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