Non-metric units

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

1 foot = 12 inches,

the relationship between square feet and square inches is

1 square foot = 144 square inches,

where 144 = 12² = 12 × 12. Similarly:

• 1 square yard = 9 square feet
• 1 square mile = 3,097,600 square yards = 27,878,400 square feet

In addition, conversion factors include:

• 1 square inch = 6.4516 square centimetres
• 1 square foot = 0.09290304 square metres
• 1 square yard = 0.83612736 square metres
• 1 square mile = 2.589988110336 square kilometres

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:^[2]

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]

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.

Dissection, parallelograms, and triangles

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

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:^[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]

Circles

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:

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

Ellipses

The formula for the area enclosed by an ellipse is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes x and y the formula is:^[2]

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

Areas of 2-dimensional figures

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]

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:^[1]

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

• 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:^[1]

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

• 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.

Bounded area between two quadratic functions

To find the bounded area between two quadratic functions, we first subtract one from the other, writing 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. By the area integral formulas above and Vieta's formula, we can obtain that^[28][29]

$${\displaystyle A={\frac {(b^{2}-4ac)^{3/2}}{6a^{2}}}={\frac {a}{6}}(\beta -\alpha )^{3},\qquad a\neq 0.}$$

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

Surface area of 3-dimensional figures

• Cone:^[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.^[6]
• 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.

General formula for surface area

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:^[8]

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