# Difference between revisions of "Shoelace Theorem"

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==Proof 1== | ==Proof 1== | ||

− | + | Lemma 1: The area of a triangle with coordinates <math>(x_1, y_1, 0)</math>, <math>(x_2, y_2, 0)</math>, and <math>(x_3, y_3, 0)</math> is <math>x_1,y_2-x_2y_1</math>. | |

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+ | Let's translate <math>(x_1, y_1, 0)</math> to the origin so that the other two points are now <math>A'(x_2-x_1, y_2-y_1, 0)</math> and <math>B'(x_3-x_1, y_3-y_1, 0)</math>. | ||

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+ | We will proceed with induction. We start by proving it is true for a triangle: | ||

+ | Let the triangle have coordinates <math>(x_1, y_1)</math> <math>(x_2, y_2)</math> and <math>(x_3, y_3)</math>. To simplify calculations let's translate <math>(x_1, y_1)</math> to the origin. | ||

+ | |||

==Proof 2== | ==Proof 2== | ||

Let <math>\Omega</math> be the set of points belonging to the polygon. | Let <math>\Omega</math> be the set of points belonging to the polygon. |

## Revision as of 13:17, 6 July 2018

The **Shoelace Theorem** is a nifty formula for finding the area of a polygon given the coordinates of its vertices.

## Theorem

Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area of is

The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.

## Proof 1

Lemma 1: The area of a triangle with coordinates , , and is .

Let's translate to the origin so that the other two points are now and .

We will proceed with induction. We start by proving it is true for a triangle: Let the triangle have coordinates and . To simplify calculations let's translate to the origin.

## Proof 2

Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result

## Problems

### Introductory

In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of .

## External Links

A good explanation and exploration into why the theorem works by James Tanton: [1] AOPS