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| \documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath,fp,xstring}
\newcommand\pythahypo[6][3]{%
Dans le triangle $#2#3#4$ rectangle en $#2$, d'après le théorème de Pythagore on a :
\FPeval\resulta{clip(#5*#5)}\FPeval\resultb{clip(#6*#6)}\FPeval\resultc{clip(\resulta+\resultb)}%
\FPeval\resultd{pow(0.5,\resultc)}%
\FPeval\resulte{clip(round(\resultd,#1))}%
\FPeval\decnum{1/pow(#1+1,10)}\FPeval\resultdiff{abs(\resultd-\resulte)}%
\FPifgt\resultdiff\decnum\let\eqsign\approx\else\let\eqsign= =\fi
\begin{align*}
#3#4^2 & = #2#3^2+#2#4^2\\
#3#4^2 & = \StrSubstitute{#5}.{{,}}^2+\StrSubstitute{#6}.{{,}}^2\\
#3#4^2 & = \StrSubstitute\resulta.{{,}}+\StrSubstitute\resultb.{{,}}\\
#3#4^2 & = \StrSubstitute\resultc.{{,}}\\
\intertext{donc :}
#3#4 & = \sqrt{\StrSubstitute\resultc.{{,}}}\\
#3#4 & \eqsign\StrSubstitute\resulte.{{,}}
\end{align*}
}
\begin{document}
\pythahypo{A}{B}{C}{24.65}{59.16}
\pythahypo[1]{A}{B}{C}{24.65}{59.16}
\end{document} |
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