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

Beispiel Nr: 07
$ \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=40^{\circ} \qquad \pi=3\frac{16}{113} \qquad r=5m \qquad \\ \\ \textbf{Rechnung:} \\ A = \frac{r^{2} \cdot \pi \cdot \alpha }{ 360} \\ \alpha=40^{\circ}\\ \pi=3\frac{16}{113}\\ r=5m\\ A = \frac{(5m)^{2} \cdot 3\frac{16}{113} \cdot 40^{\circ} }{ 360}\\\\A=8,73m^{2} \\\\\\ \small \begin{array}{|l|} \hline alpha=\\ \hline 40 ° \\ \hline 2,4\cdot 10^{3} \text{'} \\ \hline 1,44\cdot 10^{5} \text{''} \\ \hline 44\frac{4}{9} gon \\ \hline 0,698 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 8,73 m^2 \\ \hline 873 dm^2 \\ \hline 8,73\cdot 10^{4} cm^2 \\ \hline 8726646\frac{7}{18} mm^2 \\ \hline 0,0873 a \\ \hline 0,000873 ha \\ \hline \end{array}$