**Solution 1.**
Une solution avec *tikz* basé sur `\pgfmathparse` et la règle de Pythagore.
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (3.2,0);
\pgfmathparse{sqrt((5.5)^2-(3.2)^2)}
\coordinate (C) at (0,\pgfmathresult);
\draw (A)--(B)--(C)--cycle;
\end{tikzpicture}
ou (plus court, courte, mais moins élégant!) élégante!)
\begin{tikzpicture}
\pgfmathparse{sqrt((5)^2-(3)^2)}
\edef\triangle{\pgfmathresult}
\draw (0,0)--(3,0)--(0,\triangle)--cycle;
\end{tikzpicture}
**ECM**
\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (3.2,0);
\pgfmathparse{sqrt((5.5)^2-(3.2)^2)}
\coordinate (C) at (0,\pgfmathresult);
\draw (A)--(B)--(C)--cycle;
\end{tikzpicture}
ou encore
\begin{tikzpicture}
\pgfmathparse{sqrt((5)^2-(3)^2)}
\edef\triangle{\pgfmathresult}
\draw (0,0)--(3,0)--(0,\triangle)--cycle;
\end{tikzpicture}
\end{document}
**Solution 2.**
Une outre autre solution (plus élégant). élégante). Voir section `13.5 Coordinate Calculations` de la documentation.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\draw (0,0)--(3.2,0)--($sqrt((5.4)^2-(3.2)^2)*(0,1)$)--cycle;
\end{tikzpicture}
\end{document}
