# Posts Tagged ‘polygons’

## Green’s Theorem and Area of Polygons

June 4, 2014 by apnorton. 9 comments

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.

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