Geometrie-Viereck-Rechtwinkliges Trapez

• $A = \frac{a+c}{ 2}\cdot h$
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$a = \frac{2\cdot A}{ h} - c$
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$c = \frac{2\cdot A}{ h} - a$
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$h = \frac{2\cdot A}{a+c}$
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Beispiel Nr: 01
$\text{Gegeben:}\\\text{Grundlinie a} \qquad a \qquad [m] \\ \text{Höhe} \qquad h \qquad [m] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \\ \text{Gesucht:} \\\text{Grundlinie c} \qquad c \qquad [m] \\ \\ c = \frac{2\cdot A}{ h} - a\\ \textbf{Gegeben:} \\ a=4m \qquad h=5m \qquad A=16m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ c = \frac{2\cdot A}{ h} - a \\ a=4m\\ h=5m\\ A=16m^{2}\\ c = \frac{2\cdot 16m^{2}}{ 5m} - 4m\\\\c=2\frac{2}{5}m \\\\\\ \small \begin{array}{|l|} \hline a=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline h=\\ \hline 5 m \\ \hline 50 dm \\ \hline 500 cm \\ \hline 5\cdot 10^{3} mm \\ \hline 5\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline 16 m^2 \\ \hline 1,6\cdot 10^{3} dm^2 \\ \hline 1,6\cdot 10^{5} cm^2 \\ \hline 1,6\cdot 10^{7} mm^2 \\ \hline \frac{4}{25} a \\ \hline 0,0016 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline c=\\ \hline 2\frac{2}{5} m \\ \hline 24 dm \\ \hline 240 cm \\ \hline 2,4\cdot 10^{3} mm \\ \hline 2,4\cdot 10^{6} \mu m \\ \hline \end{array}$