# Area

**Area**is a quantity expressing the size of a region of space.

**Surface area**refers to the summation of the exposed sides of an object. Area (

*Cx*

^{2}) is the derivative of volume (

*Cx*

^{3}). Area is the antiderivative of length (

*Cx*

^{1}). In the case of the perfect closed curve in two dimensions, which is the circle, the area is the simple integral of the circumference. Thus, the circumference is 2πr, while the area is πr

^{2}.

Table of contents |

2 Some formulas 3 See also 4 How to define area 5 External links |

## Units

Units for measuring surface area include:

- square metre - SI derived unit
- are - 100 square metres
- hectare - 10,000 square metres
- square kilometre - 1,000,000 square metres
- square megametre - 10
^{12}square metres

- square foot (plural feet) - 0.09290304 square meters.
- square yard - 9 square feet - 0.83612736 square metres
- square perch - 30.25 square yards - 25.2928526 square metres
- acre - 160 square perches or 43,560 square feet - 4046.8564224 square metres
- square mile - 640 acres - 2.5899881103 square kilometres

## Some formulas

For a two dimensional object the area and surface area are the same:

- square or rectangle:
(where l is the length and w is the width; in the case of a square, l = w.*l × w* - circle:
(where r is the radius)*π×r*^{2} - any regular polygon:
(where P = the length of the perimeter, and a is the length of the apothem of the polygon [the distance from the center of the polygon to the center of one side])*P × a / 2* - a parallelogram:
(where the base B is any side, and the height h is the distance between the lines that the sides of length B lie on)*B × h* - a trapezoid:
(B and b are the lengths of the parallel sides, and h is the distance between the lines on which the parallel sides lie)*(B + b) × h / 2* - a triangle:
(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*B × h / 2**Heron's formula*can be used:(where a, b, c are the sides of the triangle, and s = (a + b + c)/2 is half of its perimeter)*√(s×(s-a)×(s-b)×(s-c))* - the area between the graphss of two functions is equal to the integral of one function,
*f*(*x*), minus the integral of the other function,*g*(*x*).

- cube:
, where s is the length of any side*6×(s*^{2}) - rectangular box:
, where l, w, and h are the length, width, and height of the box*2×((l × w) + (l × h) + (w × h))* - sphere:
, where π is the ratio of circumference to diameter of a circle, 3.14159..., and r is the radius of the sphere*4×π×(r*^{2}) - cylinder:
, where r is the radius of the circular base, and h is the height*2×π×r×(h + r)* - cone:
, where r is the radius of the circular base, and h is the height.*π×r×(r + √(r*^{2}+ h^{2}))

## See also

*An artist should feel free to add some example diagrams.*

## How to define area

Although area seems to be one of the basic notions in geometry it is not at all easy to define, even in the Euclidean plane. Most books avoid it, wrongly referring to self-evidence. To make area meaningful one has to define it, at the very least, on polygons in the Euclidean plane, and it can be done using the following definition:

The area of a polygon in the Euclidean plane is a positive number such that:

- The area of the unit square is equal to one.
- Equal polygons have equal area.
- If a polygon is a union of two polygons which do not have common interior points, then its area is the sum of the areas of these polygons.

But before using this definition one has to prove that such an area indeed exists.

In other words, one can also give a formula for the area of an arbitrary triangle, and then define the area of an arbitrary polygon using the idea that the area of a union of polygons (without common interior points) is the sum of the areas of its pieces. But then it is not easy to show that such area does not depend on the way you break the polygon into pieces.

Nowadays, the most standard (correct) way to introduce area is through the more advanced notion of Lebesgue measure, but one should note that in general, if one adopts the axiom of choice then it is possible to prove that there are some shapes whose Lebesgue measure cannot be meaningfully defined. Such 'shapes' (they cannot *a fortiori* be simply visualised) enter into Tarski's circle-squaring problem (and, moving to three dimensions, in the Banach-Tarski paradox). The sets involved do not arise in practical matters.

## External links