# Wanted : for more formulas to find the area of a triangle?

I know some formulas to find a triangle's area, like the ones below.

1. Is there any reference containing most triangle area formulas?

$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$ Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix} a^2 & b^2 & c^2 \end{bmatrix}\begin{bmatrix} -1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} a^2\\ b^2\\ c^2 \end{bmatrix}$$

Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles) $$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$ If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix} y_1 &z_1 &1 \\ y_2&z_2 &1 \\ y_3 &z_3 &1 \end{vmatrix}^2+\begin{vmatrix} z_1 &x_1 &1 \\ z_2&x_2 &1 \\ z_3 &x_3 &1 \end{vmatrix}^2+\begin{vmatrix} x_1 &y_1 &1 \\ x_2&y_2 &1 \\ x_3 &y_3 &1 \end{vmatrix}^2}$$ When we have 2-d coordinate $$s=\frac{1}{2}\begin{vmatrix} x_a &y_a &1 \\ x_b &y_b &1 \\ x_c &y_c & 1 \end{vmatrix}$$

In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by: $$s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$

• $A=\frac{abc}{4R}$ Jul 23, 2015 at 18:43
• $\iint_{\mathrm{Triangle}}1\,dx\,dy$. :-) Jul 23, 2015 at 18:44
• Is it possible to make reference sheet ,by this (or like this ) question ? Jul 23, 2015 at 18:45
• "Przemysław Scherwentke" what about single integral ? Jul 23, 2015 at 18:51
• @daryakhosrotash Sure, $\int_{bTriangle} xdy$ follows from his formula by Stoke's theorem. Jul 23, 2015 at 20:38

Vectors: The area of a parallelogram embedded in a three-dimensional Euclidean space can be calculated using vectors. Let vectors $AB$ and $AC$ point respectively from $A$ to $B$ and from $A$ to $C$. The area of parallelogram ABDC is then $$\left|AB \times AC\right|$$ so that the area of a triangle is half of this, giving $$A_{\text{triangle}} = \frac{1}{2} |AB \times AC|.$$

Pick's Theorem: $$A_{\text{triangle}} = i + \frac{b}{2} - 1$$ where $i$ is the number of internal lattice points of a triangle and $b$ is the number of lattice points lying on the border of the triangle. As per mathlove: We require that all the triangle's vertices are on lattice points.

• I've never seen this Pick's theorem before. Very cool! (and it might actually be useful for something i'm working on) Jul 23, 2015 at 19:04
• @PaddlingGhost en.wikipedia.org/wiki/Pick%27s_theorem - it's actually quite fascinating and I've seen it talked about mainly on mathoverflow, which makes me think it's very high-level. Enjoy the link! :-) Jul 23, 2015 at 19:06
• @Zain Patel in fact, Pick's theorem has a very elementary proof.
– qwr
Jul 23, 2015 at 20:36
• Note that when the points are defined by coordinates, the cross-product formula is very simple, fast and robust and consequently commonly used. Jul 24, 2015 at 4:03
• @Zain thanks for the link! Aug 26, 2015 at 7:49

A two part paper by Marcus Baker (1849-1903) in vols. 1 and 2 of the Annals of Mathematics, readily available online, gives $110$ such formulae (warning: the Wikipedia article on triangles states that some of them are erroneous).

A collection of formulae for the area of a plane triangle] [Part 1], Annals of Mathematics (1) 1 #6 (January 1885), 134-138. JSTOR link google-books link archive.org link

A collection of formulae for the area of a plane triangle [Part 2], Annals of Mathematics (1) 2 #1 (September 1885), 11-18. JSTOR link google-books link archive.org link

Added as an edit since I can't comment. The links to these articles have been given above. While I'm at it, here is a systematic way to derive these formulae and even find your own new ones. Without loss of generality, one can assume that the vertices are $A=(0,0)$, $B=(1,0)$ and $C=(p,q)$. One can then spend a pleasant hour computing the metric quantities involved in the identities (side lengths, trigonometric functions of the angles, lengths of medians, angle bisectors, altitudes ....) in terms of $p$ and $q$. This reduces the problem to showing that an expression in these variables reduces to $\frac q 2$ or, after squaring, to $\frac{q^2} 4$. This can often be done by hand---in cases of emergency, one can use mathematica.

• can you address it ? you say it's available online ... Aug 4, 2015 at 19:57
• Sorry, I didn't see that you'd mentioned Baker's paper already. If it's OK with you, I'd like to paste the relevant parts of my answer into yours and delete my answer. (I'm surprised anyone knew about Baker's paper, but I guess with everything old being freely available on the internet now, someone was bound to run across it.) Aug 4, 2015 at 21:14
• I was waiting for your response and didn't realize you couldn't comment. I'll put the relevant parts of my answer into yours. Feel free to modify things as you see fit. Aug 5, 2015 at 13:46

I have a little more:) $$S = 4R^2(\sin A + \sin B + \sin C)\sin\frac A2\sin\frac B2\sin\frac C2\\ S = \frac{r^2}{4}\frac{\sin A + \sin B + \sin C}{\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}\\ S = r^2\left(\cot\frac A2+\cot\frac B2 + \cot\frac C2\right)\\ S = r^2\cot\frac A2 \cot\frac B2 \cot\frac C2\\ S = 2p^2 \frac{\sin A\sin B\sin C}{(\sin A+\sin B+\sin C)^2}\\ S = 4p^2 \frac{\sin\dfrac A2 \sin\dfrac B2 \sin\dfrac C2}{\sin A+\sin B+\sin C}$$

• is $p=\frac{a+b+c}{2}$ ? Jul 25, 2015 at 15:08
• @daryakhosrotash, of course, I used your notation Jul 25, 2015 at 15:08
• Thank you , this page is going to be fantastic ! Jul 25, 2015 at 15:10
• @daryakhosrotash, I didn't included formulas which can simply constructed from these (for example, by connections with $r$ and $R$ etc.) Jul 25, 2015 at 15:13

If $W$ is the "hypotenuse-face" of a right-corner tetrahedron, and $X$, $Y$, $Z$ are the (right-triangular) "leg-faces", then

$$W^2 = X^2 + Y^2 + Z^2$$

where, yes, we are squaring areas. (This fact is actually equivalent to Heron's formula for non-obtuse triangles. You can extend it to include obtuse triangles by allowing the tetrahedron to have imaginary(!) edge-lengths at its right corner.)

More generally, if $W$, $X$, $Y$, $Z$ are the faces of a tetrahedron, and $\angle XY$ (etc) represents the dihedral angle between faces $X$ and $Y$, then we have a familiar-looking Law of Cosines:

$$W^2 = X^2 + Y^2 + Z^2 - 2 X Y \cos \angle XY - 2 Y Z \cos \angle YZ - 2 Z X \cos \angle ZX$$

(This is easily proven with vectors.) The above further implies another, more-familiar-looking Law:

\begin{align} W^2 + X^2 - 2 WX \cos\angle WX \;&=\; Y^2 + Z^2 - 2 YZ \cos\angle YZ \\ W^2 + Y^2 - 2 WY \cos\angle WY \;&=\; Z^2 + X^2 - 2 ZX \cos\angle ZX \\ W^2 + Z^2 - 2 WZ \cos\angle WZ \;&=\; X^2 + Y^2 - 2 XY \cos\angle XY \end{align}

(At the point where you say to yourself, "If there's any justice, each of these expressions should equal the square of the area of some face!", you will have inferred the existence of the tetrahedron's "pseudo-faces". But I digress ...)

Using the law of sines, one can get

$$s=\frac{1}{2}bc \sin(\alpha)=\frac{1}{2}bc \frac{a}{2R}=\frac{abc}{4R}$$

where $R$ is the radius of the circumscribed circle.

• note that : Dr. Sonnhard Graubner says this first of all Jul 26, 2015 at 13:28
• @Khosrotash comments should not be used to supply answers. This is a misuse of the comment feature. Apr 24, 2017 at 5:03

Expressing the side lengths a,b & c in term of the radii a',b' & c' of the mutually tangent circles centered on the triangle vertices (which define the Soddy circles) $$a=b'+c'\\b=a'+c'\\c=a'+b'$$give the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$

• This is a special form of Heron's formula. Jul 25, 2015 at 12:05

$$\require{cancel}$$

Some errors spotted in Baker's papers, referenced in the accepted answer.

Expression N20 is definitely wrong (even the dimension):

\begin{align} 20.\ & \cancel{\color{blue}{\frac{R\,r}{\beta_a\beta_b\beta_c}\, \left(\frac1a+\frac1b\right) \left(\frac1b+\frac1c\right) \left(\frac1c+\frac1a\right)}} , \end{align}

and most likely, the term $$\beta_a\beta_b\beta_c$$ must be in the numerator and a constant $$\tfrac12$$ is missing, so the correct form should be

\begin{align} 20.\ & \tfrac12\,{R\,r}{\beta_a\beta_b\beta_c}\, \left(\frac1a+\frac1b\right) \left(\frac1b+\frac1c\right) \left(\frac1c+\frac1a\right) . \end{align}

In the second Baker's paper, expressions N63, N80 are wrong :

\begin{align} 63.\ & \cancel{\color{blue}{R^2\sin 2A (1+\cos C)}} \end{align}

\begin{align} 80.\ & \cancel{\color{blue}{a\,(-\beta_a\,\sin \tfrac12\,A+\beta_b\,\sin \tfrac12\,B+\beta_c\,\sin \tfrac12\,C)}} \\ \quad&= \cancel{\color{blue}{2\,s\,(\beta_a\,\sin \tfrac12\,A+\beta_b\,\sin \tfrac12\,B+\beta_c\,\sin \tfrac12\,C)}} . \end{align}

Also, in N94 the two first expressions are correct: \begin{align} 94.\ & r^2\cot\tfrac12\,A+2\,R\,r\,\sin A \\ &= r_a^2\cot\tfrac12\,A-2\,R\,r_a\,\sin A , \end{align} but there are obvious typos in the third and fourth. They should be

\begin{align} &=r_b^2\cot\tfrac12\,B-2\,R\,r_b\,\sin B \\ &=r_c^2\cot\tfrac12\,C-2\,R\,r_c\,\sin C . \end{align}

I have an expression in terms of dot products of the edges. It works for triangles in any spatial dimension. It only uses dot-products for computation, so is very efficient for use on a computer. It is also quite aesthetically pleasing for its symmetry. No absolute value taken. $$\begin{split} e_0 &= v_2 - v_1\\ e_1 &= v_0 - v_2\\ e_2 &= v_1 - v_0\\ A &= \frac{1}{2} \sqrt{e_0^T e_1 \cdot e_1^Te_2 + e_1^T e_2 \cdot e_2^Te_0 + e_2^T e_0 \cdot e_0^Te_1} \end{split}$$ See meshplex.

The formula can, for example, be derived from the Gram-determinant representation (see https://math.stackexchange.com/a/1742348/36678).

• really very interesting and symmetric! Nov 6, 2019 at 21:39
• Good ole Gram-Determinant. Nov 25, 2019 at 16:51

Incircle bisectors $$d_a,d_b,d_c$$, mentioned in incircle-bisectors-and-related-measures along with the radii of corresponding incircles $$r,r_a,r_b,r_c$$ provide a whole lot of expressions for the area $$S$$ of $$\triangle ABC$$.

Recall that cevian $$AD_a$$ splits the triangle $$ABC$$ into a pair of triangles $$T_1=\triangle ABD_a$$, $$T_2=\triangle AD_aC$$ (assuming that $$A,B,C$$ are in counterclockwise order), such that the radii of their incircles are the same, $$r_{a1}=r_{a2}=r_a$$.

Similar conditions hold for cevians $$BD_b$$ and $$CD_c$$.

Given that $$a,b,c$$ are the sides of $$\triangle ABC$$, $$r$$ is the radius of its inscribed circle and $$\rho=\tfrac12(a+b+c)$$, \begin{align} |AD_a|=d_a&=\sqrt{\rho(\rho-a)} ,\\ r_a&=\frac{r}{1+\sqrt{1-\frac a\rho}} ,\\ l_a&=\frac1{r_a} \end{align}

together with similarly defined $$d_b,d_c,r_b,r_c,l_b,l_c$$, expressions for the area follows:

\begin{align} S&= \frac{d_a d_b d_c}{\sqrt{d_a^2+d_b^2+d_c^2}} \tag{1}\label{1} ,\\ &= \frac{a r_a^2}{2r_a-r} \tag{2}\label{2} ,\\ &= (d_a^2+d_b^2+d_c^2)(\tfrac r{r_a}-1)(\tfrac r{r_b}-1)(\tfrac r{r_c}-1) \tag{3}\label{3} ,\\ &= r_a\,\Big(d_a+\sqrt{d_a^2+d_b^2+d_c^2}\Big) \tag{4}\label{4} ,\\ &= \frac{r^2\,r_a\,r_b\,r_c}{(r-r_a)(r-r_b)(r-r_c)} \tag{5}\label{5} . \end{align}

Note, that \eqref{5} can be expressed in terms of just the three radii $$r_a,r_b,r_c$$ since we can express $$r$$ as \begin{align} r&= \frac{l_a+l_b+l_c+\sqrt{2\,(l_a l_b+l_b l_c+l_c l_a)-(l_a^2+l_b^2+l_c^2)}}{l_a^2+l_b^2+l_c^2} \tag{6}\label{6} \end{align}
and \eqref{5} becomes \begin{align} S&= \frac{(l_a^2+l_b^2+l_c^2)\,(l_a+l_b+l_c+\sigma)^2} {(l_a\,(l_b+l_c+\sigma)-l_b^2-l_c^2)\,(l_b\,(l_c+l_a+\sigma)-l_c^2-l_a^2)\,(l_c\,(l_a+l_b+\sigma)-l_a^2-l_b^2)} \tag{7}\label{7} , \end{align} where \begin{align} \sigma&= \sqrt{2\,l_a\,l_b+2\,l_b\,l_c+2\,l_c\,l_a-l_a^2-l_b^2-l_c^2} \tag{8}\label{8} . \end{align}

Edit

An interesting equation for $$S$$ also exists for the case of uneven bisection, that is, when the corresponding pairs of inscribed circles have different radii, \begin{align} r_{a1}&\ne r_{a2},\quad r_{b1}\ne r_{b2},\quad r_{c1}\ne r_{c2} ,\\ l_{a1}&=1/r_{a1},\quad l_{a2}=1/r_{a2} ,\\ l_{b1}&=1/r_{b1},\quad l_{b2}=1/r_{b2} ,\\ l_{c1}&=1/r_{c1},\quad l_{c2}=1/r_{c2} . \end{align}

In this case we have

\begin{align} S&= r^2\,\sqrt{\frac{r_{a1}r_{a2}r_{b1}r_{b2}r_{c1}r_{c2}}{(r-r_{a1})(r-r_{a2})(r-r_{b1})(r-r_{b2})(r-r_{c1})(r-r_{c2})}} ,\\ r&= \frac{l_{a1}+l_{a2}+l_{b1}+l_{b2}+l_{c1}+l_{c2}}{2 (l_{a1} l_{a2}+l_{b1} l_{b2}+l_{c1} l_{c2})} \\ &+ \frac{\sqrt{(l_{a1}+l_{a2}+l_{b1}+l_{b2}+l_{c1}+l_{c2})^2-8 (l_{a1} l_{a2}+l_{b1} l_{b2}+l_{c1} l_{c2})}} {2 (l_{a1} l_{a2}+l_{b1} l_{b2}+l_{c1} l_{c2})} . \end{align}

Consider $$\triangle ABC$$ and inscribed circle with radius $$r$$. Three smaller similar triangles $$\triangle AcBcC,\ \triangle AbBCb,\ \triangle ABaCa$$ are cut by tangent lines to the incircle parallel to the corresponding sides of $$\triangle ABC$$.

Given the radii $$r_a,\ r_b,\ r_c$$ of the incircles of $$\triangle ABaCa,\ \triangle AbBCb,\ \triangle AcBcC$$, respectively, the area of $$\triangle ABC$$ is found as

\begin{align} \bbox[5px,border:2px solid #C0A000]{ S= \sqrt{\frac{(r_a+r_b+r_c)^7}{r_a\,r_b\,r_c}} } \tag{1} . \end{align}

Also, inradius $$r$$, semiperimeter $$\rho$$, and circumradius $$R$$ of $$\triangle ABC$$ can be found as

\begin{align} r&=r_a+r_b+r_c \tag{2} ,\\ \rho&= \sqrt{\frac{(r_a+r_b+r_c)^5}{r_a\,r_b\,r_c}} \tag{3} ,\\ R&= \tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c} \\ &= \tfrac14\,\frac{(r_a+r_b+r_c)(r_a+r_b)(r_b+r_c)(r_c+r_a)}{r_a r_b r_c} \tag{4} . \end{align}

And the side lengths of $$\triangle ABC$$ can be found explicitly as

\begin{align} a&=r\,(r_b+r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}} ,\\ b&=r\,(r_c+r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}} ,\\ c&=r\,(r_a+r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}} . \end{align}

Two more expressions, similar to Heron’s formula for the area of $$\triangle ABC$$.

Let cevians $$AD,AE,BF,BG,CK,CL$$ be such that (see name-for-the-pair-of-constrained-triples-of-cevians for more info) \begin{align} |BD|&=|CE|=ka ,\quad |CF|=|AG|=kb ,\quad |AK|=|BL|=kc \tag{1} \end{align}

for some $$k\in\mathbb{R}$$, and let \begin{align} |AD|&=d_{a1} ,\quad |BF|=d_{b1} ,\quad |CK|=d_{c1} \tag{2} ,\\ |AE|&=d_{a2} ,\quad |BG|=d_{b2} ,\quad |CL|=d_{c2} \tag{3} ,\\ \delta_1&=\tfrac12\,(d_{a1}+d_{b1}+d_{c1}) \tag{4} ,\\ \delta_2&=\tfrac12\,(d_{a2}+d_{b2}+d_{c2}) \tag{5} . \end{align}

Then for any $$k\in\mathbb{R}$$

\begin{align} S_{ABC}&= \frac{\sqrt{\delta_1(\delta_1-d_{a1})(\delta_1-d_{b1})(\delta_1-d_{c1})}}{k^2-k+1} \tag{6} ,\\ &= \frac{\sqrt{\delta_2(\delta_2-d_{a2})(\delta_2-d_{b2})(\delta_2-d_{c2})}}{k^2-k+1} \tag{7} . \end{align}

In fact, expressions (6)-(7) also include the famous Heron’s formula as a special case when either $$k=0$$ of $$k=1$$, and the triplet of cevians becomes a triplet of the sides $$a,b,c$$.

• Wow, it was amazing ...+1 nice Dec 4, 2020 at 20:27