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Antti H S Laaksonen 2017-02-25 21:12:39 +02:00
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chapter29.tex
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@ -14,7 +14,7 @@ and our task is to calculate its area.
For example, a possible input for the problem is as follows:
\begin{center}
\begin{tikzpicture}[scale=0.45]
\begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (5,6) circle [radius=0.1];
@ -27,7 +27,7 @@ One way to approach the problem is to divide
the quadrilateral into two triangles by a straight
line between two opposite vertices:
\begin{center}
\begin{tikzpicture}[scale=0.45]
\begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (5,6) circle [radius=0.1];
@ -41,7 +41,8 @@ line between two opposite vertices:
After this, it suffices to sum the areas
of the triangles.
The area of a triangle can be calculated,
for example, using \key{Heron's formula}
for example, using \key{Heron's formula}\footnote{Heron of Alexandria
(c. 10--70) was a Greek mathematician.}
\[ \sqrt{s (s-a) (s-b) (s-c)},\]
where $a$, $b$ and $c$ are the lengths
of the triangle's sides and
@ -56,7 +57,7 @@ two arbitrary opposite vertices.
For example, in the following situation,
the division line is outside the quadrilateral:
\begin{center}
\begin{tikzpicture}[scale=0.45]
\begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (3,2) circle [radius=0.1];
@ -69,7 +70,7 @@ the division line is outside the quadrilateral:
\end{center}
However, another way to draw the line works:
\begin{center}
\begin{tikzpicture}[scale=0.45]
\begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (3,2) circle [radius=0.1];
@ -500,7 +501,8 @@ so $b$ is outside the polygon.
\section{Polygon area}
A general formula for calculating the area
of a polygon is
of a polygon\footnote{This formula is sometimes called the
\index{shoelace formula} \key{shoelace formula}.} is
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
where the vertices are
@ -571,7 +573,7 @@ along the boundary of the polygon.
\index{Pick's theorem}
\key{Pick's theorem} provides another way to calculate
\key{Pick's theorem} \cite{pic99} provides another way to calculate
the area of a polygon provided that all vertices
of the polygon have integer coordinates.
According to Pick's theorem, the area of the polygon is