Difference between revisions of "Shoelace Theorem"

Revision as of 18:39, 28 February 2010

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

Theorem

Suppose the polygon $P$ has vertices $(a_1, b_1)$ , $(a_2, b_2)$ , ... , $(a_n, b_n)$ , listed in clockwise order. Then the area of $P$ is

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

The Shoelace Theorem gets its name because if one lists the the coordinates in a column, $\begin{align*} (a_1 &, b_1) \\ (a_2 &, b_2) \\ & \vdots \\ (a_n &, b_n) \\ (a_1 &, b_1) \end{align*},$ and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.

Proof

Let $\Omega$ be the set of points belonging to the polygon. We have that $A=\int_{\Omega}\alpha,$ where $\alpha=dx\wedge dy$ . The volume form $\alpha$ is an exact form since $d\omega=\alpha$ , where $\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$ Using this substitution, we have $\int_{\Omega}\alpha=\int_{\Omega}d\omega.$ Next, we use the theorem of Green to obtain $\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$ We can write $\partial \Omega=\bigcup A(i)$ , where $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$ . With this notation, we may write $\int_{\partial\Omega}\omega=\sum_{i=1}^n\int_{A(i)}\omega.$ If we substitute for $\omega$ , we obtain $\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$ If we parameterize, we get $\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$ Performing the integration, we get $\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$ More algebra yields the result $\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$

Problems

Introductory

In right triangle $ABC$ , we have $\angle ACB=90^{\circ}$ , $AC=2$ , and $BC=3$ . Medians $AD$ and $BE$ are drawn to sides $BC$ and $AC$ , respectively. $AD$ and $BE$ intersect at point $F$ . Find the area of $\triangle ABF$ .

This article is a stub. Help us out by expanding it.

@@ Line 17: / Line 17: @@
 ==Proof==
-{{incomplete|proof}}
+Let <math>\Omega</math> be the set of points belonging to the polygon.
+We have that
+<cmath>
+A=\int_{\Omega}\alpha,
+</cmath>
+where <math>\alpha=dx\wedge dy</math>.
+The volume form <math>\alpha</math> is an exact form since <math>d\omega=\alpha</math>, where
+<cmath>
+\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}
+</cmath>
+Using this substitution, we have
+<cmath>
+\int_{\Omega}\alpha=\int_{\Omega}d\omega.
+</cmath>
+Next, we use the theorem of Green to obtain
+<cmath>
+\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.
+</cmath>
+We can write <math>\partial \Omega=\bigcup A(i)</math>, where <math>A(i)</math> is the line
+segment from <math>(x_i,y_i)</math> to <math>(x_{i+1},y_{i+1})</math>. With this notation,
+we may write
+<cmath>
+\int_{\partial\Omega}\omega=\sum_{i=1}^n\int_{A(i)}\omega.
+</cmath>
+If we substitute for <math>\omega</math>, we obtain
+<cmath>
+\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.
+</cmath>
+If we parameterize, we get
+<cmath>
+\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}.
+</cmath>
+Performing the integration, we get
+<cmath>
+\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)].
+</cmath>
+More algebra yields the result
+<cmath>
+\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).
+</cmath>
 == Problems ==

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