Area of a triangle

A graphic derivation of the formula Template:Math that avoids the usual procedure of doubling the area of the triangle and then halving it.

In geometry, calculating the area of a triangle is an elementary and widely encountered problem. The best known and simplest formula for the area of an arbitrary triangle is $$\text{area}=\frac{1}{2}\times \text{base}\times \text{height},$$ where "base" means the length of any side of the triangle, and "height" means the shortest distance between the line of the base and the corresponding apex (opposite corner of the triangle).

Although simple, this formula is only useful if the height can be readily found, which is not always the case. For example, the land surveyor of a triangular field might find it relatively easy to measure the length of each side, but relatively difficult to construct a 'height'. Various methods may be used in practice, depending on what is known about the triangle. Other frequently used formulas for the area of a triangle use trigonometry, side lengths (Heron's formula), vectors, coordinates, line integrals, Pick's theorem, or other properties.

Basic formula

The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height.

The area of a plane figure (two-dimensional shape) is a quantity which describes its size or extent. Conventionally, area is measured in units based on a square whose side is one unit long (for instance, a square meter). When using such units, the area of a rectangle can be found by multiplying together the lengths of two adjacent sides, sometimes called the base, Template:Mvar, and height, Template:Mvar. For example, a rectangle which is 2 meters long and 5 meters tall has area 10 square meters.

A right triangle, one with two sides (called the legs) that are perpendicular with a right angle between them, is half of a rectangle with the two perpendicular sides as its base and height, bisected along a diagonal. Thus the area Template:Mvar of a right triangle can be found by halving the area of the corresponding rectangle; that is, by multiplying the lengths of the two perpendicular sides and dividing the result by two: $T=b\times \frac{1}{2}h.$ Taking either leg of the right triangle to be the base of the triangle, the corresponding altitude of the triangle is the other leg.

Any other triangle, choosing an arbitrary side to be its base, is likewise half of a parallelogram, again bisected along a diagonal, which can then be cut into pieces and rearranged into a rectangle with the same base length and the same height; the triangle's area Template:Mvar can therefore be calculated using the same formula, $T=\frac{1}{2}bh.$ For a general triangle, the altitude does not coincide with any of the sides, but is a new line segment perpendicular to the base with one of its endpoints at the apex (opposite corner) and the other endpoint on the straight line containing the base. The length of this line segment is called the height or altitude of the triangle, and is the shortest distance between the apex and the line containing the base.

Orange triangles Template:Math share a base Template:Mvar and area. The locus of their apex Template:Mvar is a line (dashed green) parallel to the base.

One consequence of the triangle area formula is the analog in the plane of Lexell's theorem about spherical triangles: all of the triangles with a fixed base side, a fixed area, and the apex on the same side of the base line have their apex on a specific line parallel to the base.

Calculations from angles

Applying trigonometry to find the altitude Template:Mvar.

The area Template:Mvar of a triangle can be found through the application of trigonometry.

Knowing SAS (side-angle-side)

Using the labels in the image on the right, the height or altitude is $h=a\sin \gamma .$ Substituting this in the area formula $T=\frac{1}{2}bh$ derived above, the area of the triangle can be expressed as: $$T=\frac{1}{2}ab\sin \gamma =\frac{1}{2}bc\sin \alpha =\frac{1}{2}ca\sin \beta$$ Where Template:Mvar is the line Template:Mvar, Template:Mvar is the line Template:Mvar, Template:Mvar is the line Template:Mvar, Template:Mvar is the interior angle at Template:Mvar, Template:Mvar is the interior angle at Template:Mvar, Template:Mvar is the interior angle at Template:Mvar.

Furthermore, since $\sin \alpha =\sin (\pi -\alpha )=\sin (\beta +\gamma ),$ and similarly for the other two angles: $$T=\frac{1}{2}ab\sin (\alpha +\beta )=\frac{1}{2}bc\sin (\beta +\gamma )=\frac{1}{2}ca\sin (\gamma +\alpha )$$

Knowing AAS (angle-angle-side)

Since $\cot \gamma =-\cot (\pi -\gamma )=-\cot (\alpha +\beta ),$ and similarly for the other two angles: $$T=\frac{b^2}{2(\cot \alpha -\cot (\alpha +\beta ))}=\frac{b^2\sin \alpha \sin (\alpha +\beta )}{2\sin \beta },$$

and analogously if the known side is Template:Mvar or Template:Mvar.

Knowing ASA (angle-side-angle)

$$T=\frac{a^2}{2(\cot \beta +\cot \gamma )}=\frac{a^2\sin \beta \sin \gamma }{2\sin (\beta +\gamma )},$$

and analogously if the known side is Template:Mvar or Template:Mvar.

Using side lengths (Heron's formula) Template:Anchor

A triangle's shape is uniquely determined by the lengths of the sides, so its metrical properties, including area, can be described in terms of those lengths. By Heron's formula, $$T=\sqrt{s(s-a)(s-b)(s-c)}$$

where $s=\frac{1}{2}(a+b+c)$ is the semiperimeter, or half of the triangle's perimeter.

Three other equivalent ways of writing Heron's formula are $$\begin{array}{cc}T& =\frac{1}{4}\sqrt{(a^2+b^2+c^2{)}^2-2(a^4+b^4+c^4)}\\ & =\frac{1}{4}\sqrt{2(a^2{b}^2+a^2{c}^2+b^2{c}^2)-(a^4+b^4+c^4)}\\ & =\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}.\end{array}$$

Formulas resembling Heron's formula

Three formulas have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides Template:Mvar, Template:Mvar, and Template:Mvar respectively as Template:Mvar, Template:Mvar, and Template:Mvar and their semi-sum $\sigma =\frac{1}{2}(m_a+m_b+m_c),$ we have $$T=\frac{4}{3}\sqrt{\sigma (\sigma -m_a)(\sigma -m_b)(\sigma -m_c)}.$$

Next, denoting the altitudes from sides Template:Mvar, Template:Mvar, and Template:Mvar respectively as Template:Mvar, Template:Mvar, and Template:Mvar, and denoting the semi-sum of the reciprocals of the altitudes as $H=\frac{1}{2}(h_a^{-1}+h_b^{-1}+h_c^{-1}),$ we have $$T^{-1}=4\sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}.$$

And denoting the semi-sum of the angles' sines as $S=\frac{1}{2}(\sin \alpha +\sin \beta +\sin \gamma ),$ we have $$T=D^2\sqrt{S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}$$

where Template:Mvar is the diameter of the circumcircle: $D=\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }.$

Using vectors

The area of triangle Template:Math is half of the area of a parallelogram: $$T=\frac{1}{2}\Vert (𝐛-𝐚)\wedge (𝐜-𝐚)\Vert =\frac{1}{2}\Vert 𝐚\wedge 𝐛+𝐛\wedge 𝐜+𝐜\wedge 𝐚\Vert ,$$

where Template:Math, Template:Math, and Template:Math are vectors to the triangle's vertices from any arbitrary origin point, so that Template:Math and Template:Math are the translation vectors from vertex Template:Mvar to each of the others, and Template:Math is the wedge product. If vertex Template:Mvar is taken to be the origin, this simplifies to Template:Nobr.

The oriented relative area of a parallelogram in any affine space, a type of bivector, is defined as Template:Math where Template:Math and Template:Math are translation vectors from one vertex of the parallelogram to each of the two adjacent vertices. In Euclidean space, the magnitude of this bivector is a well-defined scalar number representing the area of the parallelogram. (For vectors in three-dimensional space, the bivector-valued wedge product has the same magnitude as the vector-valued cross product, but unlike the cross product, which is only defined in three-dimensional Euclidean space, the wedge product is well-defined in an affine space of any dimension.)

The area of triangle Template:Math can also be expressed in terms of dot products. Taking vertex Template:Mvar to be the origin and calling translation vectors to the other vertices Template:Math and Template:Math,

$$T=\frac{1}{2}\sqrt{{𝐛}^2{𝐜}^2-(𝐛\cdot 𝐜{)}^2},$$

where for any Euclidean vector Template:Nobr. This area formula can be derived from the previous one using the elementary vector identity Template:Nobr

In two-dimensional Euclidean space, for a vector Template:Math with coordinates Template:Math and vector Template:Math with coordinates Template:Math, the magnitude of the wedge product is $$\Vert 𝐛\wedge 𝐜\Vert =|x_B{y}_C-x_C{y}_B|.$$

(See the following section.)

Using coordinates

If vertex Template:Mvar is located at the origin Template:Math of a Cartesian coordinate system and the coordinates of the other two vertices are given by Template:Math and Template:Math, then the area can be computed as Template:Frac times the absolute value of the determinant $$T=\frac{1}{2}|det\left(\begin{array}{cc}{x}_B& x_C\\ y_B& y_C\end{array}\right)|=\frac{1}{2}|x_B{y}_C-x_C{y}_B|.$$

For three general vertices, the equation is: $$T=\frac{1}{2}|det\left(\begin{array}{ccc}{x}_A& x_B& x_C\\ y_A& y_B& y_C\\ 1& 1& 1\end{array}\right)|=\frac{1}{2}|x_A{y}_B-x_A{y}_C+x_B{y}_C-x_B{y}_A+x_C{y}_A-x_C{y}_B|,$$

which can be written as $$T=\frac{1}{2}|(x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y_A)|.$$

If the points are labeled sequentially in the counterclockwise direction, the above determinant expressions are positive and the absolute value signs can be omitted. The above formula is known as the shoelace formula or the surveyor's formula.

If we locate the vertices in the complex plane and denote them in counterclockwise sequence as Template:Math, Template:Math, and Template:Math, and denote their complex conjugates as Template:Mvar, Template:Mvar, and Template:Mvar, then the formula $$T=\frac{i}{4}\left|\begin{array}{ccc}a& \overline{a}& 1\\ b& \overline{b}& 1\\ c& \overline{c}& 1\end{array}\right|$$

is equivalent to the shoelace formula.

In three dimensions, the area of a general triangle Template:Math, Template:Math and Template:Math) is the Pythagorean sum of the areas of the respective projections on the three principal planes (i.e. Template:Math, Template:Math and Template:Math): $$T=\frac{1}{2}\sqrt{{\left|\begin{array}{ccc}{x}_A& x_B& x_C\\ y_A& y_B& y_C\\ 1& 1& 1\end{array}\right|}^2+{\left|\begin{array}{ccc}{y}_A& y_B& y_C\\ z_A& z_B& z_C\\ 1& 1& 1\end{array}\right|}^2+{\left|\begin{array}{ccc}{z}_A& z_B& z_C\\ x_A& x_B& x_C\\ 1& 1& 1\end{array}\right|}^2}.$$

Using line integrals

The area within any closed curve, such as a triangle, is given by the line integral around the curve of the algebraic or signed distance of a point on the curve from an arbitrary oriented straight line Template:Mvar. Points to the right of Template:Mvar as oriented are taken to be at negative distance from Template:Mvar, while the weight for the integral is taken to be the component of arc length parallel to Template:Mvar rather than arc length itself.

This method is well suited to computation of the area of an arbitrary polygon. Taking Template:Mvar to be the Template:Mvar-axis, the line integral between consecutive vertices Template:Math and Template:Math is given by the base times the mean height, namely $$\frac{1}{2}(x_{i+1}-x_i)(y_i+y_{i+1}).$$ The sign of the area is an overall indicator of the direction of traversal, with negative area indicating counterclockwise traversal. The area of a triangle then falls out as the case of a polygon with three sides.

While the line integral method has in common with other coordinate-based methods the arbitrary choice of a coordinate system, unlike the others it makes no arbitrary choice of vertex of the triangle as origin or of side as base. Furthermore, the choice of coordinate system defined by Template:Mvar commits to only two degrees of freedom rather than the usual three, since the weight is a local distance (e.g. Template:Math in the above) whence the method does not require choosing an axis normal to Template:Mvar.

When working in polar coordinates it is not necessary to convert to Cartesian coordinates to use line integration, since the line integral between consecutive vertices Template:Math and Template:Math of a polygon is given directly by $$\frac{1}{2}{r}_i{r}_{i+1}\sin ({\theta }_{i+1}-{\theta }_i).$$ This is valid for all values of Template:Mvar, with some decrease in numerical accuracy when Template:Math is many orders of magnitude greater than Template:Pi. With this formulation negative area indicates clockwise traversal, which should be kept in mind when mixing polar and cartesian coordinates. Just as the choice of Template:Mvar-axis (Template:Math) is immaterial for line integration in cartesian coordinates, so is the choice of zero heading (Template:Math) immaterial here.

Using Pick's theorem

See Pick's theorem for a technique for finding the area of any arbitrary lattice polygon (one drawn on a grid with vertically and horizontally adjacent lattice points at equal distances, and with vertices on lattice points).

The theorem states: $$T=I+\frac{1}{2}B-1$$

where Template:Mvar is the number of internal lattice points and Template:Mvar is the number of lattice points lying on the border of the polygon.

Other area formulas

Proof without words that the area of a triangle equals the product of its inradius and its semiperimeter

Numerous other area formulas exist, such as $$T=rs,$$

where Template:Mvar is the inradius, and Template:Mvar is the semiperimeter (in fact, this formula holds for all tangential polygons), andTemplate:Rp

$$T=r_a(s-a)=r_b(s-b)=r_c(s-c)$$

where Template:Mvar, Template:Mvar, Template:Mvar are the radii of the excircles tangent to sides Template:Mvar, Template:Mvar, Template:Mvar respectively.

We also have

$$T=\frac{1}{2}{D}^2(\sin \alpha )(\sin \beta )(\sin \gamma )$$

and $$T=\frac{abc}{2D}=\frac{abc}{4R}$$

for circumdiameter Template:Mvar; and $$T=\frac{1}{4}(\tan \alpha )(b^2+c^2-a^2)$$

for angle Template:Math.

The area can also be expressed as $$T=\sqrt{r{r}_a{r}_b{r}_c}.$$

In 1885, Baker gave a collection of over a hundred distinct area formulas for the triangle. These include: $$\begin{array}{cc}T& =\frac{1}{2}\sqrt[3]{abc{h}_a{h}_b{h}_c},\\ & =\frac{1}{2}\sqrt{ab{h}_a{h}_b},\\ & =\frac{a+b}{2(h_a^{-1}+h_b^{-1})},\\ & =\frac{R{h}_b{h}_c}{a}\end{array}$$

for circumradius (radius of the circumcircle) Template:Mvar, and $$T=\frac{h_a{h}_b}{2\sin \gamma }.$$

Upper bound on the area

The area Template:Mvar of any triangle with perimeter Template:Mvar satisfies

$$T\le \frac{p^2}{12\sqrt{3}},$$

with equality holding if and only if the triangle is equilateral.Template:Rp

Other upper bounds on the area Template:Mvar are given byTemplate:Rp

$$4\sqrt{3}T\le a^2+b^2+c^2$$

and

$$4\sqrt{3}T\le \frac{9abc}{a+b+c},$$

both with equality holding if and only if the triangle is equilateral.

History

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.

Bisecting the area

There are infinitely many lines that bisect the area of a triangle. Three of them are the medians, which are the only area bisectors that go through the centroid. Three other area bisectors are parallel to the triangle's sides.

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter. There can be one, two, or three of these for any given triangle.

See also

• Area of a circle
• Area of a quadrilateral
• Congruence of triangles