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: 03
$\begin{array}{l} \text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \\ \text{Gesucht:} \\\text{Radius} \qquad r \qquad [m] \\ \\ r = \sqrt{\frac{A\cdot 360}{\alpha \cdot \pi }}\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad A=15m^{2} \qquad \alpha=90^{\circ} \qquad \\ \\ \textbf{Rechnung:} \\ r = \sqrt{\frac{A\cdot 360}{\alpha \cdot \pi }} \\ \pi=3\frac{16}{113}\\ A=15m^{2}\\ \alpha=90^{\circ}\\ r = \sqrt{\frac{15m^{2}\cdot 360}{90^{\circ} \cdot 3\frac{16}{113} }}\\\\r=4,37m \\\\ \small \begin{array}{|l|} \hline A=\\ \hline 15 m^2 \\ \hline 1,5\cdot 10^{3} dm^2 \\ \hline 1,5\cdot 10^{5} cm^2 \\ \hline 1,5\cdot 10^{7} mm^2 \\ \hline \frac{3}{20} a \\ \hline 0,0015 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline alpha=\\ \hline 90 ° \\ \hline 5,4\cdot 10^{3} \text{'} \\ \hline 3,24\cdot 10^{5} \text{''} \\ \hline 100 gon \\ \hline 1,57 rad \\ \hline \end{array} \small \begin{array}{|l|} \hline r=\\ \hline 4,37 m \\ \hline 43,7 dm \\ \hline 437 cm \\ \hline 4,37\cdot 10^{3} mm \\ \hline 4,37\cdot 10^{6} \mu m \\ \hline \end{array} \end{array}$