Examen Local

📅 January 07, 2024   |   👁️ Views: 284




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\documentclass[12pt,a4paper]{article}
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\usepackage{amsmath,amsfonts,amssymb}
\usepackage{multirow}
\usepackage{tabularx,tabulary}
\usepackage{graphicx}

\usepackage{tikz}
\usetikzlibrary{calc}

\pagestyle{empty}

\begin{document}
\newcolumntype{Y}{>{\centering\arraybackslash}X}
\begin{center}
    \begin{tabularx}{\linewidth}{@{}|Y|Y|Y|@{}}
        \hline
    \multirow{2}{*}{Collège Tariq Ben Ziad 2}
    & \multirow{2}{*}{Examen local (Durée: 2h)}
    & \multirow{2}{*}{Année scolaire: 2023-2024} \\
    & & \\
        \hline
    \end{tabularx}
    \begin{tabularx}{\linewidth}{@{}|c|X|@{}}
        \hline
    & \\
        & \underline{\textbf{Exercice 1:}(5pts)} \\
        1x5 & Calculer et simplifier \\
    & \hspace*{1cm}
        $A=3\sqrt{9}-2\sqrt{16}+\sqrt{25}$\hspace*{0.2cm};\hspace*{0.2cm}
        $B=2\sqrt{2}-\sqrt{50}$\hspace*{0.2cm};\hspace*{0.2cm}
        $C=\frac{1}{1-\sqrt{3}}+\frac{1}{1+\sqrt{3}}$\hspace*{0.2cm};\hspace*{0.2cm}
        \\
        & \\
        & \hspace*{1cm}
        $D=(2-\sqrt{5})^2+(\sqrt{5}-2)(\sqrt{5}+2)$\hspace*{0.2cm};\hspace*{0.2cm}
        $E=(\frac{-3}{5})^{-1}-[1-(\frac{3}{2})^{-1}]^{2}$ \\
        & \\
        & \underline{\textbf{Exercice 2:}(3pts)} \\
    1 & 1 - a) Comparer $3\sqrt{5}$ et $2\sqrt{10}$ \\
    0.5& \hspace*{0.2cm} - b) En déduir la comparaison de $5+3\sqrt{5}$ et $3+2\sqrt{10}$ \\
        & \\
    1.5& 2 - Soit $x$ un nombre reel tel que: \hspace*{0.5cm} $-1 \le x < 3$\\
    & \hspace*{1cm} Encadrer $x+3$\hspace*{0.2cm}; $x-2$\hspace*{0.2cm}; $3x$ \\
        & \\
        & \underline{\textbf{Exercice 3:}(4pts)} \\
    & \begin{minipage}[t]{0.6\textwidth}
        Soit la figure suivante telle que:\\
        $AB=8$\hspace*{0.2cm}; $AC=6$\hspace*{0.2cm}; $BC=10$\\
        \hspace*{0.2cm}1) Montrer que le triangle $ABC$ est rectangle en $A$\\
        \hspace*{0.2cm}2) Calculer $\tan A\hat{B}C$\\
        Soit $E$ un point du segment $[BC]$ tel que $BE=4$.\\
        la droite perpondiculaire à $(BC)$ en $E$ coupe le segment $[AB]$ au point $F$ \\
      \end{minipage}%
      \begin{minipage}[t]{0.4\textwidth}
        \begin{tikzpicture}[baseline={(current bounding box.north)}]
            % Define the coordinates
            \coordinate[label=above:$A$] (A) at (3.5,2);
            \coordinate[label=above left:$B$] (B) at (0,2);
            \coordinate[label=below:$C$] (C) at (3.5,0.5);

            % Draw the triangle ABC
            \draw (A) -- (B) -- (C) -- cycle;

            % Additional points and lines
            \coordinate[label=above:$F$] (F) at (2.5,2);
            \coordinate[label=below left:$E$] (E) at (2.1,1.1);

            % Dashed line from F to E
            \draw (E) -- (F);

            % Draw the right-angle mark
            \draw ($(E)!0.10!(F)$) -- ++(155:0.12) -- ++(-113:0.11);
            \draw (A) --++(-0.15,0) -- ++(0,-0.15) -- ++(0.15,0);

        \end{tikzpicture}%
      \end{minipage}\\
    1 & 3) Montrer que $EF=3$ \\
    1 & 4) Calculer $BF$ \\
    & \\
        & \underline{\textbf{Exercice 4:}(4pts)} \\
    1x2& 1) $x$ la mesure d'un angle aigu tel que $\cos x = \frac{1}{2}$. Calculer $\sin x$ et $\tan x$\\
    1& 2) $a$ la mesure d'un angle aigu. Simplifier: \hspace*{0.5cm} $A=(\cos a -2\sin a)(\cos a +2\sin a)+5\sin^{2}a$\\
    1& 3) Simplifier puis calculer: $B=\cos^{2}50+\sin^{2}20+\cos^{2}40+\sin^{2}70$\\
    & \\
    & \underline{\textbf{Exercice 5:}(2pts)} \\
        & \begin{minipage}[t]{0.6\textwidth}
            Soit la figure suivante telle que:\\
            $AB=6$\hspace*{0.2cm}; $AE=2$\hspace*{0.2cm}; $AF=3$\hspace*{0.2cm}; $FC=6$ et $EF=2.5$\\
            \\
            \hspace*{0.2cm}1) Montrer que $(EF)//(BC)$\\
            \hspace*{0.2cm}2) Calculer $BC$\\
          \end{minipage}%
          \begin{minipage}[t]{0.4\textwidth}
            \begin{tikzpicture}[baseline={(current bounding box.north)}]
                % Define the coordinates
                \coordinate[label=below:$B$] (B) at (0,0);
                \coordinate[label=above:$A$] (A) at (2,2);
                \coordinate[label=below:$C$] (C) at (4,0);

                % Draw the triangle ABC
                \draw (A) -- (B) -- (C) -- cycle;

                % Additional points F and E
                \coordinate[label=above left:$E$] (E) at ($(A)!0.55!(B)$);
                \coordinate[label=above right:$F$] (F) at ($(A)!0.55!(C)$);

                % Draw the line (EF)
                \draw (E) -- (F);
            \end{tikzpicture}%
          \end{minipage}\\
    & \underline{\textbf{Exercice 6:}(2pts)} \\
        & \begin{minipage}[t]{0.6\textwidth}
            Soit la figure suivante telle que: $A\hat OC=36°$\\
            \\
            \hspace*{0.2cm}2) Calculer $A\hat MB$ et $A\hat NC$\\

          \end{minipage}%
          \begin{minipage}[t]{0.4\textwidth}
            \begin{tikzpicture}[baseline={(current bounding box.north)}]
               % Define the coordinates
                \coordinate[label=below:$O$] (O) at (0,0);
                \coordinate[label=above:$A$] (A) at (120:1.5);
                \coordinate[label=above right:$B$] (B) at (30:1.5);
                \coordinate[label=above:$C$] (C) at (155:1.5);
                \coordinate[label=below right:$M$] (M) at (-30:1.5);
                \coordinate[label=below left:$N$] (N) at (-120:1.5);

                \coordinate[label=above:{\fontsize{3}{4}\selectfont$36^\circ$}] (K) at (160:0.5);

                % Draw the circle
                \draw (O) circle (1.5);

                % Draw radii and chord
                \draw (O) -- (A);
                \draw (O) -- (B);
                \draw (O) -- (C);
                \draw (B) -- (M);
                \draw (A) -- (M);
                \draw (A) -- (N);
                \draw (C) -- (N);


                % Draw right-angle mark for angle AOB
                \draw ($(O)!0.10!(B)$) -- ++(120:0.15) -- ++(-150:0.15) ;
                \draw ($(O)!0.30!(C)$) arc (180:90:0.2);



            \end{tikzpicture}%
          \end{minipage}\\
        & \\
    & \hfill Bonne chance\\
        \hline
    \end{tabularx}
\end{center}
\end{document}

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