Geometrie-Viereck-Raute



Beispiel Nr: 07
$ \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=1\frac{2}{3}m^{2} \qquad e=\frac{4}{5}m \qquad \\ \\ \textbf{Rechnung:} \\ f = \frac{2\cdot A}{ e} \\ A=1\frac{2}{3}m^{2}\\ e=\frac{4}{5}m\\ f = \frac{2\cdot 1\frac{2}{3}m^{2}}{ \frac{4}{5}m}\\\\f=4\frac{1}{6}m \\\\\\ \small \begin{array}{|l|} \hline A=\\ \hline 1\frac{2}{3} m^2 \\ \hline 166\frac{2}{3} dm^2 \\ \hline 16666\frac{2}{3} cm^2 \\ \hline 1666666\frac{2}{3} mm^2 \\ \hline \frac{1}{60} a \\ \hline 0,000167 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline \frac{4}{5} m \\ \hline 8 dm \\ \hline 80 cm \\ \hline 800 mm \\ \hline 8\cdot 10^{5} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline f=\\ \hline 4\frac{1}{6} m \\ \hline 41\frac{2}{3} dm \\ \hline 416\frac{2}{3} cm \\ \hline 4166\frac{2}{3} mm \\ \hline 4166666\frac{2}{3} \mu m \\ \hline \end{array}$