index.tex

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\newcounter{numeroexo}
\newcommand{\exo}{\par\noindent\stepcounter{numeroexo}
        \hspace{-.25cm}\Ovalbox{\textbf{Exercice \arabic{numeroexo}}}\quad}
 
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\textcolor{red}{\textsl{(sur #1 points)}}\quad
\textcolor{blue}{\textsc{#2 }\\*[.1cm]}}
 
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\begin{document}
{\bf Nom :  \hspace*{5cm} Prénom : \hfill{5$^{\textrm{e}}$\hspace*{2pc}}\par
\vspace{6pt}
\centerline{
        \shadowbox{\Large{Devoir en classe}}
}
\vspace{6pt}
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%exercice 
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\exo\pts{9}{Priorité opératoire} 
\vspace{-6pt}
\begin{enumerate}
\item Souligne le calcul par lequel il faut commencer, puis effectue le calcul des expressions suivantes :\par
\begin{tabular}{p{4,5cm}p{4,5cm}p{4,5cm}}
$7\times(0,3\times10+7)$ & $12\times(54\div3-13)$ & $42-27+13$ \\
&&\\
&&\\
&&\\
\end{tabular}
\item Transforme les expressions suivantes en utilisant le signe "$\div$" \\
\begin{tabular}{p{4.5cm}p{4.5cm}p{4.5cm}}
\Large $\frac{43-12}{7}=$ & \Large $\frac{43}{12-7}=$ & $43-$ \Large $\frac{12}{7}=$\\
&&\\
\end{tabular}
\item Calcule les expressions suivantes sachant que $a=7$, $b=9$ et $c=3$ \\
\begin{tabular}{p{4cm}p{4cm}p{4cm}}
\large $10ab=$ & \large $4b-5a=$ & \large $4a-\frac{b}{c}=$\\
&&\\
\end{tabular}
\end{enumerate}
 
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%exercice suivant
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\exo\pts{3}{Comparaison} 
Sans effectuer les divisions, comparer en expliquant : \par
\begin{tabular}{p{5cm}p{5cm}p{5cm}}
        \Large $\frac{5}{6} \cdots \frac{2}{3}$ & \Large $\frac{18}{17} \cdots \frac{8}{9}$ & \Large $\frac{15}{19} \cdots \frac{5}{6}$ 
\end{tabular}
\\
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%exercice suivant
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\exo\pts{4}{Calcul fractionnaire} 
Calculer et donner le résultat sous la forme d'une fraction plus simple : \\
\begin{tabular}{b{8cm}b{8cm}}
        \Large $\frac{2}{5}+\frac{3}{5}=$ & \Large $\frac{6}{8}-\frac{1}{4}=$ \\
        &\\
        \Large $\frac{3}{2}\times\frac{3}{5}=$ & \Large $\frac{40}{28} \times\frac{7}{10}=$ 
\end{tabular}
\\ 
\vspace{.3cm}
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%exercice suivant
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\exo\pts{3}{Fractions} 
Dans les égalités suivantes, les signes $+$, $-$ et $\times$ ont été remplacés par des étoiles. Les retrouver. \\
\begin{tabular}{p{5cm}p{5cm}p{5cm}}
        \Large $\frac{3\star3}{2}\star\frac{1}{6}=\frac{14}{3}$ & \Large $\frac{3\star3}{2}\star\frac{1}{6}=\frac{17}{6}$ & \Large $\frac{3\star3}{2}\star\frac{1}{6}=\frac{1}{2}$\\
        &&\\
        &&\\
\end{tabular}
 
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%exercice suivant
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\exo\pts{2}{Calcul prioritaire} 
Calculer puis simplifier :  \\
{\Large $\frac54- \frac8{24} \times \frac6{16}=$}
\vspace{.3cm}
 
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%exercice suivant
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\exo\pts{4}{Calcul d'angles} 
Pour la figure ci-dessous, on précise que $(AB)$ est la médiatrice de [IJ] et que $\widehat{KIJ}=48$\degres.\\
$$\includegraphics[width=6cm]{devoirrevision5e.1}$$
\begin{enumerate}
    \item Complète les phrases suivantes :\\
  $(AB)$ est la médiatrice de [IJ] donc B est le {\Large\phantom{milieu}} de $[IJ]$. De plus $(AB)$ est {\large\phantom{perpendiculaire}} à $[IJ]$. Comme $A$ appartient à la médiatrice de [IJ], il est à la {\large\phantom{même distance}} des points I et J. Donc IA = {\Large\phantom{IB}} .
\item Quel est la nature du triangle $AIJ$? 
  \item Calculer l'angle $\widehat{IAK}$.
\end{enumerate}
\vspace{1.5cm}
 
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%exercice suivant
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\exo\pts{5}{Constructions} 
\vspace{-.5cm}
\begin{enumerate}
  \item Construire un triangle $ABC$ rectangle en $A$ tel que $BC=10$~cm et $\widehat{ABC}=72$\degres. 
  \item Calculer l'angle $\widehat{BCA}$.\vspace{1.5cm}
   \item Construire le cercle circonscrit au triangle $ABC$.
\end{enumerate}
 
\end{document}