Area

After completing this section, you should be able to:

1. Calculate the area of triangles.
2. Calculate the area of quadrilaterals.
3. Calculate the area of other polygons.
4. Calculate the area of circles.

The formula for the area of a triangle is given as follows.

For example, consider the triangle in .

The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area: \[A=\frac{1}{2}(4)(5)=\frac{1}{2}(20)=10\ {\text{cm}}^{2}\]

In , the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then \[A=\frac{1}{2}(7)(3.5)=12.25\ {\text{cm}}^{2}.\]

A quadrilateral is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle.

In , a \(12\ \text{in}\times 12\ \text{in}\) grid is represented with twelve \(1\ \text{in}\times 1\ \text{in}\) squares across each row, and twelve \(1\ \text{in}\times 1\ \text{in}\) squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area, we have a formula. Thus, the formula for the area of a square, where \(s=\text{length of a side}\), is \(A=s⋅s\). The area of the square in is \(A=12\ \text{in}\times 12\ \text{in}=144\ {\text{in}}^{2}.\)

Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in has width equal to 5 in and length equal to 12 in. The area is \(A=5(12)=60\ {\text{in}}^{2}.\)

Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area?

The area of a parallelogram can be found using the formula for the area of a triangle. Notice in , if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram: \[\begin{array}{lll}A & = & 2(\frac{1}{2}bh) \\ A & = & bh\end{array}\]

For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be \(A=(10)(5)=50\ {\text{in}}^{2}.\)

Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases \(a\) and \(b\) and height \(h\) is given here.

For example, find the area of the trapezoid in that has base \(a\) equal to 8 cm, base \(b\) equal to 6 cm, and height equal to 6 cm.

The area is \(A=\frac{1}{2}(6)(6+8)=42\ {\text{cm}}^{2}\).

The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals.

For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is \(220\ {\text{cm}}^{2},\) and the measure of \({d}_{2}=11,\) find the measure of \({d}_{1}.\) To solve this problem, we input the known values into the formula and solve for the unknown.

We have that \[\begin{array}{lll}220 & = & \frac{11{d}_{1}}{2} \\ 220(2) & = & 11{d}_{1} \\ \frac{440}{11} & = & {d}_{1}=40\end{array}\]

To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem \(a\) of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius \(r\) of a regular polygon is also a line segment that starts at the center but extends to a vertex.

For example, consider the regular hexagon shown in with a side length of 4 cm, and the apothem measures \(a=2\sqrt{3}.\)

We have the perimeter, \(p=6(4)=24\ \text{cm}.\) We have the apothem as \(a=2\sqrt{3}.\) Then, the area is: \[\begin{array}{lll}A & = & \frac{1}{2}(2\sqrt{3})(24) \\ & = & 24\sqrt{3}=41.57\ {\text{cm}}^{2}\end{array}\]

Often, we have the need to change the units of one or more items in a problem to find a solution. For example, suppose you are purchasing new carpet for a room measured in feet, but carpeting is sold in terms of yards. You will have to convert feet to yards to purchase the correct amount of carpeting. Or, you may need to convert centimeters to inches, or feet to meters. In each case, it is essential to use the correct equivalency.

Just as the circumference of a circle includes the number \(\pi ,\) so does the formula for the area of a circle. Recall that \(\pi\) is a non-terminating, non-repeating decimal number: \(\pi =3.14159\ldots\). It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area.

For example, to find the of the circle with radius equal to 3 cm, as shown in , is found using the formula \(A=\pi {r}^{2}.\)

We have \[\begin{array}{lll}A & = & \pi {r}^{2} \\ & = & \pi {(3)}^{2} \\ & = & 9\pi =28.27\ {\text{cm}}^{2}\end{array}\]

Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems.

Condensed: the full section is in OpenStax Contemporary Mathematics.

• The area \(A\) of a triangle is found with the formula \(A=\frac{1}{2}bh,\) where \(b\) is the base and \(h\) is the height.
• The area of a parallelogram is found using the formula \(A=bh,\) where \(b\) is the base and \(h\) is the height.
• The area of a rectangle is found using the formula \(A=lw,\) where \(l\) is the length and \(w\) is the width.
• The area of a trapezoid is found using the formula \(A=\frac{1}{2}h({b}_{1}+{b}_{2}),\) where \(h\) is the height, \({b}_{1}\) is the length of one base, and \({b}_{2}\) is the length of the other base.
• The area of a rhombus is found using the formula \(A=\frac{{d}_{1}{d}_{2}}{2},\) where \({d}_{1}\) is the length of one diagonal and \({d}_{2}\) is the length of the other diagonal.
• The area of a regular polygon is found using the formula \(A=\frac{1}{2}ap,\) where \(a\) is the apothem and \(p\) is the perimeter.
• The area of a circle is found using the formula \(A=\pi {r}^{2},\) where \(r\) is the radius.

The area of a triangle is given as \(A=\frac{1}{2}bh,\) where \(b\) represents the base and \(h\) represents the height.

The formula for the area of a square is \(A=s⋅s\) or \(A={s}^{2}.\)

The area of a rectangle is given as \(A=lw.\)

The area of a parallelogram is \(A=bh.\)

The formula for the area of a trapezoid is given as \(A=\frac{1}{2}h(a+b).\)

The area of a rhombus is found using one of these formulas:

• \(A=\frac{{d}_{1}{d}_{2}}{2},\) where \({d}_{1}\) and \({d}_{2}\) are the diagonals.
• \(A=\frac{1}{2}bh,\) where \(b\) is the base and \(h\) is the height.

The area of a regular polygon is found with the formula \(A=\frac{1}{2}ap,\) where \(a\) is the apothem and \(p\) is the perimeter.

Condensed: the full section is in OpenStax Contemporary Mathematics.

Area is the measure of a region's size on a surface. The area of a plane region or plane area refers to the area of a shape or planar lamina, while surface area refers to the area of an open surface or the boundary 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 analogue of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept). Two different regions may have the same area (as in squaring the circle); by synecdoche, "area" sometimes is used to refer to the region, as in a "polygonal area".

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 boundaries, calculus is usually required to compute the area. In all, 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 if one supposes the axiom of choice. 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.

An approach to defining what is meant by "area" is through axioms. "Area" can be defined as a function from a collection M of a special kinds 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.

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.

Calculation of the area of a square whose length and width are 1 metre:

1 metre × 1 metre = 1 m

and so, a rectangle with different sides (say length of 3 metres and width of 2 metres) has an area in square units that can be calculated as:

3 metres × 2 metres = 6 m. 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.

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.

• 1 killa = 1 acre
• 1 ghumaon = 1 acre
• 1 kanal = 0.125 acre (1 acre = 8 kanal)
• 1 decimal = 48.4 square yards
• 1 chatak = 180 square feet

Condensed: the full section is in Wikipedia.

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).

In 300 BCE Greek mathematician Euclid proved that the area of a triangle is half that of a parallelogram with the same base and height in his book Elements of Geometry. Heron 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 CE Aryabhata, a 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.

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 \((x_i, y_i) (i=0, 1, ..., n-1)\) of whose n vertices are known, the area is given by the surveyor's formula:

\[A = \frac{1}{2} \Biggl\vert \sum_{i = 0}^{n - 1}( x_i y_{i + 1} - x_{i + 1} y_i) \Biggr\vert\]

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

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:

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.

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).

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,

\(4\pi A \le L^2,\)

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).