Import work from year 2013-2014
This commit is contained in:
Benjamin Bertrand
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3e/DS/DS_131017/10_thales_arithm.pdf
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3e/DS/DS_131017/10_thales_arithm.pdf
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3e/DS/DS_131017/10_thales_arithm.tex
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3e/DS/DS_131017/10_thales_arithm.tex
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\documentclass[a4paper,10pt]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classDS}
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\usepackage{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/2013_2014}
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% Title Page
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\titre{2}
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% \quatreC \quatreD \troisB \troisPro
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\classe{\troisB}
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\date{17 octobre 2013}
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\duree{1 heure}
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\sujet{%{{infos.subj%}}}
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% DS DSCorr DM DMCorr Corr
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\typedoc{DS}
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\begin{document}
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\maketitle
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Le barème est donné à titre indicatif, il pourra être modifié.
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\begin{Exo}
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Dans les marais salants, le sel récolté est stocké en tas sur une surface plance. On admet que le tas de sel a la forme d'un cône de révolution. La situation peut être modélisé par les deux dessins ci dessous.
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\begin{minipage}{0.5\textwidth}
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\includegraphics[scale=0.25]{./fig/cone3D.pdf}
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\end{minipage}
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\begin{minipage}{0.5\textwidth}
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\includegraphics[scale=0.25]{./fig/cone_cote.pdf}
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\end{minipage}
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\begin{enumerate}
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\item Démontrer que la hauteur de ce cône de sel est égale à 2,5 mètres.
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\item À l'aide de la formule $V_{cone} = \frac{\pi \times rayon^2 \times hauteur}{3}$, determiner, en $m^3$, le volume de sel contenu dans le cône. Arrondir le résultat au $m^3$ près.
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\end{enumerate}
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\end{Exo}
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\begin{Exo}
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Les deux coônes de récolution de rayons $KA$ et $IB$ sont opposés par le sommet.
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\begin{minipage}{0.5\textwidth}
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\includegraphics[scale=0.25]{./fig/sablier.pdf}
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\end{minipage}
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\begin{minipage}{0.5\textwidth}
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Les droites $(AB)$ et $(KI)$ se coupent en $S$ et $(BI)$ est parallèle à $(KA)$. On donne $KA = 4,5cm$, $KS = 6cm$ et $SI = 4cm$.
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Calculer $BI$.
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\end{minipage}
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\end{Exo}
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\begin{Exo}
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\begin{enumerate}
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\item Donner la liste des diviseurs de 207 et 138.
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\item Simplifier la fraction $\frac{207}{138}$.
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\item En écrivant les premiers multiples de 11, 2 et 3, trouver le plus petit multiple commun de ces trois chiffres.
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\item Calculer $A = \frac{2}{11} + \frac{3}{2} + \frac{4}{3}$ et donner le résultat sous la forme d'une fraction irreductible.
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\end{enumerate}
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\end{Exo}
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\begin{Exo}
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Calculer en détaillant les étapes et donner le résultat sous forme d'une fraction irreductible.
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\begin{eqnarray*}
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A = \frac{ %{{random.lst_int(2,100,4)|join("\\times")%}} }{%{{random.lst_int(2,100,3)|join("\\times")%}}} &\quad &
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B = %{{ random.frac() %}} + %{{ random.frac() %}} \times %{{ random.frac() %}} \\
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C = \left( %{{ random.frac() %}} - %{{ random.frac() %}} \right) \times %{{ random.frac() %}} & \quad &
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D = %{{ random.frac() %}} \div \left( %{{ random.frac() %}} + %{{ random.frac() %}} \right)
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\end{eqnarray*}
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\end{Exo}
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "master"
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%%% End:
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BIN
3e/DS/DS_131017/10_thales_arithm_1.pdf
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3e/DS/DS_131017/10_thales_arithm_1.pdf
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Binary file not shown.
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3e/DS/DS_131017/10_thales_arithm_1.tex
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3e/DS/DS_131017/10_thales_arithm_1.tex
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\documentclass[a4paper,10pt]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classDS}
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||||
\usepackage{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/2013_2014}
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||||
|
||||
% Title Page
|
||||
\titre{2}
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||||
% \quatreC \quatreD \troisB \troisPro
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||||
\classe{\troisB}
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||||
\date{17 octobre 2013}
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\duree{1 heure}
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\sujet{1}
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% DS DSCorr DM DMCorr Corr
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\typedoc{DS}
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||||
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||||
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||||
\begin{document}
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\maketitle
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||||
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Le barème est donné à titre indicatif, il pourra être modifié.
|
||||
|
||||
|
||||
\begin{Exo}
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||||
Les deux coônes de révolution de rayons $KA$ et $IB$ sont opposés par le sommet.
|
||||
|
||||
\begin{minipage}{0.5\textwidth}
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||||
\includegraphics[scale=0.25]{./fig/sablier.pdf}
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||||
\end{minipage}
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
Les droites $(AB)$ et $(KI)$ se coupent en $S$ et $(BI)$ est parallèle à $(KA)$. On donne $KA = 4,5cm$, $KS = 6cm$ et $SI = 4cm$.
|
||||
|
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Calculer $BI$.
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||||
\end{minipage}
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||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
Dans les marais salants, le sel récolté est stocké en tas sur une surface plane. On admet que le tas de sel a la forme d'un cône de révolution. La situation peut être modélisée par les deux dessins ci-dessous.
|
||||
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
\includegraphics[scale=0.25]{./fig/cone3D.pdf}
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||||
\end{minipage}
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||||
\begin{minipage}{0.5\textwidth}
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\includegraphics[scale=0.25]{./fig/cone_cote.pdf}
|
||||
\end{minipage}
|
||||
\begin{enumerate}
|
||||
\item Démontrer que la hauteur de ce cône de sel est égale à 2,5 mètres.
|
||||
\item À l'aide de la formule $V_{cone} = \frac{\pi \times rayon^2 \times hauteur}{3}$, determiner, en $m^3$, le volume de sel contenu dans le cône. Arrondir le résultat au $m^3$ près.
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\end{enumerate}
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\end{Exo}
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\begin{Exo}
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\begin{enumerate}
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\item Donner la liste des diviseurs de 152 et 798.
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\item Simplifier la fraction $\frac{152}{798}$.
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\item En écrivant les premiers multiples de 7, 2 et 3, trouver le plus petit multiple commun de ces trois chiffres.
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\item Calculer $A = \frac{2}{7} + \frac{3}{2} + \frac{4}{3}$ et donner le résultat sous la forme d'une fraction irreductible.
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\end{enumerate}
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\end{Exo}
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\begin{Exo}
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Calculer en détaillant les étapes et donner le résultat sous forme d'une fraction irreductible.
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\begin{eqnarray*}
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A = \frac{ 96 \times 60 \times 92 \times22 }{5 \times 92 \times 50} &\quad &
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B = \frac{ 9 }{ 2 } + \frac{ 13 }{ 7 } \times \frac{ 8 }{ 3 } \\
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C = \left( \frac{ 14 }{ 5 } - \frac{ 19 }{ 9 } \right) \times \frac{ 2 }{ 3 } & \quad &
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D = \frac{ 9 }{ 2 } \div \left( \frac{ 19 }{ 9 } + \frac{ 9 }{ 3 } \right)
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\end{eqnarray*}
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\end{Exo}
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\end{document}
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%%% Local Variables:
|
||||
%%% mode: latex
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%%% TeX-master: "master"
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%%% End:
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||||
BIN
3e/DS/DS_131017/10_thales_arithm_2.pdf
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BIN
3e/DS/DS_131017/10_thales_arithm_2.pdf
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Binary file not shown.
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3e/DS/DS_131017/10_thales_arithm_2.tex
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76
3e/DS/DS_131017/10_thales_arithm_2.tex
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@@ -0,0 +1,76 @@
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\documentclass[a4paper,10pt]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classDS}
|
||||
\usepackage{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/2013_2014}
|
||||
|
||||
% Title Page
|
||||
\titre{2}
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||||
% \quatreC \quatreD \troisB \troisPro
|
||||
\classe{\troisB}
|
||||
\date{17 octobre 2013}
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||||
\duree{1 heure}
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||||
\sujet{2}
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||||
% DS DSCorr DM DMCorr Corr
|
||||
\typedoc{DS}
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||||
|
||||
|
||||
\begin{document}
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||||
\maketitle
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||||
|
||||
Le barème est donné à titre indicatif, il pourra être modifié.
|
||||
|
||||
\begin{Exo}
|
||||
Dans les marais salants, le sel récolté est stocké en tas sur une surface plane. On admet que le tas de sel a la forme d'un cône de révolution. La situation peut être modélisé par les deux dessins ci-dessous.
|
||||
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
\includegraphics[scale=0.25]{./fig/cone3D.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
\includegraphics[scale=0.25]{./fig/cone_cote.pdf}
|
||||
\end{minipage}
|
||||
\begin{enumerate}
|
||||
\item Démontrer que la hauteur de ce cône de sel est égale à 2,5 mètres.
|
||||
\item À l'aide de la formule $V_{cone} = \frac{\pi \times rayon^2 \times hauteur}{3}$, déterminer, en $m^3$, le volume de sel contenu dans le cône. Arrondir le résultat au $m^3$ près.
|
||||
\end{enumerate}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
Les deux cônes de révolution de rayons $KA$ et $IB$ sont opposés par le sommet.
|
||||
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
\includegraphics[scale=0.25]{./fig/sablier.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
Les droites $(AB)$ et $(KI)$ se coupent en $S$ et $(BI)$ est parallèle à $(KA)$. On donne $KA = 4,5cm$, $KS = 6cm$ et $SI = 4cm$.
|
||||
|
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Calculer $BI$.
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||||
\end{minipage}
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||||
\end{Exo}
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||||
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||||
\begin{Exo}
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||||
\begin{enumerate}
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||||
\item Donner la liste des diviseurs de 207 et 138.
|
||||
\item Simplifier la fraction $\frac{207}{138}$.
|
||||
\item En écrivant les premiers multiples de 11, 2 et 3, trouver le plus petit multiple commun de ces trois chiffres.
|
||||
\item Calculer $A = \frac{2}{11} + \frac{3}{2} + \frac{4}{3}$ et donner le résultat sous la forme d'une fraction irreductible.
|
||||
\end{enumerate}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
Calculer en détaillant les étapes et donner le résultat sous forme d'une fraction irreductible.
|
||||
|
||||
\begin{eqnarray*}
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A = \frac{ 85\times37\times86\times10 }{20\times74\times85} &\quad &
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B = \frac{ 4 }{ 8 } + \frac{ 15 }{ 6 } \times \frac{ 20 }{ 3 } \\
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C = \left( \frac{ 18 }{ 7 } - \frac{ 18 }{ 5 } \right) \times \frac{ 19 }{ 2 } & \quad &
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D = \frac{ 18 }{ 8 } \div \left( \frac{ 9 }{ 8 } + \frac{ 7 }{ 4 } \right)
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\end{eqnarray*}
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\end{Exo}
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\end{document}
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%%% Local Variables:
|
||||
%%% mode: latex
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||||
%%% TeX-master: "master"
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||||
%%% End:
|
||||
BIN
3e/DS/DS_131017/fig/cone3D.pdf
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3e/DS/DS_131017/fig/cone3D.pdf
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3e/DS/DS_131017/fig/cone3D.svg
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3e/DS/DS_131017/fig/cone3D.svg
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31
3e/DS/DS_131017/index.rst
Normal file
31
3e/DS/DS_131017/index.rst
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@@ -0,0 +1,31 @@
|
||||
Notes sur un DS pour les 3e
|
||||
###########################
|
||||
|
||||
:date: 2013-10-17
|
||||
:modified: 2014-07-01
|
||||
:tags: DS, Géométrie
|
||||
:category: 3e
|
||||
:authors: Benjamin Bertrand
|
||||
:summary: Pas de résumé, note créée automatiquement parce que je ne l'avais pas bien fait...
|
||||
|
||||
|
||||
|
||||
`Lien vers 10_thales_arithm_1.tex <10_thales_arithm_1.tex>`_
|
||||
|
||||
`Lien vers 10_thales_arithm.pdf <10_thales_arithm.pdf>`_
|
||||
|
||||
`Lien vers number_rotation.py <number_rotation.py>`_
|
||||
|
||||
`Lien vers 10_thales_arithm_1.pdf <10_thales_arithm_1.pdf>`_
|
||||
|
||||
`Lien vers 10_thales_arithm_2.pdf <10_thales_arithm_2.pdf>`_
|
||||
|
||||
`Lien vers 10_thales_arithm_2.tex <10_thales_arithm_2.tex>`_
|
||||
|
||||
`Lien vers 10_thales_arithm.tex <10_thales_arithm.tex>`_
|
||||
|
||||
`Lien vers fig/cone_cote.pdf <fig/cone_cote.pdf>`_
|
||||
|
||||
`Lien vers fig/cone3D.pdf <fig/cone3D.pdf>`_
|
||||
|
||||
`Lien vers fig/sablier.pdf <fig/sablier.pdf>`_
|
||||
98
3e/DS/DS_131017/number_rotation.py
Executable file
98
3e/DS/DS_131017/number_rotation.py
Executable file
@@ -0,0 +1,98 @@
|
||||
#!/usr/bin/env python
|
||||
# encoding: utf-8
|
||||
|
||||
import jinja2, random, os
|
||||
import sys
|
||||
import optparse
|
||||
|
||||
def randfloat(approx = 1, low = 0, up = 10):
|
||||
""" return a random number between low and up with approx floating points """
|
||||
ans = random.random()
|
||||
ans = ans*(up - low) + low
|
||||
ans = round(ans, approx)
|
||||
return ans
|
||||
|
||||
random.randfloat = randfloat
|
||||
|
||||
def gaussRandomlist(mu = 0, sigma = 1, size = 10, manip = lambda x:x):
|
||||
""" return a list of a gaussian sample """
|
||||
ans = []
|
||||
for i in range(size):
|
||||
ans += [manip(random.gauss(mu,sigma))]
|
||||
return ans
|
||||
|
||||
random.gaussRandomlist = gaussRandomlist
|
||||
|
||||
def gaussRandomlist_strInt(mu = 0, sigma = 1, size = 10):
|
||||
return gaussRandomlist(mu, sigma, size, manip = lambda x: str(int(x)))
|
||||
|
||||
random.gaussRandomlist_strInt = gaussRandomlist_strInt
|
||||
|
||||
def lst_int(min = 2, max = 100, size = 1):
|
||||
""" return a list of integers """
|
||||
return [random.randint(min, max) for i in range(size)]
|
||||
|
||||
random.lst_int = lst_int
|
||||
|
||||
def frac(irred = 0):
|
||||
"""Return a latex fraction"""
|
||||
a = random.randint(2,20)
|
||||
b = random.randint(2,10)
|
||||
while a == b:
|
||||
a = random.randint(2,20)
|
||||
b = random.randint(2,10)
|
||||
|
||||
return "\\frac{{ {a} }}{{ {b} }}".format(a=a,b=b)
|
||||
|
||||
random.frac = frac
|
||||
|
||||
report_renderer = jinja2.Environment(
|
||||
block_start_string = '%{',
|
||||
block_end_string = '%}',
|
||||
variable_start_string = '%{{',
|
||||
variable_end_string = '%}}',
|
||||
loader = jinja2.FileSystemLoader(os.path.abspath('.'))
|
||||
)
|
||||
|
||||
|
||||
def main(options):
|
||||
template = report_renderer.get_template(options.template)
|
||||
|
||||
if options.output:
|
||||
output_basename = options.output
|
||||
else:
|
||||
tpl_base = os.path.splitext(options.template)[0]
|
||||
output_basename = tpl_base + "_"
|
||||
|
||||
|
||||
for subj in range(options.num_subj):
|
||||
subj = subj+1
|
||||
dest = output_basename + str(subj) + '.tex'
|
||||
with open( dest, 'w') as f:
|
||||
f.write(template.render(random = random, infos = {"subj" : subj}))
|
||||
os.system("pdflatex " + dest)
|
||||
|
||||
if not options.dirty:
|
||||
os.system("rm *.aux *.log")
|
||||
|
||||
if __name__ == '__main__':
|
||||
|
||||
|
||||
parser = optparse.OptionParser()
|
||||
parser.add_option("-t","--tempalte",action="store",type="string",dest="template", help="File with template")
|
||||
parser.add_option("-o","--output",action="store",type="string",dest="output",help="Base name for output (without .tex or any extension))")
|
||||
parser.add_option("-n","--number_subjects", action="store",type="int", dest="num_subj", default = 2, help="The number of subjects to make")
|
||||
parser.add_option("-d","--dirty", action="store_true", dest="dirty", help="Do not clean after compilation")
|
||||
|
||||
(options, args) = parser.parse_args()
|
||||
|
||||
if not options.template:
|
||||
print("I need a template!")
|
||||
sys.exit(0)
|
||||
|
||||
main(options)
|
||||
|
||||
# -----------------------------
|
||||
# Reglages pour 'vim'
|
||||
# vim:set autoindent expandtab tabstop=4 shiftwidth=4:
|
||||
# cursor: 16 del
|
||||
Reference in New Issue
Block a user