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: 04
$ \text{Gegeben:}\\\text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Radius} \qquad r \qquad [m] \\ \\ \text{Gesucht:} \\\text{Fläche} \qquad A \qquad [m^{2}] \\ \\ A = \frac{r^{2} \cdot \pi \cdot \alpha }{ 360}\\ \textbf{Gegeben:} \\ \alpha=30^{\circ} \qquad \pi=3\frac{16}{113} \qquad r=5m \qquad \\ \\ \textbf{Rechnung:} \\ A = \frac{r^{2} \cdot \pi \cdot \alpha }{ 360} \\ \alpha=30^{\circ}\\ \pi=3\frac{16}{113}\\ r=5m\\ A = \frac{(5m)^{2} \cdot 3\frac{16}{113} \cdot 30^{\circ} }{ 360}\\\\A=6,54m^{2} \\\\\\ \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 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 6,54 m^2 \\ \hline 654 dm^2 \\ \hline 6,54\cdot 10^{4} cm^2 \\ \hline 6544984\frac{19}{24} mm^2 \\ \hline 0,0654 a \\ \hline 0,000654 ha \\ \hline \end{array}$