Geometrie-Viereck-Raute



Beispiel Nr: 05
$ \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=\frac{1}{3}m \qquad A=\frac{3}{4}m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ e = \frac{2\cdot A}{ f} \\ f=\frac{1}{3}m\\ A=\frac{3}{4}m^{2}\\ e = \frac{2\cdot \frac{3}{4}m^{2}}{ \frac{1}{3}m}\\\\e=4\frac{1}{2}m \\\\\\ \small \begin{array}{|l|} \hline f=\\ \hline \frac{1}{3} m \\ \hline 3\frac{1}{3} dm \\ \hline 33\frac{1}{3} cm \\ \hline 333\frac{1}{3} mm \\ \hline 333333\frac{1}{3} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline \frac{3}{4} m^2 \\ \hline 75 dm^2 \\ \hline 7,5\cdot 10^{3} cm^2 \\ \hline 7,5\cdot 10^{5} mm^2 \\ \hline 0,0075 a \\ \hline 7,5\cdot 10^{-5} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 4\frac{1}{2} m \\ \hline 45 dm \\ \hline 450 cm \\ \hline 4,5\cdot 10^{3} mm \\ \hline 4,5\cdot 10^{6} \mu m \\ \hline \end{array}$