Geometrie-Viereck-Raute

Beispiel Nr: 11
$\text{Gegeben:}\\\text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Diagonale e} \qquad e \qquad [m] \\ \\ \text{Gesucht:} \\\text{Diagonale f} \qquad f \qquad [m] \\ \\ f = \frac{2\cdot A}{ e}\\ \textbf{Gegeben:} \\ A=1\frac{1}{5}m^{2} \qquad e=1\frac{1}{2}m \qquad \\ \\ \textbf{Rechnung:} \\ f = \frac{2\cdot A}{ e} \\ A=1\frac{1}{5}m^{2}\\ e=1\frac{1}{2}m\\ f = \frac{2\cdot 1\frac{1}{5}m^{2}}{ 1\frac{1}{2}m}\\\\f=1\frac{3}{5}m \\\\\\ \small \begin{array}{|l|} \hline A=\\ \hline 1\frac{1}{5} m^2 \\ \hline 120 dm^2 \\ \hline 1,2\cdot 10^{4} cm^2 \\ \hline 1,2\cdot 10^{6} mm^2 \\ \hline 0,012 a \\ \hline 0,00012 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline 1\frac{1}{2} m \\ \hline 15 dm \\ \hline 150 cm \\ \hline 1,5\cdot 10^{3} mm \\ \hline 1,5\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline f=\\ \hline 1\frac{3}{5} m \\ \hline 16 dm \\ \hline 160 cm \\ \hline 1,6\cdot 10^{3} mm \\ \hline 1,6\cdot 10^{6} \mu m \\ \hline \end{array}$