Geometrie-Kreis-Kreissektor (Grad)

$A = \frac{r^{2} \cdot \pi \cdot \alpha }{ 360}$
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$r = \sqrt{\frac{A\cdot 360}{\alpha \cdot \pi }}$
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$\alpha = \frac{A\cdot 360}{r^{2} \cdot \pi }$
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$b = \frac{2\cdot r\cdot \pi \cdot \alpha }{ 360}$
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$r = \frac{b\cdot 360}{\alpha \cdot \pi \cdot 2}$
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$\alpha = \frac{b\cdot 360}{r\cdot \pi \cdot 2}$
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Beispiel Nr: 06
$\begin{array}{l} \text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \\ \text{Gesucht:} \\\text{Radius} \qquad r \qquad [m] \\ \\ r = \sqrt{\frac{A\cdot 360}{\alpha \cdot \pi }}\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad A=95m^{2} \qquad \alpha=30^{\circ} \qquad \\ \\ \textbf{Rechnung:} \\ r = \sqrt{\frac{A\cdot 360}{\alpha \cdot \pi }} \\ \pi=3\frac{16}{113}\\ A=95m^{2}\\ \alpha=30^{\circ}\\ r = \sqrt{\frac{95m^{2}\cdot 360}{30^{\circ} \cdot 3\frac{16}{113} }}\\\\r=19m \\\\ \small \begin{array}{|l|} \hline A=\\ \hline 95 m^2 \\ \hline 9,5\cdot 10^{3} dm^2 \\ \hline 9,5\cdot 10^{5} cm^2 \\ \hline 9,5\cdot 10^{7} mm^2 \\ \hline \frac{19}{20} a \\ \hline 0,0095 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline alpha=\\ \hline 30 ° \\ \hline 1,8\cdot 10^{3} \text{'} \\ \hline 1,08\cdot 10^{5} \text{''} \\ \hline 33\frac{1}{3} gon \\ \hline 0,524 rad \\ \hline \end{array} \small \begin{array}{|l|} \hline r=\\ \hline 19 m \\ \hline 190 dm \\ \hline 1,9\cdot 10^{3} cm \\ \hline 1,9\cdot 10^{4} mm \\ \hline 1,9\cdot 10^{7} \mu m \\ \hline \end{array} \end{array}$