Area

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.

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

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:[14]

• 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:[14]

• 1 barn = 10⁻²⁸ square meters.

The barn is commonly used in describing the cross-sectional area of interaction in nuclear physics.[14]

In India,

• 20 dhurki = 1 dhur
• 20 dhur = 1 khatha
• 20 khata = 1 bigha
• 32 khata = 1 acre

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,[15] 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.[16]

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 πr² 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.[17] In 1794, French mathematician Adrien-Marie Legendre proved that π² is irrational; this also proves that π is irrational.[18] 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.[17]^{: p. 196}

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,[19] 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.[20]

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.

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.

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.

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.

For a non-self-intersecting (simple) polygon, the Cartesian coordinates ${\displaystyle (x_{i},y_{i})}$ (i=0, 1, ..., n-1) of whose n vertices are known, the area is given by the surveyor's formula:[21]

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

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

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][22]

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:[6][2][23]

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.

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.

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:[2]

A = bh  (parallelogram).

A parallelogram split into 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]

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

Similar arguments can be used to find area formulas for the trapezoid[24] as well as more complicated polygons.[25]

A circle can be divided into sectors which rearrange to form an approximate parallelogram.

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 an 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 πr²:[2]

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.[26]

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:

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

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:[2]

${\displaystyle A=\pi xy.}$

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 (see: developable surfaces). 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 Archimedes in his work On the Sphere and Cylinder. The formula is:[7]

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.

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

• A triangle: ${\displaystyle {\tfrac {1}{2}}Bh}$ (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: ${\displaystyle {\sqrt {s(s-a)(s-b)(s-c)}}}$ where a, b, c are the sides of the triangle, and ${\displaystyle s={\tfrac {1}{2}}(a+b+c)}$ is half of its perimeter.[2] If an angle and its two included sides are given, the area is ${\displaystyle {\tfrac {1}{2}}ab\sin(C)}$ 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 ${\displaystyle {\tfrac {1}{2}}(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3})}$. 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: ${\displaystyle i+{\frac {b}{2}}-1}$, 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.[27]

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:[6]

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

• 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):

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

• An area bounded by a function ${\displaystyle r=r(\theta )}$ expressed in polar coordinates is:[6]

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

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

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

or the z-component of

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

(For details, see Green's theorem § Area calculation.) This is the principle of the planimeter mechanical device.

To find the bounded area between two quadratic functions, we subtract one from the other to write the difference as

${\displaystyle f(x)-g(x)=ax^{2}+bx+c=a(x-\alpha )(x-\beta )}$

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

${\displaystyle \Delta =b^{2}-4ac.}$

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[28][29]

${\displaystyle A={\frac {\Delta {\sqrt {\Delta }}}{6a^{2}}}={\frac {a}{6}}(\beta -\alpha )^{3},\qquad a\neq 0.}$

The above remains valid if one of the bounding functions is linear instead of quadratic.

• Cone:[30] ${\displaystyle \pi r\left(r+{\sqrt {r^{2}+h^{2}}}\right)}$, where r is the radius of the circular base, and h is the height. That can also be rewritten as ${\displaystyle \pi r^{2}+\pi rl}$[30] or ${\displaystyle \pi r(r+l)\,\!}$ where r is the radius and l is the slant height of the cone. ${\displaystyle \pi r^{2}}$ is the base area while ${\displaystyle \pi rl}$ is the lateral surface area of the cone.[30]
• Cube: ${\displaystyle 6s^{2}}$, where s is the length of an edge.[7]
• Cylinder: ${\displaystyle 2\pi r(r+h)}$, where r is the radius of a base and h is the height. The ${\displaystyle 2\pi r}$ can also be rewritten as ${\displaystyle \pi d}$, where d is the diameter.
• Prism: ${\displaystyle 2B+Ph}$, where B is the area of a base, P is the perimeter of a base, and h is the height of the prism.
• pyramid: ${\displaystyle B+{\frac {PL}{2}}}$, 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: ${\displaystyle 2(\ell w+\ell h+wh)}$, where ${\displaystyle \ell }$ is the length, w is the width, and h is the height.

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

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

An even more general formula for the area of the graph of a parametric surface in the vector form ${\displaystyle \mathbf {r} =\mathbf {r} (u,v),}$ where ${\displaystyle \mathbf {r} }$ is a continuously differentiable vector function of ${\displaystyle (u,v)\in D\subset \mathbb {R} ^{2}}$ is:[9]

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

Additional common formulas for area:

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

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

The areas of irregular (and thus arbitrary) polygons can be calculated using the "Surveyor's formula" (shoelace formula).[26]

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,

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

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 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).

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. [31]