$\newcommand{\mvec}[1]{\mathbf{#1}}\newcommand{\gvec}[1]{\boldsymbol{#1}}\definecolor{eqcol2}{RGB}{114,0,172}\definecolor{eqcol1}{RGB}{172,0,114}\definecolor{eqcol3}{RGB}{0,172,114}\definecolor{eqcol4}{RGB}{230,190,120}$In this article, we shall derive a formula that computes the area of a convex polygon. That is, given the vertices $\mvec{v_0}, \mvec{v_1},\dots,\mvec{v_n}$ of some convex polygon $P$, we want to compute its area, which we denote $A(P)$. We will use the below polygon for illustrations:

It is easy to compute the area of a triangle. Therefore, our strategy will be to first triangulate $P$, and then calculate the sum of the areas of all the triangles. We define $\mvec{o}=(o_x,o_y,o_z)$ to be the centroid of the polygon. Since $P$ is convex, we can simply triangulate $P$ by drawing edges from $\mvec{o}$ to all the vertices. This algorithm yields the triangulation $(\mvec{o},\mvec{v_0},\mvec{v_1}), (\mvec{o},\mvec{v_1},\mvec{v_2}),\dots, (\mvec{o},\mvec{v_{n-1}},\mvec{v_{n}}),(\mvec{o},\mvec{v_{n}},\mvec{v_{0}})$. The below figure illustrates such a triangulation.

In the above case, the triangulation is \begin{equation*} (\mvec{o},\mvec{v_0},\mvec{v_1}),(\mvec{o},\mvec{v_1},\mvec{v_2}),(\mvec{o},\mvec{v_2},\mvec{v_3}),(\mvec{o},\mvec{v_3},\mvec{v_4}),(\mvec{o},\mvec{v_4},\mvec{v_5}), (\mvec{o},\mvec{v_5},\mvec{v_0}). \end{equation*} Next, a formula for computing the area of a single triangle is necessary.

Let us compute the area of the above triangle $(\mvec{v_0},\mvec{v_1},\mvec{v_2})$, and let us call it $T$. It is well-known that the magnitude of the cross product computes the area of a parallelogram. More precisely, $||\mvec{a} \times \mvec{b}||$ is the area of the parallelogram spanned by $\mvec{a}$ and $\mvec{b}$. But this means that $||(\mvec{v_1}-\mvec{v_0}) \times (\mvec{v_2}-\mvec{v_0})||$ is the area of the parallelogram to the right. We can also see that the triangle on the left side of the dashed diagonal is simply $T$, and the triangle on the right side is $T$ again, but this time mirrored along the diagonal. To conclude, the area of the parallelogram is twice the area of $T$. Therefore, \begin{equation*} A(T) = \frac{1}{2}||(\mvec{v_1}-\mvec{v_0}) \times (\mvec{v_2}-\mvec{v_0})||. \end{equation*} For the cross product to be defined, $\mvec{v_0}$,$\mvec{v_1}$ and $\mvec{v_2}$ must be three-dimensional vectors. So the $z$-component is set to $0$, so that we have $\mvec{v_0} = (x_0,y_0,0)$, $\mvec{v_1} = (x_1,y_1,0)$, $\mvec{v_2} = (x_2,y_2,0)$. Simplifying the expression for $A(T)$ gives \begin{align*} A(T) = \frac{1}{2}||(\mvec{v_1}-\mvec{v_0}) \times (\mvec{v_2}-\mvec{v_0})|| =\\ \frac{1}{2} \left|\left|\begin{bmatrix} 0 \\ 0 \\ (x_1-x_0)(y_2-y_0) - (x_2-x_0)(y_1-y_0) \end{bmatrix}\right|\right| =\\ \frac{1}{2} (x_1 y_2 - x_1 y_0 - x_0 y_2 - x_2 y_1 + x_2 y_0 + x_0 y_1). \end{align*} The above formula may appear slightly unwieldy, but if we use it to compute the polygon area $A(P)$, a rather elegant formula falls out. As familiar, $A(P)$ is simply the sum of the areas of the triangles of its triangulation. We have already found this triangulation, so $A(P)$ is \begin{equation} A(P) = A(\mvec{o},\mvec{v_0},\mvec{v_1}) + A(\mvec{o},\mvec{v_1},\mvec{v_2}) + \dots + A(\mvec{o},\mvec{v_{n-1}},\mvec{v_{n}}) + A(\mvec{o},\mvec{v_{n}},\mvec{v_{0}}) = \sum_{i=0}^{n} A(\mvec{o},\mvec{v_{i}},\mvec{v_{i+1}}), \label{eq:apsum} \end{equation} where we have defined $\mvec{v_{n+1}} = \mvec{v_{0}}$, for the purpose of handling the wraparound from $\mvec{v_n}$ to $\mvec{v_0}$. Further, $A(\mvec{v_0},\mvec{v_1},\mvec{v_2})$ is the area of the triangle with vertices $\mvec{v_0}$, $\mvec{v_1}$, and $\mvec{v_2}$.

Notice that the expression $\eqref{eq:apsum}$ is a large sum of terms. Term number $j$ is $A(\mvec{o},\mvec{v_{j}}, \mvec{v_{j+1}})$. Let now $j$ represent the number of an arbitrary term in the sum, so that it can be any of the terms. We will examine the terms with numbers $j-1$, $j$ and $j+1$, and calculate their sum: \begin{align*} A(\mvec{o},\mvec{v_{j-1}},\mvec{v_{j}}) + A(\mvec{o},\mvec{v_{j}},\mvec{v_{j+1}}) + A(\mvec{o},\mvec{v_{j+1}},\mvec{v_{j+2}}) = \\ \frac{1}{2}( \color{black}{x_{j-1} y_{j}} \color{black}{- x_{j-1} o_y} \color{eqcol1}{- o_x y_{j}} \color{black}{- x_{j} y_{j-1}} \color{eqcol2}{+ x_{j} o_y} \color{black}{+ o_x y_{j-1}}\color{black}{+}\\ \color{black}{x_{j} y_{j+1}} \color{eqcol2}{- x_{j} o_y} \color{eqcol3}{- o_x y_{j+1}} \color{black}{- x_{j+1} y_{j}} \color{eqcol4}{+ x_{j+1} o_y} \color{eqcol1}{+ o_x y_{j}}\color{black}{+}\\ \color{black}{x_{j+1} y_{j+2}} \color{eqcol4}{- x_{j+1} o_y} \color{black}{- o_x y_{j+2}} \color{black}{- x_{j+2} y_{j+1}} \color{black}{+ x_{j+2} o_y} \color{eqcol3}{+ o_x y_{j+1}}). \end{align*} It can be seen that for the term $j$(the middle term) only $\frac{1}{2}(x_{j} y_{j+1} - x_{j+1}y_{j})$ is remaining. The rest was cancelled out, as indicated by the colors. Recall that $j$ is an arbitrary index that represents an arbitrary term in $\eqref{eq:apsum}$, and thus the same cancellation will occur for all the terms in the sum for $A(P)$. So we have \begin{equation*} A(P) = \sum_{i=0}^{n} A(\mvec{o},\mvec{v_{i}},\mvec{v_{i+1}}) = \frac{1}{2}\sum_{i=0}^{n} (x_{i} y_{i+1}- x_{i+1}y_{i}). \end{equation*} This concludes our derivation. We have arrived at the famous Shoelace formula. The attractive part of this derivation is that it requires no knowledge of calculus; the only utilized tools were some elementary geometry and vector algebra. If, on the other hand, the powerful tools of calculus are at our disposal, the Shoelace formula can be found using Green's theorem with relative ease.

Source code that implements the derived Shoelace formula is included in a gist.