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: 02
$\begin{array}{l} \text{Gegeben:}\\\text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Diagonale e} \qquad e \qquad [m] \\ \\ \text{Gesucht:} \\\text{Diagonale f} \qquad f \qquad [m] \\ \\ f = \frac{2\cdot A}{ e}\\ \textbf{Gegeben:} \\ A=1m^{2} \qquad e=4m \qquad \\ \\ \textbf{Rechnung:} \\ f = \frac{2\cdot A}{ e} \\ A=1m^{2}\\ e=4m\\ f = \frac{2\cdot 1m^{2}}{ 4m}\\\\f=\frac{1}{2}m \\\\\\ \small \begin{array}{|l|} \hline A=\\ \hline 1 m^2 \\ \hline 100 dm^2 \\ \hline 10^{4} cm^2 \\ \hline 10^{6} mm^2 \\ \hline \frac{1}{100} a \\ \hline 0,0001 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline f=\\ \hline \frac{1}{2} m \\ \hline 5 dm \\ \hline 50 cm \\ \hline 500 mm \\ \hline 5\cdot 10^{5} \mu m \\ \hline \end{array} \end{array}$