіца ) ‎ भोजपुरी Български Boarisch Bosanski Brezhoneg Català Чӑвашла Cebuano Čeština ChiShona Cymraeg Dansk Deutsch ދިވެހިބަސް Dolnoserbski Eesti Ελληνικά Español Esperanto Euskara فارسی Føroyskt Français Frysk Gaelg Gàidhlig Galego 贛 語 ગુજરાતી 客家 語 / Hak - kâ - ngî 한국어 Hawaiʻi Հայերեն हिन्दी Hornjoserbsce Hrvatski Ido Ilokano Bahasa Indonesia Interlingua Ирон Íslenska Italiano עברית Basa Jawa ಕನ್ನಡ ქართული Қазақша Kiswahili Kurdî ລາວ Latina Latviešu Lëtzebuergesch Lietuvių Limburgs Lingála Magyar मैथिली Македонски Malagasy മലയാളം मराठी მარგალური مازِرونی Bahasa Melayu Mìng - dĕ̤ng - ngṳ̄ Mirandés မြန်မာဘာသာ Nederlands Nedersaksies नेपाली 日本 語 Нохчийн Nordfriisk Norsk Norsk nynorsk Occitan Олык марий Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ Pälzisch پنجابی Patois ភាសា ខ្មែរ Plattdüütsch Polski Português Română Runa Simi Русский संस्कृतम् Scots Shqip Sicilianu Simple English Slovenčina Slovenščina Словѣньскъ / ⰔⰎⰑⰂⰡⰐⰠⰔⰍⰟ Ślůnski Soomaaliga کوردی Српски / srpski Srpskohrvatski / српскохрватски Basa Sunda Suomi Svenska Tagalog தமிழ் Татарча / tatarça తెలుగు ไทย Тоҷикӣ Türkçe Українська اردو Tiếng Việt Võro West - Vlams Winaray Wolof 吴语 ייִדיש Yorùbá 粵語 Zeêuws Žemaitėška 中文 Edit links This page was last edited on 11 September 2017, at 12 : 56. 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Question: unit of measure for area of a triangle
Answer: square metre