# Author Archive

## Green’s Theorem and Area of Polygons

A common method used to find the area of a polygon is to break the polygon into smaller shapes of known area. For example, one can separate the polygon below into two triangles and a rectangle:

By breaking this composite shape into smaller ones, the area is at hand: $$\begin{align}A_1 &= bh = 5\cdot 2 = 10 \\ A_2 = A_3 &= \frac{bh}{2} = \frac{2\cdot 1}{2} = 1 \\ A_{total} &= A_1+A_2+A_3 = 12\end{align}$$

Unfortunately, this approach can be difficult for a person to use when they cannot physically (or mentally) *see* the polygon, such as when a polygon is given as a list of many vertices.

** Formula **

Happily, there is a formula for the area of any simple polygon that only requires knowledge of the coordinates of each vertex. It is as follows:
$$A = \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \tag{1}$$
*(Where \({n}\) is the number of vertices, \({(x_k, y_k)}\) is the \({k}\)-th point when labelled in a counter-clockwise manner, and \({(x_{n+1}, y_{n+1}) = (x_0, y_0)}\); that is, the starting vertex is found both at the start and end of the list of vertices.)*

It should be noted that the formula is not “symmetric” with respect to the signs of the \({x}\) and \({y}\) coordinates. This can be explained by considering the “negative areas” incurred when adding the signed areas of the triangles with vertices \({(0,0)-(x_k, y_k)-(x_{k+1}, y_{k+1})}\).

In the next sections, I derive this formula using Green’s Theorem, show an example of its use, and provide some applications.