Modifié le 3 Mai 2009 à 21 h 01.
%@Auteur: François Meria
%@Dif:3
On donne les résultats suivants en rappelant que $a^2$
signifie $a\times a$ :
\begin{center}
$\begin{array}{|*{11}{c|}}
\hline
a & 1 & 2 & 3 & 4 & {\bf 5} & 6 & 7 & 8 & {\bf 9} & {\bf 10} \\
\hline
a^2 & 1 & 4 & 9 & 16 & {\bf 25} & 36 & 49 & 64 & {\bf 81} & {\bf 100} \\
\hline
\end{array}$
\par\vspace{5mm}\par
$\begin{array}{|*{11}{c|}}
\hline
a & 11 & 12 & {\bf 13} & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline
a^2 & 121 & 144 & {\bf 169} & 196 & 225 & 256 & 289 & 324 & 361 & 400 \\
\hline
\end{array}$
\end{center}
Dans chacun des cas suivants, calculer la longueur inconnue :
\begin{multicols}{3}
\begin{myenumerate}
\item\subitem{}\par
\begin{tabular}{lc}
$ \left\{
\begin{array}{l}
AC=3 \\
AB=4 \\
BC=x\\
\end{array}
\right. $ & \psset{unit=0.75}
\begin{pspicture}(0,+1.2)(2,3.2)
\pstTriangle[PointSymbol=none](0,0){A}(2,0){C}(0,3){B}
\pstRightAngle{C}{A}{B}
\put(1.2,1.6){$x$}
\end{pspicture}
\end{tabular}
\par\vspace{1cm}\par
\item\subitem{}\par
\begin{tabular}{lc}
$ \left\{
\begin{array}{l}
AC=6 \\
AB=8 \\
BC=x\\
\end{array}
\right. $ & \psset{unit=0.75}
\begin{pspicture}(0,+1.2)(2,3.2)
\pstTriangle[PointSymbol=none](0,0){A}(2,0){C}(0,3){B}
\pstRightAngle{C}{A}{B}
\put(1.2,1.6){$x$}
\end{pspicture}
\end{tabular}
\par\vspace{1cm}\par
\item\subitem{}\par
\begin{tabular}{lc}
$ \left\{
\begin{array}{l}
AC=4,8 \\
AB=1,4 \\
BC=x\\
\end{array}
\right. $ & \psset{unit=0.75}
\begin{pspicture}(0,1.2)(2,3.2)
\pstTriangle[PointSymbol=none](0,1){A}(3.2,1){C}(0,2.5){B}
\pstRightAngle{C}{A}{B}
\put(1.4,2.1){$x$}
\end{pspicture}
\end{tabular}
\par\vspace{1cm}\par
\item\subitem{}\par
\begin{tabular}{lc}
$ \left\{
\begin{array}{l}
AC=12 \\
AB=5 \\
BC=x\\
\end{array}
\right. $ & \psset{unit=0.75}
\begin{pspicture}(0,1.2)(2,3.2)
%\psgrid
\pstTriangle[PointSymbol=none](0,1){A}(3.2,1){C}(0,2.5){B}
\pstRightAngle{C}{A}{B}
\put(1.4,2.1){$x$}
\end{pspicture}
\end{tabular}
\par\vspace{1cm}\par
\item\subitem{}\par
\begin{tabular}{lc}
$ \left\{
\begin{array}{l}
AC=x \\
AB=40 \\
BC=41\\
\end{array}
\right. $ & \psset{unit=0.75}
\begin{pspicture}(0,+1.2)(2,3.2)
\pstTriangle[PointSymbol=none](0,0){A}(2,0){C}(0,3){B}
\pstRightAngle{C}{A}{B}
\put(0.95,-0.4){$x$}
\end{pspicture}
\end{tabular}
\par\vspace{1cm}\par
\item\subitem{}\par
\begin{tabular}{lc}
$ \left\{
\begin{array}{l}
AC=1 \\
AB=2 \\
BC=x\\
\end{array}
\right. $ & \psset{unit=0.75}
\begin{pspicture}(0,+1.2)(2,3.2)
\pstTriangle[PointSymbol=none](0,0){A}(2,0){C}(0,3){B}
\pstRightAngle{C}{A}{B}
\put(1.2,1.6){$x$}
\end{pspicture}
\end{tabular}
\end{myenumerate}
\end{multicols}