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

  1. Is there any reference containing most triangle area formulas?
  2. If you know more, please add them as an answer

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

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 $$ s=\frac{1}{2}\begin{vmatrix} x_a &y_a &1 \\ x_b &y_b &1 \\ x_c &y_c & 1 \end{vmatrix}$$ enter image description here

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: $$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$

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    $\begingroup$ $A=\frac{abc}{4R}$ $\endgroup$ Commented Jul 23, 2015 at 18:43
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    $\begingroup$ $\iint_{\mathrm{Triangle}}1\,dx\,dy$. :-) $\endgroup$ Commented Jul 23, 2015 at 18:44
  • $\begingroup$ Is it possible to make reference sheet ,by this (or like this ) question ? $\endgroup$
    – Khosrotash
    Commented Jul 23, 2015 at 18:45
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    $\begingroup$ @daryakhosrotash Sure, $\int_{bTriangle} xdy$ follows from his formula by Stoke's theorem. $\endgroup$ Commented Jul 23, 2015 at 20:38
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    $\begingroup$ $$s=pr$$($r$ is inradius.) Surprisingly, I haven't seen it mentioned in the question/comments/answers. $\endgroup$ Commented Aug 14, 2022 at 4:53

13 Answers 13

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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.

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  • $\begingroup$ I've never seen this Pick's theorem before. Very cool! (and it might actually be useful for something i'm working on) $\endgroup$ Commented Jul 23, 2015 at 19:04
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    $\begingroup$ @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! :-) $\endgroup$
    – Zain Patel
    Commented Jul 23, 2015 at 19:06
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    $\begingroup$ @Zain Patel in fact, Pick's theorem has a very elementary proof. $\endgroup$
    – qwr
    Commented Jul 23, 2015 at 20:36
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    $\begingroup$ Note that when the points are defined by coordinates, the cross-product formula is very simple, fast and robust and consequently commonly used. $\endgroup$
    – Keith
    Commented Jul 24, 2015 at 4:03
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    $\begingroup$ @Zain thanks for the link! $\endgroup$
    – Aneek
    Commented Aug 26, 2015 at 7:49
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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.

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  • $\begingroup$ can you address it ? you say it's available online ... $\endgroup$
    – Khosrotash
    Commented Aug 4, 2015 at 19:57
  • $\begingroup$ 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.) $\endgroup$ Commented Aug 4, 2015 at 21:14
  • $\begingroup$ 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. $\endgroup$ Commented Aug 5, 2015 at 13:46
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I have a little more:) $$ S = 4R^2(\sin A + \sin B + \sin C)\sin\frac A2\sin\frac B2\sin\frac C2\\ S = \frac{r^2}{4}\frac{\sin A + \sin B + \sin C}{\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}\\ S = r^2\left(\cot\frac A2+\cot\frac B2 + \cot\frac C2\right)\\ S = r^2\cot\frac A2 \cot\frac B2 \cot\frac C2\\ S = 2p^2 \frac{\sin A\sin B\sin C}{(\sin A+\sin B+\sin C)^2}\\ S = 4p^2 \frac{\sin\dfrac A2 \sin\dfrac B2 \sin\dfrac C2}{\sin A+\sin B+\sin C} $$

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

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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.

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  • $\begingroup$ note that : Dr. Sonnhard Graubner says this first of all $\endgroup$
    – Khosrotash
    Commented Jul 26, 2015 at 13:28
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    $\begingroup$ @Khosrotash comments should not be used to supply answers. This is a misuse of the comment feature. $\endgroup$
    – miracle173
    Commented Apr 24, 2017 at 5:03
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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 $$s=\sqrt{a'b'c'(a'+b'+c')}$$

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  • $\begingroup$ This is a special form of Heron's formula. $\endgroup$
    – wythagoras
    Commented Jul 25, 2015 at 12:05
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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}

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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).

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  • $\begingroup$ really very interesting and symmetric! $\endgroup$
    – G Cab
    Commented Nov 6, 2019 at 21:39
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    $\begingroup$ Good ole Gram-Determinant. $\endgroup$ Commented Nov 25, 2019 at 16:51
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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}

enter image description here

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}

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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 $\triangle ABC$.

enter image description here

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}

Related answer.

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

enter image description here

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 $a,b,c$.

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  • $\begingroup$ Wow, it was amazing ...+1 nice $\endgroup$
    – Khosrotash
    Commented Dec 4, 2020 at 20:27
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In light of the answers which have appeared after mine (under the moniker ring), I thought it might be of interest to expand on it. This is because there is a general principle at work here: given three quantities associated with a triangle, any additional quantity (in particular, area) can be directly computed from them. Of course, in this generality, it is only a guideline, not in any sense a theorem, but if used with some common sense, it can lead to interesting results. A typical case is Heron´s formula which shows how you can calculate the area from the side-lengths of the triangle.

One can give a formal definition of the concept of a quantity in this context but it suffices for our purposes that it is a number which can be computed from the coordinates of the vertices of a given triangle (side lengths, angles, lengths of medians, area, perimeter, ...).

We illustrate the method by using it to rediscover and prove Heron´s formula. The basic method is the following (we use the notion $A_1$, $A_2$, $A_3$ to denote the vertices of a triangle, $e_{23}$, $e_{31}$, $e_{12}$ the corresponding side-lengths). The basic ingredient is the so-called $p,q$ method--we choose coordinates so that the vertices are $(0,0)$, $(e_{12},0)$ and $(p,q)$. We now compute the basic quantities--the side-lengths--in terms of $p$, $q$ and $e_{12}$. This leads to the equations: $$e_{12}=e_{12},\,(p-e_{12})^2+q^2=e_{32}^2,\,p^2+q^2=e_{31}^2.$$ (we have included the first equation which is, of course, trivial--this will not be the case in general). These equations can easily solved by hand to obtain $p$ and $q$ in terms of the side lengths. We can then compute the area in terms of the latter using the formula $A=\frac 12 q\, e_{12}$. If the reader carries out the computations, he will find that he has proved Heron´s formula. Note that as we have $p$ and $q$, we can compute any triangle quantities in terms of the side-lengths. For example the angle at $A_1$ is the inverse tangent of $\frac q p$.

In order to illustrate this method, we sketch a proof of a generalisation of the generalisation of Heron given in the answer by rove. With the above notion we consider points $B_1$, $B_2$ and $B_3$ on the sides opposite to the corresponding vertices, whereby

$$\frac{|A_2 B_1|}{|A_2 A_3|}=\lambda, \, \frac{|A_3 B_2|}{|A_3 A_1|}=\mu, \, \frac{|A_1 B_3|}{|A_2 A_1|}=\nu.$$

Our intention is to compute the area in terms of the lengths $A A_{1}$ (and, of course, of the parameters),etc. If we denote them by $e_1$, $e_2$ and $e_3$, then, as above, we get equations $$\frac 1{\lambda^2}((1-\lambda)^2 e_{12}^2+2\lambda(1-\lambda)e_{12}p+\lambda^2 p^2)+q^2=\frac 1{e_1^2}{\lambda^2},$$ $$\frac 1{(1-\mu)^2}((1-\mu)^2 -2(1-\mu)p =e_{12}^2 p^2 e_{12}^2)+q^2=\frac 1{e_2^2}{(1-\mu)^2}$$ and $$p^2-2 \nu p+ \nu^2 e_{12}^2+q^2=e_3^2. $$ These can be solved by hand (eliminate the terms $p^2+q^2$ to reduce to linear equations in $p$ and $q$). The resulting formula is rather gruesome and probably not too interesting as a theorem. However, we have answered the principle question of whether the area can be computed from the initial data. The complicated formula indicates that it will not be a simple task to work out a synthetic proof. Of course, it contains as special case, Heron, the latter´s generalisation to the median lengths and to the lengths of the feet of the angle bisectors (together with those of other cevians associated to known triangle centres of which there are now over 40,000!).

This sketch should suffice to show how to solve the problem of expressing the area (and, indeed, any other triangle quantity) in terms of three given ones. Express the given quantities in terms of $p$, $q$ and $e_{12}$ and solve the resulting equations for the latter. At least in principle--there is no guarantee that in a given situation the resulting equations can be solved by hand or, say,by using Mathematica. I have, however, tried it out in many concrete cases and if one chooses quantities which are sensible and simple enough, it leads to a myriad of formulae of various levels of interest.

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