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$ A = \frac{1}{2}\cdot e\cdot f $
$ e = \frac{2\cdot A}{ f} $
$ f = \frac{2\cdot A}{ e} $
Geometrie-Viereck-Raute
$A = \frac{1}{2}\cdot e\cdot f$
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$e = \frac{2\cdot A}{ f}$
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$f = \frac{2\cdot A}{ e}$
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Beispiel Nr: 08
$\begin{array}{l}
\text{Gegeben:}\\\text{Fläche} \qquad A \qquad [m^{2}] \\
\text{Diagonale e} \qquad e \qquad [m] \\
\\ \text{Gesucht:} \\\text{Diagonale f} \qquad f \qquad [m] \\
\\ f = \frac{2\cdot A}{ e}\\ \textbf{Gegeben:} \\ A=0,002m^{2} \qquad e=\frac{2}{5}m \qquad \\ \\ \textbf{Rechnung:} \\
f = \frac{2\cdot A}{ e} \\
A=0,002m^{2}\\
e=\frac{2}{5}m\\
f = \frac{2\cdot 0,002m^{2}}{ \frac{2}{5}m}\\\\f=\frac{1}{100}m
\\\\\\ \small \begin{array}{|l|} \hline A=\\ \hline 0,002 m^2 \\ \hline \frac{1}{5} dm^2 \\ \hline 20 cm^2 \\ \hline 2\cdot 10^{3} mm^2 \\ \hline 2\cdot 10^{-5} a \\ \hline 2\cdot 10^{-7} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline \frac{2}{5} m \\ \hline 4 dm \\ \hline 40 cm \\ \hline 400 mm \\ \hline 4\cdot 10^{5} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline f=\\ \hline \frac{1}{100} m \\ \hline \frac{1}{10} dm \\ \hline 1 cm \\ \hline 10 mm \\ \hline 10^{4} \mu m \\ \hline \end{array} \end{array}$