Geometrie-Viereck-Raute



Beispiel Nr: 04
$ \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=12m \qquad A=14m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ e = \frac{2\cdot A}{ f} \\ f=12m\\ A=14m^{2}\\ e = \frac{2\cdot 14m^{2}}{ 12m}\\\\e=2\frac{1}{3}m \\\\\\ \small \begin{array}{|l|} \hline f=\\ \hline 12 m \\ \hline 120 dm \\ \hline 1,2\cdot 10^{3} cm \\ \hline 1,2\cdot 10^{4} mm \\ \hline 1,2\cdot 10^{7} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline 14 m^2 \\ \hline 1,4\cdot 10^{3} dm^2 \\ \hline 1,4\cdot 10^{5} cm^2 \\ \hline 1,4\cdot 10^{7} mm^2 \\ \hline \frac{7}{50} a \\ \hline 0,0014 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 2\frac{1}{3} m \\ \hline 23\frac{1}{3} dm \\ \hline 233\frac{1}{3} cm \\ \hline 2333\frac{1}{3} mm \\ \hline 2333333\frac{1}{3} \mu m \\ \hline \end{array}$