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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    $r = \frac{b\cdot 360}{\alpha \cdot \pi \cdot 2}$
    1 2 3 4
    $\alpha = \frac{b\cdot 360}{r\cdot \pi \cdot 2}$
    1 2 3 4

Beispiel Nr: 01
$ \text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Radius} \qquad r \qquad [m] \\ \\ \text{Gesucht:} \\\text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \\ \alpha = \frac{A\cdot 360}{r^{2} \cdot \pi }\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad A=5m^{2} \qquad r=6m \qquad \\ \\ \textbf{Rechnung:} \\ \alpha = \frac{A\cdot 360}{r^{2} \cdot \pi } \\ \pi=3\frac{16}{113}\\ A=5m^{2}\\ r=6m\\ \alpha = \frac{5m^{2}\cdot 360}{(6m)^{2} \cdot 3\frac{16}{113} }\\\\\alpha=15,9^{\circ} \\\\ \small \begin{array}{|l|} \hline A=\\ \hline 5 m^2 \\ \hline 500 dm^2 \\ \hline 5\cdot 10^{4} cm^2 \\ \hline 5\cdot 10^{6} mm^2 \\ \hline \frac{1}{20} a \\ \hline 0,0005 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline r=\\ \hline 6 m \\ \hline 60 dm \\ \hline 600 cm \\ \hline 6\cdot 10^{3} mm \\ \hline 6\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline alpha=\\ \hline 15,9 ° \\ \hline 955 \text{'} \\ \hline 5,73\cdot 10^{4} \text{''} \\ \hline 17,7 gon \\ \hline \frac{5}{18} rad \\ \hline \end{array}$