Geometrie-Viereck-Raute

$A = \frac{1}{2}\cdot e\cdot f$
1 2 3 4 5 6 7 8 9 10 11 12
$e = \frac{2\cdot A}{ f}$
1 2 3 4 5 6 7 8 9 10 11 12
$f = \frac{2\cdot A}{ e}$
1 2 3 4 5 6 7 8 9 10 11 12
Beispiel Nr: 03
$\begin{array}{l} \text{Gegeben:}\\\text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Diagonale e} \qquad e \qquad [m] \\ \\ \text{Gesucht:} \\\text{Diagonale f} \qquad f \qquad [m] \\ \\ f = \frac{2\cdot A}{ e}\\ \textbf{Gegeben:} \\ A=\frac{1}{2}m^{2} \qquad e=4m \qquad \\ \\ \textbf{Rechnung:} \\ f = \frac{2\cdot A}{ e} \\ A=\frac{1}{2}m^{2}\\ e=4m\\ f = \frac{2\cdot \frac{1}{2}m^{2}}{ 4m}\\\\f=\frac{1}{4}m \\\\\\ \small \begin{array}{|l|} \hline A=\\ \hline \frac{1}{2} m^2 \\ \hline 50 dm^2 \\ \hline 5\cdot 10^{3} cm^2 \\ \hline 5\cdot 10^{5} mm^2 \\ \hline 0,005 a \\ \hline 5\cdot 10^{-5} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline f=\\ \hline \frac{1}{4} m \\ \hline 2\frac{1}{2} dm \\ \hline 25 cm \\ \hline 250 mm \\ \hline 2,5\cdot 10^{5} \mu m \\ \hline \end{array} \end{array}$