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: 11
$\begin{array}{l} \text{Gegeben:}\\\text{Diagonale f} \qquad f \qquad [m] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \\ \text{Gesucht:} \\\text{Diagonale e} \qquad e \qquad [m] \\ \\ e = \frac{2\cdot A}{ f}\\ \textbf{Gegeben:} \\ f=1\frac{1}{5}m \qquad A=1\frac{1}{2}m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ e = \frac{2\cdot A}{ f} \\ f=1\frac{1}{5}m\\ A=1\frac{1}{2}m^{2}\\ e = \frac{2\cdot 1\frac{1}{2}m^{2}}{ 1\frac{1}{5}m}\\\\e=2\frac{1}{2}m \\\\\\ \small \begin{array}{|l|} \hline f=\\ \hline 1\frac{1}{5} m \\ \hline 12 dm \\ \hline 120 cm \\ \hline 1,2\cdot 10^{3} mm \\ \hline 1,2\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline 1\frac{1}{2} m^2 \\ \hline 150 dm^2 \\ \hline 1,5\cdot 10^{4} cm^2 \\ \hline 1,5\cdot 10^{6} mm^2 \\ \hline 0,015 a \\ \hline 0,00015 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 2\frac{1}{2} m \\ \hline 25 dm \\ \hline 250 cm \\ \hline 2,5\cdot 10^{3} mm \\ \hline 2,5\cdot 10^{6} \mu m \\ \hline \end{array} \end{array}$