Geometrie-Kreis-Kreisring

• $A = (r_{a} ^{2} - r_{i} ^{2} )\cdot\pi$
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$r_{a} = \sqrt{\frac{A}{\pi } + r_{i} ^{2} }$
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$r_{i} = \sqrt{r_{a} ^{2} - \frac{A}{\pi } }$
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Beispiel Nr: 01
$\text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Radius (außerer Kreis)} \qquad r_{a} \qquad [m] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \\ \text{Gesucht:} \\\text{Radius (innerer Kreis)} \qquad r_{i} \qquad [m] \\ \\ r_{i} = \sqrt{r_{a} ^{2} - \frac{A}{\pi } }\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad r_{a}=5m \qquad A=6m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ r_{i} = \sqrt{r_{a} ^{2} - \frac{A}{\pi } } \\ \pi=3\frac{16}{113}\\ r_{a}=5m\\ A=6m^{2}\\ r_{i} = \sqrt{5m ^{2} - \frac{6m^{2}}{3\frac{16}{113} } }\\\\r_{i}=4,81m \\\\ \small \begin{array}{|l|} \hline ra=\\ \hline 5 m \\ \hline 50 dm \\ \hline 500 cm \\ \hline 5\cdot 10^{3} mm \\ \hline 5\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline 6 m^2 \\ \hline 600 dm^2 \\ \hline 6\cdot 10^{4} cm^2 \\ \hline 6\cdot 10^{6} mm^2 \\ \hline \frac{3}{50} a \\ \hline 0,0006 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline ri=\\ \hline 4,81 m \\ \hline 48,1 dm \\ \hline 481 cm \\ \hline 4,81\cdot 10^{3} mm \\ \hline 4,81\cdot 10^{6} \mu m \\ \hline \end{array}$