# MATHS BITE: Shoelace Theorem

The Shoelace theorem is a useful formula for finding the area of a polygon when we know the coordinates of its vertices. The formula was described by Meister in 1769, and then by Gauss in 1795.

### Formula

Let’s suppose that a polygon P has vertices (a1, b1), (a2, b2), …, (an, bn), in clockwise order. Then the area of P is given by

$$\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|$$

The name of this theorem comes from the fact that if you were to list the coordinates in a column and mark the pairs to be multiplied, then the image looks like laced-up shoes.

### Proof

(Note: this proof is taken from artofproblemsolving.)

Let $\Omega$ be the set of points that belong to the polygon. Then

$$A=\int_{\Omega}\alpha,$$

where $\alpha=dx\wedge dy$.

Note that the volume form $\alpha$ is an exact form since $d\omega=\alpha$, where

$$\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$$

Substitute this in to give us

$$\int_{\Omega}\alpha=\int_{\Omega}d\omega.$$

and then use Stokes’ theorem (a key theorem in vector calculus) to obtain

$$\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$$

where

$\partial \Omega=\bigcup A(i)$

and $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$, i.e.  is the boundary of the polygon.

Next we substitute for $\omega$:

$$\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$$

Parameterising this expression gives us

$$\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$$

Then, by integrating this we obtain

$$\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$$

This then yields, after further manipulation, the shoelace formula:

$$\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$

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