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.


*

*Is there any reference containing most triangle area formulas?  

*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 :$$(4s)^2=\begin{bmatrix}
a^2 &  b^2 & c^2  
\end{bmatrix}\begin{bmatrix}
-1 & 1  & 1\\ 
1 &  -1 & 1\\ 
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\ 
b^2\\ 
c^2
\end{bmatrix}$$

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

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}
