Geometrie-Viereck-Raute

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