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: $$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
Solution 1:
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.
Solution 2:
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.
Solution 3:
$s=pr$ where $p=\frac{a+b+c}{2}$ and $r$ is the radius of the inscribed circle.
$s=\sqrt{r\cdot r_a\cdot r_b\cdot r_c}$ where $r_a,r_b,r_c$ are the exradii of excircles.