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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
            
        
           $\begin{array}{l} 
      \text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\
      \text{Radius (außerer Kreis)} \qquad r_{a} \qquad [m] \\
      \text{Radius (innerer Kreis)} \qquad r_{i} \qquad [m] \\
      \\ \text{Gesucht:} \\\text{Fläche} \qquad A \qquad [m^{2}] \\
     \\ A = (r_{a} ^{2}  - r_{i} ^{2} )\cdot\pi\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad r_{a}=14m \qquad r_{i}=6m \qquad \\ \\ \textbf{Rechnung:} \\
      A = (r_{a} ^{2}  - r_{i} ^{2} )\cdot \pi \\
      \pi=3\frac{16}{113}\\
      r_{a}=14m\\
      r_{i}=6m\\
      A = (14m ^{2}  - 6m ^{2} )\cdot 3\frac{16}{113}\\\\A=503m^{2}
    \\\\ \small \begin{array}{|l|} \hline ra=\\  \hline 14 m  \\  \hline 140 dm  \\  \hline 1,4\cdot 10^{3} cm  \\  \hline 1,4\cdot 10^{4} mm  \\  \hline 1,4\cdot 10^{7} \mu m  \\ \hline \end{array} \small \begin{array}{|l|} \hline ri=\\  \hline 6 m  \\  \hline 60 dm  \\  \hline 600 cm  \\  \hline 6\cdot 10^{3} mm  \\  \hline 6\cdot 10^{6} \mu m  \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\  \hline 503 m^2  \\  \hline 5,03\cdot 10^{4} dm^2  \\  \hline 5026548\frac{8}{25} cm^2  \\  \hline 5,03\cdot 10^{8} mm^2  \\  \hline 5\frac{3}{113} a  \\  \hline 0,0503 ha  \\ \hline \end{array}  \end{array}$