Geometrie-Viereck-Raute



Beispiel Nr: 10
$ \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}{2}m \qquad A=\frac{1}{5}m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ e = \frac{2\cdot A}{ f} \\ f=1\frac{1}{2}m\\ A=\frac{1}{5}m^{2}\\ e = \frac{2\cdot \frac{1}{5}m^{2}}{ 1\frac{1}{2}m}\\\\e=\frac{4}{15}m \\\\\\ \small \begin{array}{|l|} \hline f=\\ \hline 1\frac{1}{2} m \\ \hline 15 dm \\ \hline 150 cm \\ \hline 1,5\cdot 10^{3} mm \\ \hline 1,5\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline \frac{1}{5} m^2 \\ \hline 20 dm^2 \\ \hline 2\cdot 10^{3} cm^2 \\ \hline 2\cdot 10^{5} mm^2 \\ \hline 0,002 a \\ \hline 2\cdot 10^{-5} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline \frac{4}{15} m \\ \hline 2\frac{2}{3} dm \\ \hline 26\frac{2}{3} cm \\ \hline 266\frac{2}{3} mm \\ \hline 266666\frac{2}{3} \mu m \\ \hline \end{array}$