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
4
4.1
4.1.1 Rectangles 4.1.2 Dissection, parallelograms, and triangles 4.2
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
4.4 List of formulas 4.5 Relation of area to perimeter 4.6 Fractals 5
Formal definition[edit] 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.[10] Units[edit] 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.[11] 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.[3] Conversions[edit] 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 x 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 x 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[edit] 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 = 0.09290304 square metres 1 square yard = 0.83612736 square metres 1 square mile = 2.589988110336 square kilometres Other units including historical[edit] 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:[11] 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:[11] 1 barn = 10−28 square meters. The barn is commonly used in describing the cross sectional area of interaction in nuclear physics.[11] In India, 20 Dhurki = 1 Dhur 20 Dhur = 1 Khatha 20 Khata = 1 Bigha 32 Khata = 1 Acre History[edit]
( 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:[18] 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[edit] 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:[2][19] 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:[1][2][20] A = s2 (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[edit]
Main articles:
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:[2] A = 1 2 b h displaystyle A= frac 1 2 bh (triangle). Similar arguments can be used to find area formulas for the
trapezoid[21] as well as more complicated polygons.[22]
A circle can be divided into sectors which rearrange to form an approximate parallelogram. Main article:
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. The limit of the areas of the approximate parallelograms is exactly πr2, which is the area of the circle.[23] 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;=;2int _ -r ^ r sqrt r^ 2 -x^ 2 ,dx;=;pi r^ 2 . Ellipses[edit]
Main article:
A = π x y . displaystyle A=pi xy. Surface area[edit] Main article: Surface area
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 cannot be
flattened out. The formula for the surface area of a sphere was first
obtained by
A = 4πr2 (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[edit] Areas of 2-dimensional figures[edit] 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
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.[2] If an angle and its two included sides are given, the area is 1 2 a b sin ( C ) displaystyle tfrac 1 2 absin(C) where C is the given angle and a and b are its included sides.[2] 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 (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: 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.[24]
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:[1] 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:[1] A = 1 2 ∫ r 2 d θ . displaystyle A= 1 over 2 int r^ 2 ,dtheta . 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[edit] 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[25][26] 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 aneq 0. The above remains valid if one of the bounding functions is linear
instead of quadratic.
Cone:[27] π r ( r + r 2 + h 2 ) displaystyle pi rleft(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 [27] 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.[27] cube: 6 s 2 displaystyle 6s^ 2 , where s is the length of an edge.[6] cylinder: 2 π r ( r + h ) displaystyle 2pi 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[edit] 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 Dsubset 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 Dsubset mathbb R ^ 2 is:[7] 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[edit] 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[1] 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[2] 1 2 a b sin ( C ) displaystyle tfrac 1 2 absin(C),! a displaystyle a and b displaystyle b are any two sides, and C displaystyle C is the angle between them. Triangle[1] 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(2pi /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 Lcdot 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[2] 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 2pi 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 2pi rh,! r displaystyle r and h displaystyle h are the radius and height, respectively. Total surface area of a sphere[6] 4 π r 2
or π d 2 displaystyle 4pi 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[6] 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[6] 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.
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".[23] Relation of area to perimeter[edit] 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 4pi Aleq 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 πr2 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[edit]
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. [28]
π 3 3 displaystyle frac pi 3 sqrt 3 , is larger than that of any non-equilateral triangle.[32] 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.[30] See also[edit]
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#
References[edit] ^ a b c d e f g h Weisstein, Eric W. "Area". Wolfram MathWorld.
Archived from the original on 5 May 2012. Retrieved 3 July 2012.
^ a b c d e f g h i j k "
A = π r 2 . displaystyle A=pi r^ 2 .
^ a b 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.
^ Eves, Howard (1990), An Introduction to the History of Mathematics
(6th ed.), Saunders, p. 121, ISBN 0-03-029558-0
^ Heath, Thomas L. (1921). A History of Greek
External links[edit] Find more aboutAreaat's sister projects Definitions from Wiktionary Media from Wikimedia Commons Data from Wikidata v t e
Linear/translational quantities Angular/rotational quantities Dimensions 1 L L2 Dimensions 1 1 1 T time: t s absement: A m s T time: t s 1 distance: d, position: r, s, x, displacement m area: A m2 1 angle: θ, angular displacement: θ rad solid angle: Ω rad2, sr T−1 frequency: f s−1, Hz speed: v, velocity: v m s−1 kinematic viscosity: ν, specific angular momentum: h m2 s−1 T−1 frequency: f s−1, Hz angular speed: ω, angular velocity: ω rad s−1 T−2 acceleration: a m s−2 T−2 angular acceleration: α rad s−2 T−3 jerk: j m s−3 T−3 angular jerk: ζ rad s−3 M mass: m kg ML2 moment of inertia: I kg m2 MT−1 momentum: p, impulse: J kg m s−1, N s action: 𝒮, actergy: ℵ kg m2 s−1, J s ML2T−1 angular momentum: L, angular impulse: ΔL kg m2 s−1 action: 𝒮, actergy: ℵ kg m2 s−1, J s MT−2 force: F, weight: Fg kg m s−2, N energy: E, work: W kg m2 s−2, J ML2T−2 torque: τ, moment: M kg m2 s−2, N m energy: E, work: W kg m2 s−2, J MT−3 yank: Y kg m s−3, N s−1 power: P kg m2 s−3, W ML2T−3 rotatum: P kg m2 s−3, N m s−1 power: P kg m2 s−3, W Authority control |