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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: 07
$\begin{array}{l}
\text{Gegeben:}\\\text{Diagonale f} \qquad f \qquad [m] \\
\text{Fläche} \qquad A \qquad [m^{2}] \\
\\ \text{Gesucht:} \\\text{Diagonale e} \qquad e \qquad [m] \\
\\ e = \frac{2\cdot A}{ f}\\ \textbf{Gegeben:} \\ f=1\frac{2}{3}m \qquad A=\frac{4}{5}m^{2} \qquad \\ \\ \textbf{Rechnung:} \\
e = \frac{2\cdot A}{ f} \\
f=1\frac{2}{3}m\\
A=\frac{4}{5}m^{2}\\
e = \frac{2\cdot \frac{4}{5}m^{2}}{ 1\frac{2}{3}m}\\\\e=\frac{24}{25}m
\\\\\\ \small \begin{array}{|l|} \hline f=\\ \hline 1\frac{2}{3} m \\ \hline 16\frac{2}{3} dm \\ \hline 166\frac{2}{3} cm \\ \hline 1666\frac{2}{3} mm \\ \hline 1666666\frac{2}{3} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline \frac{4}{5} m^2 \\ \hline 80 dm^2 \\ \hline 8\cdot 10^{3} cm^2 \\ \hline 8\cdot 10^{5} mm^2 \\ \hline \frac{1}{125} a \\ \hline 8\cdot 10^{-5} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline \frac{24}{25} m \\ \hline 9\frac{3}{5} dm \\ \hline 96 cm \\ \hline 960 mm \\ \hline 9,6\cdot 10^{5} \mu m \\ \hline \end{array} \end{array}$