<s>[INST] 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:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
 [/INST] 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}
Related answer.
</s>