Geometrie-Viereck-Drachen

$A = \frac{1}{2}\cdot e\cdot f$
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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
$\begin{array}{l} \text{Gegeben:}\\\text{Diagonale f} \qquad f \qquad [m] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \\ \text{Gesucht:} \\\text{Diagonale e} \qquad e \qquad [m] \\ \\ e = \frac{2\cdot A}{ f}\\ \textbf{Gegeben:} \\ f=0,002m \qquad A=\frac{2}{5}m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ e = \frac{2\cdot A}{ f} \\ f=0,002m\\ A=\frac{2}{5}m^{2}\\ e = \frac{2\cdot \frac{2}{5}m^{2}}{ 0,002m}\\\\e=400m \\\\\\ \small \begin{array}{|l|} \hline f=\\ \hline 0,002 m \\ \hline \frac{1}{50} dm \\ \hline \frac{1}{5} cm \\ \hline 2 mm \\ \hline 2\cdot 10^{3} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline \frac{2}{5} m^2 \\ \hline 40 dm^2 \\ \hline 4\cdot 10^{3} cm^2 \\ \hline 4\cdot 10^{5} mm^2 \\ \hline 0,004 a \\ \hline 4\cdot 10^{-5} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 400 m \\ \hline 4\cdot 10^{3} dm \\ \hline 4\cdot 10^{4} cm \\ \hline 4\cdot 10^{5} mm \\ \hline 4\cdot 10^{8} \mu m \\ \hline \end{array} \end{array}$