Import work from year 2013-2014
This commit is contained in:
Benjamin Bertrand
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4e/DS/4eD/03_pyth_litt/03_pyth_litt_1.pdf
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4e/DS/4eD/03_pyth_litt/03_pyth_litt_1.tex
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\documentclass[a4paper,12pt]{/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{7}
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% \quatreC \quatreD \troisB \troisPro
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\classe{\quatreC}
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\date{20 mars 2014}
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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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%\printanswers
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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. Des points sont réservés à présentation.
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\begin{questions}
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\question[4]
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Des électriciens veulent poser un câble électrique entre deux poteaux. Le sommet du premier poteaux se trouve à 5m du sol alors que le sommet du deuxième se trouve à 8m. Les deux poteaux sont séparés de 15m.
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\begin{parts}
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\part Faire un schéma de la situation.
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\begin{solution}
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\begin{center}
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\includegraphics[scale=0.4]{./fig/poteau}
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\end{center}
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\end{solution}
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\part Quelle est la longueur de câble devront-ils prévoir s'ils veulent relier le sommet des deux poteaux?
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\begin{solution}
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D'après le dessin, on remarque que $AB = 8 - 5 = 3m$. \\
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D'après le dessin, on a le triangle $ABC$ rectangle en $A$ donc d'après le théorème de Pythagore, on a
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\begin{eqnarray*}
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BC^2 &=& AB^2 + AC^2 \\
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BC^2 &=& 3^2 + 15^2 \\
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BC^2 &=& 9 + 225 \\
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BC^2 &=& 234 \\
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BC &=& \sqrt{234} \approx 15,3
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\end{eqnarray*}
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Donc la tyrolienne fait 15,3m de long.
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\end{solution}
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\end{parts}
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\question[6]
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On veut construire un local de la forme suivante:
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\begin{center}
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\includegraphics[scale=0.2]{./fig/local}
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\end{center}
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Les pièces utilisés pour la construction sont choisis de tel sorte que
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\begin{eqnarray*}
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AF = EB = DC \hspace{2cm} AB = EF \hspace{2cm} BC = ED = GH
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\end{eqnarray*}
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\begin{parts}
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\part Pour s'assurer que le local est bien droit, On mesure $BD$ et on trouve $BD = 25m$.
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\begin{subparts}
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\subpart Démontrer que $BCD$ est un triangle rectangle.
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\begin{solution}
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D'une part, $BC^2 + DC^2 = 24^2 + 7^2 = 576 + 49 = 625$.
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D'autre part, $BD^2 = 25^2 = 625$.
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Donc on a $BD^2 = BD^2 + DC^2$ donc d'après le réciproque du théorème de Pythagore, le triangle $BCD$ est rectangle en $C$.
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\end{solution}
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\subpart Démontrer que $BEDC$ est un rectangle.
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\begin{solution}
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Comme $EB = DC$ et que $ED = BC$, le quadrilatère $EDCB$ est un parallélogramme. Or si un parallélogramme a un angle droit, c'est un rectangle. Donc $EDCB$ est un rectangle.
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\end{solution}
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\end{subparts}
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\part On veut installer des panneaux solaires sur le toit.
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\begin{subparts}
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\subpart Calculer la distance $GE$.
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\begin{solution}
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On sait que le triangle $FGE$ est un triangle rectangle en $G$ donc d'après le théorème de Pythagore, on a
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\begin{eqnarray*}
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FE^2 &=& FG^2 + GE^2 \\
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FG^2 &=& FE^2 - GE^2 \\
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FG^2 &=& 5^2 - 2^2 \\
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FG^2 &=& 25 - 4 \\
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FG^2 &=& 21 \\
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FG &=& \sqrt{21} \approx 4,6
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\end{eqnarray*}
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Donc $GE = 4,6m$.
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\end{solution}
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\subpart Quelle est l'aire du toit du local?
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\begin{solution}
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\begin{eqnarray*}
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\mathcal{A} = GE \times ED = 4,6 \times 24 \approx 110
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\end{eqnarray*}
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L'aire du toit est $110m^2$
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\end{solution}
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\end{subparts}
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\end{parts}
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\question[4]
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Voici un programme de calcul.
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\fbox{\colorbox{base2}{
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\begin{minipage}[h]{0.4\textwidth}
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\textbf{Programme A} \\ Choisir un nombre \\ Multiplier par 3 \\ Ajouter 4 \\ Multiplier par 4 \\ Enlever 16
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\end{minipage}
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}
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}
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\begin{parts}
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\part Montrer que si l'on applique le programme à -1 on trouve -12.
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\begin{solution}
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\begin{eqnarray*}
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-1 \stackrel{\times3}{\longrightarrow} -3 \stackrel{+4}{\longrightarrow} 1 \stackrel{\times4}{\longrightarrow} 4 \stackrel{-16}{\longrightarrow} -12
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\end{eqnarray*}
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\end{solution}
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\part Appliquer le programme à 3.
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\begin{solution}
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\begin{eqnarray*}
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3 \stackrel{\times3}{\longrightarrow} 9 \stackrel{+4}{\longrightarrow} 13 \stackrel{\times4}{\longrightarrow} 52 \stackrel{-16}{\longrightarrow} 36
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\end{eqnarray*}
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\end{solution}
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\part Appliquer le programme à $x$. Montrer que l'on trouve $(3x + 4)\times 4 - 16$.
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\begin{solution}
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\begin{eqnarray*}
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x \stackrel{\times3}{\longrightarrow} 3x \stackrel{+4}{\longrightarrow} 3x+4 \stackrel{\times4}{\longrightarrow} (3x+4)\times4 \stackrel{-16}{\longrightarrow} (3x+4)\times 4 - 16
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\end{eqnarray*}
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\end{solution}
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\part Développer l'expression trouvée à la question précédente.
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\begin{solution}
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\begin{eqnarray*}
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(3x+4)\times4 - 16 & = & 3x\times4 + 4\times 4 - 16 \\
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&=& 12x + 16 - 16 \\
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&=& 12x
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\end{eqnarray*}
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\end{solution}
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\part Si le programme ne faisait qu'une seule transformation, quelle serait elle?
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\begin{solution}
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D'après la question précédente, appliquer tout le programme à $x$ revient à multiplier $x$ par 12. Donc si le programme ne faisait qu'une seule transformation, ce serait de multiplier par 12.
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\end{solution}
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\end{parts}
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\question[5]
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Voici une expression: \hspace{2cm} $A = 6(2x - 1) $
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\begin{parts}
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\part Évaluer $A$ pour $x = 4$.
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\begin{solution}
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On remplace $x$ par 4 dans l'expression de $A$.
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\begin{eqnarray*}
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A &=& 6 \times ( 2 \times 3 - 1 ) \\
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A &=& 6 \times ( 6 - 1 ) \\
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A &=& 6 \times 5 \\
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A &=& 30
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\end{eqnarray*}
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\end{solution}
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\part Développer puis réduire $A$.
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\begin{solution}
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\begin{eqnarray*}
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A & = & 6(2x - 1) \\
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A & = & 6\times 2x + 6 \times (-1) \\
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A & = & 12x - 6
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\end{eqnarray*}
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\end{solution}
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\end{parts}
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\question
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\exo{Bonus}
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Voici deux expressions.
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\begin{eqnarray*}
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B = 2(2x -4) + 4x(3 + 5x) \hspace{2cm} C = -(3x + 7) - 5x + 4
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\end{eqnarray*}
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\begin{parts}
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\part Évaluer $B$ pour $x = 2$.
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\part Développer puis réduire $B$ et $C$.
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\end{parts}
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\end{questions}
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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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4e/DS/4eD/03_pyth_litt/03_pyth_litt_1_corr.pdf
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4e/DS/4eD/03_pyth_litt/03_pyth_litt_2.pdf
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4e/DS/4eD/03_pyth_litt/03_pyth_litt_2.tex
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4e/DS/4eD/03_pyth_litt/03_pyth_litt_2.tex
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\documentclass[a4paper,12pt]{/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{7}
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% \quatreC \quatreD \troisB \troisPro
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\classe{\quatreC}
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\date{20 mars 2014}
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\duree{1 heure}
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\sujet{2}
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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. Des points sont réservés à présentation.
|
||||
|
||||
\begin{questions}
|
||||
|
||||
|
||||
\question[4]
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Des électriciens veulent poser un câble électrique entre deux poteaux. Le sommet du premier poteaux se trouve à 5m du sol alors que le sommet du deuxième se trouve à 8m. Les deux poteaux sont séparés de 15m.
|
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\begin{parts}
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\part Faire un schéma de la situation.
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\part Quelle est la longueur de câble devront-ils prévoir s'ils veulent relier le sommet des deux poteaux?
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\end{parts}
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\question[6]
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On veut construire un local de la forme suivante:
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\begin{center}
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\includegraphics[scale=0.2]{./fig/local}
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\end{center}
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Les pièces utilisés pour la construction sont choisis de tel sorte que
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\begin{eqnarray*}
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AF = EB = DC \hspace{2cm} AB = EF \hspace{2cm} BC = ED = GH
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\end{eqnarray*}
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\begin{parts}
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\part Pour s'assurer que le local est bien droit, On mesure $BD$ et on trouve $BD = 25m$.
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\begin{subparts}
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\subpart Démontrer que $BCD$ est un triangle rectangle.
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\begin{solution}
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D'une part, $BC^2 + DC^2 = 24^2 + 7^2 = 576 + 49 = 625$.
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D'autre part, $BD^2 = 25^2 = 625$.
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Donc on a $BD^2 = BD^2 + DC^2$ donc d'après le réciproque du théorème de Pythagore, le triangle $BCD$ est rectangle en $C$.
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\end{solution}
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\subpart Démontrer que $BEDC$ est un rectangle.
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\begin{solution}
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Comme $EB = DC$ et que $ED = BC$, le quadrilatère $EDCB$ est un parallélogramme. Or si un parallélogramme a un angle droit, c'est un rectangle. Donc $EDCB$ est un rectangle.
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\end{solution}
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\end{subparts}
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\part On veut installer des panneaux solaires sur le toit.
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\begin{subparts}
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\subpart Calculer la distance $GE$.
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\subpart Quelle est l'aire du toit du local?
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\end{subparts}
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\end{parts}
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\question[4]
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Voici un programme de calcul.
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\fbox{\colorbox{base2}{
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\begin{minipage}[h]{0.4\textwidth}
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\textbf{Programme A} \\ Choisir un nombre \\ Multiplier par -4 \\ Enlever 2 \\ Multiplier par 5 \\ Ajouter 10
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\end{minipage}
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}
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}
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\begin{minipage}[h]{0.5\textwidth}
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\begin{parts}
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\part Montrer que si l'on applique le programme à 2 on trouve -40.
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\part Appliquer le programme à 3.
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\part Appliquer le programme à $x$. Montrer que l'on trouve $(-4x - 2)\times 5 + 10$.
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\part Développer l'expression trouvée à la question précédente.
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\part Si le programme ne faisait qu'une seule transformation, quelle serait elle?
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\end{parts}
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\end{minipage}
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\question[5]
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Voici une expression: \hspace{2cm} $A = 4(3x - 1) $
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\begin{parts}
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\part Évaluer $A$ pour $x = 2$.
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\part Développer puis réduire $A$.
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\end{parts}
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\question
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\exo{Bonus}
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Voici deux expressions.
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\begin{eqnarray*}
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B = 4(-2x +4) + 2x(3 + x) \hspace{2cm} C = -(2x + 2) - 5x + 4
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\end{eqnarray*}
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\begin{parts}
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\part Évaluer $B$ pour $x = 2$.
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\part Développer puis réduire $B$ et $C$.
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\end{parts}
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\end{questions}
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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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4e/DS/4eD/03_pyth_litt/03_pyth_litt_revis.pdf
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4e/DS/4eD/03_pyth_litt/03_pyth_litt_revis.tex
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\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
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% Title Page
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\title{Révision Pythagore, calcul littéral - Exercices}
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\author{}
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\date{}
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\fancyhead[L]{Quatrième}
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\fancyhead[C]{\Thetitle}
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\fancyhead[R]{\thepage}
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\begin{document}
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\thispagestyle{fancy}
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\section{Théorème de Pythagore et réciproque}
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\begin{Exo}
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$ABC$ triangle rectangle en $B$ tel que $AB = 2$ et $BC = 6$. Calculer $AC$.
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\end{Exo}
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\begin{Exo}
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$EFG$ triangle rectangle en $F$ tel que $EF = 5,3$ et $EG = 5,9$.Calculer $FG$.
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\end{Exo}
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\begin{Exo}
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$IJK$ est un triangle tel que $IJ = 4,8$,$IK = 1,4$ et $JK = 5$.
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Le triangle $IJK$ est-il rectangle? S'il est rectangle quel est l'angle droit et l'hypoténuse?
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\end{Exo}
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\begin{Exo}
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$LMN$ est un triangle tel que $LM = 1,8$,$MN = 14,4$ et $NL = 14$.
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Le triangle $LMN$ est-il rectangle? S'il est rectangle quel est l'angle droit et l'hypoténuse?
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\end{Exo}
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\begin{Exo}
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Un terrain de foot (rectangulaire) mesure 60m de largeur et 90m de longueur.
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\begin{enumerate}
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\item Faire un dessin à main levée.
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\item Calculer la longueur de la diagonale de ce terrain.
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\end{enumerate}
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\end{Exo}
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\begin{Exo}
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Une tyrolienne part du sommet d'un arbre à 20m de hauteur pour arriver sur une plateforme à 10m de hauteur. La distance entre le pied de l'arbre et le pied de la plateforme est de 50m.
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\begin{enumerate}
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\item Faire un schéma représentant la situation.
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\item Quelle est la longueur de la tyrolienne?
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\end{enumerate}
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\end{Exo}
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\newpage
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\section{Calcul littéral}
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\begin{Exo}
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Évaluer les expressions ci-dessous pour les valeurs indiquées à coté.
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\begin{eqnarray*}
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A = 2x -1& \mbox{ avec } & x = 3 \\
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B = 2(- y - 1)& \mbox{ avec } & y = 8 \\
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C = (3x + 1)(4 - 2x) & \mbox{ avec } & x = -3 \\
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D = 2x - 1& \mbox{ avec } & x = \frac{3}{5} \\
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\end{eqnarray*}
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\end{Exo}
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\begin{Exo}
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Réduire les expressions suivantes:
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\begin{eqnarray*}
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A & = & 2x + 4x + 3x + 1 + 3 \\
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B & = & 4 \times 2x - 7 \times 3 \\
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C & = & 7x - 4x + 1 - 3 \\
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D & = & 2x + 4 - 3x + 3 \\
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E & = & 4\times 2x + 4\times2 - 3x + 6 \\
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\end{eqnarray*}
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\end{Exo}
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\begin{Exo}
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Relier les formes factorisées avec la forme développées qui lui est égale.
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\begin{minipage}[h]{0.2\textwidth}
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\flushright
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\textbf{Forme factorisée}
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$3(2x - 1) \qquad \bullet$ \\[0.5cm]
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$(-3x + 4)\times 2 \qquad \bullet$ \\[0.5cm]
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||||
$x(3x + 1) \qquad \bullet$ \\[0.5cm]
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$5(-x - 9) \qquad \bullet$ \\[0.5cm]
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\end{minipage}
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\hspace{1cm}
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\begin{minipage}[h]{0.2\textwidth}
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\textbf{Forme développée}
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\begin{itemize}
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\item $4x^2$
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||||
\item $6x - 1$
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||||
\item $-6x + 8$
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||||
\item $-5x - 45$
|
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29
4e/DS/4eD/03_pyth_litt/index.rst
Normal file
29
4e/DS/4eD/03_pyth_litt/index.rst
Normal file
@@ -0,0 +1,29 @@
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Notes sur 03 pyth litt
|
||||
######################
|
||||
|
||||
:date: 2014-07-01
|
||||
:modified: 2014-07-01
|
||||
:tags: DS
|
||||
:category: 4e
|
||||
:authors: Benjamin Bertrand
|
||||
:summary: Pas de résumé, note créée automatiquement parce que je ne l'avais pas bien fait...
|
||||
|
||||
|
||||
|
||||
`Lien vers 03_pyth_litt_1.pdf <03_pyth_litt_1.pdf>`_
|
||||
|
||||
`Lien vers 03_pyth_litt_1.tex <03_pyth_litt_1.tex>`_
|
||||
|
||||
`Lien vers 03_pyth_litt_2.tex <03_pyth_litt_2.tex>`_
|
||||
|
||||
`Lien vers 03_pyth_litt_revis.pdf <03_pyth_litt_revis.pdf>`_
|
||||
|
||||
`Lien vers 03_pyth_litt_2.pdf <03_pyth_litt_2.pdf>`_
|
||||
|
||||
`Lien vers 03_pyth_litt_1_corr.pdf <03_pyth_litt_1_corr.pdf>`_
|
||||
|
||||
`Lien vers 03_pyth_litt_revis.tex <03_pyth_litt_revis.tex>`_
|
||||
|
||||
`Lien vers fig/poteau.pdf <fig/poteau.pdf>`_
|
||||
|
||||
`Lien vers fig/local.pdf <fig/local.pdf>`_
|
||||
Reference in New Issue
Block a user