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$ A = (r_{a} ^{2} - r_{i} ^{2} )\cdot\pi $
$ r_{a} = \sqrt{\frac{A}{\pi } + r_{i} ^{2} } $
$ r_{i} = \sqrt{r_{a} ^{2} - \frac{A}{\pi } } $
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: 02
$\begin{array}{l}
\text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\
\text{Radius (außerer Kreis)} \qquad r_{a} \qquad [m] \\
\text{Radius (innerer Kreis)} \qquad r_{i} \qquad [m] \\
\\ \text{Gesucht:} \\\text{Fläche} \qquad A \qquad [m^{2}] \\
\\ A = (r_{a} ^{2} - r_{i} ^{2} )\cdot\pi\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad r_{a}=1\frac{2}{5}m \qquad r_{i}=\frac{1}{6}m \qquad \\ \\ \textbf{Rechnung:} \\
A = (r_{a} ^{2} - r_{i} ^{2} )\cdot \pi \\
\pi=3\frac{16}{113}\\
r_{a}=1\frac{2}{5}m\\
r_{i}=\frac{1}{6}m\\
A = (1\frac{2}{5}m ^{2} - \frac{1}{6}m ^{2} )\cdot 3\frac{16}{113}\\\\A=6,07m^{2}
\\\\ \small \begin{array}{|l|} \hline ra=\\ \hline 1\frac{2}{5} m \\ \hline 14 dm \\ \hline 140 cm \\ \hline 1,4\cdot 10^{3} mm \\ \hline 1,4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline ri=\\ \hline \frac{1}{6} m \\ \hline 1\frac{2}{3} dm \\ \hline 16\frac{2}{3} cm \\ \hline 166\frac{2}{3} mm \\ \hline 166666\frac{2}{3} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline 6,07 m^2 \\ \hline 607 dm^2 \\ \hline 6,07\cdot 10^{4} cm^2 \\ \hline 6,07\cdot 10^{6} mm^2 \\ \hline 0,0607 a \\ \hline 0,000607 ha \\ \hline \end{array} \end{array}$