Difference between revisions of "Shoelace Theorem"

(Theorem)
Line 18: Line 18:
 
==Proof==
 
==Proof==
 
{{incomplete|proof}}
 
{{incomplete|proof}}
 +
 +
== Problems ==
 +
=== Introductory ===
 +
In right triangle <math>ABC</math>, we have <math>\angle ACB=90^{\circ}</math>, <math>AC=2</math>, and <math>BC=3</math>. [[Medians]] <math>AD</math> and <math>BE</math> are drawn to sides <math>BC</math> and <math>AC</math>, respectively. <math>AD</math> and <math>BE</math> intersect at point <math>F</math>. Find the area of <math>\triangle ABF</math>.
 +
  
 
{{stub}}
 
{{stub}}

Revision as of 09:46, 3 May 2009

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 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

Template:Incomplete

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.

Invalid username
Login to AoPS