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

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 ...)
