2013-2014/4e/Geometrie/Thales/activite_decouverte/act_dec.tex
2017-06-16 09:46:40 +03:00

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\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
\usepackage{tikz}
\usetikzlibrary{calc}
% Title Page
\title{Thalès - Travail de groupe}
\author{}
\date{}
\fancyhead[L]{Quatrième}
\fancyhead[C]{\Thetitle}
\fancyhead[R]{\thepage}
\begin{document}
\thispagestyle{fancy}
Dans tous les dessins suivants, $ABC$ et $AMN$ sont deux triangles tels que $M$ est un point de $[AB]$, $N$ un point de $[AC]$ et $(MN)//(BC)$. Dans chacunes des figures, les longueurs de quatres côtés sont données. (Les mesures sur les dessins ne sont pas respectées)
\textbf{Conjecturer les valeurs des longueurs des deux cotés manquants pour remplir les tableaux des longueurs.} Justifier quand c'est possible les valeurs trouvées.
\begin{itemize}
\item \textbf{Cas 1}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (M) at (3,0);
\coordinate (B) at (6,0);
\coordinate (C) at (6,4);
\coordinate (N) at (3,2);
\draw (A) node [left] {$A$}
-- (M) node [below] {$M$} node[midway, below] {$4$}
-- (B) node [right] {$B$}
-- (C) node [right] {$C$}
-- (N) node [above] {$N$}
-- (A) node [midway, sloped, above] {5};
\draw (N) -- (M) node[midway, right] {3};
\draw[<->] ($(A)+(0,-0.5)$) -- ($(B) + (0,-0.5)$) node [midway, below] {$8$};
\end{tikzpicture}
\begin{tabular}{|c|*{3}{p{2cm}|}}
\hline
Triangle $AMN$ & AM = & AN = & MN = \\
\hline
Triangle $ABC$ & AB = & AC = & BC = \\
\hline
\end{tabular}
\item \textbf{Cas 2}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (M) at (2.3,0);
\coordinate (B) at (9.2,0);
\coordinate (C) at (8,4);
\coordinate (N) at (2,1);
\draw (A) node [left] {$A$}
-- (M) node [below] {$M$} node[midway, below] {$1$}
-- (B) node [right] {$B$}
-- (C) node [right] {$C$}
-- (N) node [above] {$N$}
-- (A) node [midway, sloped, above] {5};
\draw (N) -- (M) node[midway, right] {3};
\draw[<->] ($(A)+(0,-0.5)$) -- ($(B) + (0,-0.5)$) node [midway, below] {4};
\draw (A) --++ (2.3,0) --++ (2.3,0) node {$|$} --++ (2.3,0) node {$|$};
\end{tikzpicture}
\begin{tabular}{|c|*{3}{p{2cm}|}}
\hline
Triangle $AMN$ & AM = & AN = & MN = \\
\hline
Triangle $ABC$ & AB = & AC = & BC = \\
\hline
\end{tabular}
\item \textbf{Cas 3}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (M) at (3,0);
\coordinate (B) at (6,0);
\coordinate (C) at (6.8,4);
\coordinate (N) at (3.4,2);
\draw (A) node [left] {$A$}
-- (M) node [below] {$M$} node[midway, below] {$4$}
-- (B) node [right] {$B$}
-- (C) node [right] {$C$}
-- (N) node [above] {$N$}
-- (A) node [midway, sloped, above] {5};
\draw (N) -- (M) node[midway, right] {4.4};
\draw[<->] ($(A)+(-0.5,+0.5)$) -- ($(C) + (0,+0.75)$) node [midway, above, sloped] {$10$};
\end{tikzpicture}
\begin{tabular}{|c|*{3}{p{2cm}|}}
\hline
Triangle $AMN$ & AM = & AN = & MN = \\
\hline
Triangle $ABC$ & AB = & AC = & BC = \\
\hline
\end{tabular}
\item \textbf{Cas 4}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (M) at (3,0);
\coordinate (B) at (6,0);
\coordinate (C) at (6,4);
\coordinate (N) at (3,2);
\draw (A) node [left] {$A$}
-- (M) node [below] {$M$} node[midway, below] {$4$}
-- (B) node [right] {$B$}
-- (C) node [right] {$C$} node [midway, right] {15}
-- (N) node [above] {$N$}
-- (A) node [midway, sloped, above] {5};
\draw (N) -- (M) node[midway, right] {3};
\end{tikzpicture}
\begin{tabular}{|c|*{3}{p{2cm}|}}
\hline
Triangle $AMN$ & AM = & AN = & MN = \\
\hline
Triangle $ABC$ & AB = & AC = & BC = \\
\hline
\end{tabular}
\end{itemize}
\end{document}
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