Area of a triangle

Ingeometry,calculating theareaof atriangleis an elementary problem encountered often in many different situations. The best known and simplest formula is$T=bh/2,$wherebis thelengthof thebaseof the triangle, andhis theheightoraltitudeof the triangle. The term "base" denotes any side, and "height" denotes the length of a perpendicular from the vertex opposite the base onto the line containing the base.Euclidproved that the area of a triangle is half that of a parallelogram with the same base and height in his bookElementsin 300 BCE.In 499 CEAryabhata,used this illustrated method in theAryabhatiya(section 2.6).

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

A graphic derivation of the formula$T={\frac {h}{2}}b$that avoids the usual procedure of doubling the area of the triangle and then halving it.

Although simple, this formula is only useful if the height can be readily found, which is not always the case. For example, theland surveyorof 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.

History

Heron of Alexandriafound what is known asHeron's formulafor 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 thatArchimedesknew the formula over two centuries earlier,and sinceMetricais 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 300 BCE Greek mathematicianEuclidproved that the area of a triangle is half that of a parallelogram with the same base and height in his bookElements of Geometry.

In 499Aryabhata,a greatmathematician-astronomerfrom the classical age ofIndian mathematicsandIndian astronomy,expressed the area of a triangle as one-half the base times the height in theAryabhatiya.

A formula equivalent to Heron's was discovered by the Chinese independently of the Greeks. It was published in 1247 inShushu Jiuzhang( "Mathematical Treatise in Nine Sections"), written byQin Jiushao.

Using trigonometry

Applying trigonometry to find the altitudeh.

The height of a triangle can be found through the application oftrigonometry.

Knowing SAS (side-angle-side)

Using the labels in the image on the right, the altitude ish=asin$\gamma$.Substituting this in the formula$T={\tfrac {1}{2}}bh$derived above, the area of the triangle can be expressed as:

$T={\tfrac {1}{2}}ab\sin \gamma ={\tfrac {1}{2}}bc\sin \alpha ={\tfrac {1}{2}}ca\sin \beta$

(where α is the interior angle atA,β is the interior angle atB,$\gamma$is the interior angle atCandcis the lineAB).

Furthermore, since sin α = sin (π− α) = sin (β +$\gamma$), and similarly for the other two angles:

$T={\tfrac {1}{2}}ab\sin(\alpha +\beta )={\tfrac {1}{2}}bc\sin(\beta +\gamma )={\tfrac {1}{2}}ca\sin(\gamma +\alpha ).$

Knowing AAS (angle-angle-side)

$T={\frac {b^{2}(\sin \alpha )(\sin(\alpha +\beta ))}{2\sin \beta }},$

and analogously if the known side isaorc.

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 isborc.

Using side lengths (Heron's formula)

Main article:Heron's formula

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. ByHeron's formula,

$T={\sqrt {s(s-a)(s-b)(s-c)}}$

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

Three other equivalent ways of writing Heron's formula are

${\begin{aligned}T&={\tfrac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}\\[5mu]&={\tfrac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}\\[5mu]&={\tfrac {1}{4}}{\sqrt {(a+b-c)(a-b+c)(-a+b+c)(a+b+c)}}.\end{aligned}}$

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 sidesa,b,andcrespectively asm_a,m_b,andm_cand their semi-sum(m_a+m_b+m_c)/2as σ, we have

$T={\tfrac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.$

Next, denoting the altitudes from sidesa,b,andcrespectively ash_a,h_b,andh_c,and denoting the semi-sum of the reciprocals of the altitudes as$H=(h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1})/2$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 asS= [(sin α) + (sin β) + (sin γ)]/2,we have

$T=D^{2}{\sqrt {S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}}$

whereDis the diameter of thecircumcircle:$D={\tfrac {a}{\sin \alpha }}={\tfrac {b}{\sin \beta }}={\tfrac {c}{\sin \gamma }}.$

Using vectors

The area of triangle ABC is half of the area of aparallelogram:

$T={\tfrac {1}{2}}{\bigl \|}(\mathbf {b} -\mathbf {a} )\wedge (\mathbf {c} -\mathbf {a} ){\bigr \|}={\tfrac {1}{2}}{\bigl \|}\mathbf {b} \wedge \mathbf {c} +\mathbf {b} \wedge \mathbf {c} +\mathbf {c} \wedge \mathbf {a} {\bigr \|},$

where⁠$\mathbf {a}$⁠,⁠$\mathbf {b}$⁠,and⁠$\mathbf {c}$⁠arevectorsto the triangle's vertices from any arbitrary origin point, so that⁠$\mathbf {b} -\mathbf {a}$⁠and⁠$\mathbf {c} -\mathbf {b}$⁠are thetranslation vectorsfrom vertex⁠$A$⁠to each of the others, and⁠$\wedge$⁠is thewedge product.If vertex⁠$A$⁠is taken to be the origin, this simplifies to${\tfrac {1}{2}}\|\mathbf {b} \wedge \mathbf {c} \|$.

The oriented relativearea of a parallelogramin any affine space, a type ofbivector,is defined as⁠$\mathbf {u} \wedge \mathbf {v}$⁠where⁠$\mathbf {u}$⁠and⁠$\mathbf {v}$⁠are translation vectors from one vertex of the parallelogram to each of the two adacent 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-valuedcross 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 triangleABCcan also be expressed in terms ofdot products.Taking vertex⁠$A$⁠to be the origin and calling translation vectors to the other vertices⁠$\mathbf {b}$⁠and⁠$\mathbf {c}$⁠,

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

where for any Euclidean vector$\mathbf {v} ^{2}=\|\mathbf {v} \|^{2}=\mathbf {v} \cdot \mathbf {v}$.This area formula can be derived from the previous one using the elementary vector identity$\mathbf {u} ^{2}\mathbf {v} ^{2}=(\mathbf {u} \cdot \mathbf {v} )^{2}+\|\mathbf {u} \wedge \mathbf {v} \|^{2}$.

In two-dimensional Euclidean space, for a vector⁠$\mathbf {b}$⁠with coordinates⁠$(x_{B},y_{B})$⁠and vector⁠$\mathbf {c}$⁠with coordinates⁠$(x_{C},y_{C})$⁠,the magnitude of the wedge product is

$\|\mathbf {b} \wedge \mathbf {c} \|=|x_{B}y_{C}-x_{C}y_{B}|.$

(See the following section.)

Using coordinates

If vertexAis located at the origin (0, 0) of aCartesian coordinate systemand the coordinates of the other two vertices are given byB= (x_B,y_B)andC= (x_C,y_C),then the area can be computed as1⁄2times theabsolute valueof thedeterminant

$T={\tfrac {1}{2}}\left|\det {\begin{pmatrix}x_{B}&x_{C}\\y_{B}&y_{C}\end{pmatrix}}\right|={\tfrac {1}{2}}|x_{B}y_{C}-x_{C}y_{B}|.$

For three general vertices, the equation is:

$T={\tfrac {1}{2}}\left|\det {\begin{pmatrix}x_{A}&x_{B}&x_{C}\\y_{A}&y_{B}&y_{C}\\1&1&1\end{pmatrix}}\right|={\tfrac {1}{2}}{\big |}x_{A}y_{B}-x_{A}y_{C}+x_{B}y_{C}-x_{B}y_{A}+x_{C}y_{A}-x_{C}y_{B}{\big |},$

which can be written as

$T={\tfrac {1}{2}}{\big |}(x_{A}-x_{C})(y_{B}-y_{A})-(x_{A}-x_{B})(y_{C}-y_{A}){\big |}.$

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 theshoelace formulaor the surveyor's formula.

If we locate the vertices in the complex plane and denote them in counterclockwise sequence asa=x_A+y_Ai,b=x_B+y_Bi,andc=x_C+y_Ci,and denote their complex conjugates as${\bar {a}}$,${\bar {b}}$,and${\bar {c}}$,then the formula

$T={\frac {i}{4}}{\begin{vmatrix}a&{\bar {a}}&1\\b&{\bar {b}}&1\\c&{\bar {c}}&1\end{vmatrix}}$

is equivalent to the shoelace formula.

In three dimensions, the area of a general triangleA= (x_A,y_A,z_A),B= (x_B,y_B,z_B)andC= (x_C,y_C,z_C) is thePythagorean sumof the areas of the respective projections on the three principal planes (i.e.x= 0,y= 0 andz= 0):

$T={\tfrac {1}{2}}{\sqrt {{\begin{vmatrix}x_{A}&x_{B}&x_{C}\\y_{A}&y_{B}&y_{C}\\1&1&1\end{vmatrix}}^{2}+{\begin{vmatrix}y_{A}&y_{B}&y_{C}\\z_{A}&z_{B}&z_{C}\\1&1&1\end{vmatrix}}^{2}+{\begin{vmatrix}z_{A}&z_{B}&z_{C}\\x_{A}&x_{B}&x_{C}\\1&1&1\end{vmatrix}}^{2}}}.$

Using line integrals

The area within any closed curve, such as a triangle, is given by theline integralaround the curve of the algebraic or signed distance of a point on the curve from an arbitrary oriented straight lineL.Points to the right ofLas oriented are taken to be at negative distance fromL,while the weight for the integral is taken to be the component of arc length parallel toLrather than arc length itself.

This method is well suited to computation of the area of an arbitrarypolygon.TakingLto be thex-axis, the line integral between consecutive vertices (x_i,y_i) and (x_i+1,y_i+1) is given by the base times the mean height, namely(x_i+1−x_i)(y_i+y_i+1)/2.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 byLcommits to only two degrees of freedom rather than the usual three, since the weight is a local distance (e.g.x_i+1−x_iin the above) whence the method does not require choosing an axis normal toL.

When working inpolar coordinatesit is not necessary to convert toCartesian coordinatesto use line integration, since the line integral between consecutive vertices (r_i,θ_i) and (r_i+1,θ_i+1) of a polygon is given directly byr_ir_i+1sin(θ_i+1− θ_i)/2.This is valid for all values of θ, with some decrease in numerical accuracy when |θ| is many orders of magnitude greater than π. With this formulation negative area indicates clockwise traversal, which should be kept in mind when mixing polar and cartesian coordinates. Just as the choice ofy-axis (x= 0) is immaterial for line integration in cartesian coordinates, so is the choice of zero heading (θ = 0) immaterial here.

Using Pick's theorem

SeePick's theoremfor a technique for finding the area of any arbitrarylattice 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:

$T=I+{\tfrac {1}{2}}B-1$

where$I$is the number of internal lattice points andBis the number of lattice points lying on the border of the polygon.

Other area formulas

Numerous other area formulas exist, such as

$T=r\cdot s,$

whereris theinradius,andsis thesemiperimeter(in fact, this formula holds foralltangential polygons), and

$T=r_{a}(s-a)=r_{b}(s-b)=r_{c}(s-c)$

where$r_{a},\,r_{b},\,r_{c}$are the radii of theexcirclestangent to sidesa, b, crespectively.

We also have

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

and

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

for circumdiameterD;and

$T={\tfrac {1}{4}}(\tan \alpha )(b^{2}+c^{2}-a^{2})$

for angle α ≠ 90°.

The area can also be expressed as

$T={\sqrt {rr_{a}r_{b}r_{c}}}.$

In 1885, Bakergave a collection of over a hundred distinct area formulas for the triangle. These include:

$T={\tfrac {1}{2}}{\sqrt[{3}]{abch_{a}h_{b}h_{c}}},$

$T={\tfrac {1}{2}}{\sqrt {abh_{a}h_{b}}},$

$T={\frac {a+b}{2(h_{a}^{-1}+h_{b}^{-1})}},$

$T={\frac {Rh_{b}h_{c}}{a}}$

for circumradius (radius of the circumcircle)R,and

$T={\frac {h_{a}h_{b}}{2\sin \gamma }}.$

Upper bound on the area

The areaTof any triangle with perimeterpsatisfies

$T\leq {\tfrac {p^{2}}{12{\sqrt {3}}}},$

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

Other upper bounds on the areaTare given by

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

and

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

both again holding if and only if the triangle is equilateral.

Bisecting the area

There are infinitely manylines 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
• Congruence of triangles