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$ d= 2\cdot r $
$ r= \frac{d}{2} $
$ A = r^{2} \cdot \pi $
$ r = \sqrt{\frac{A}{\pi }} $
$ U = 2\cdot r\cdot \pi $
$ r = \frac{ U}{2\cdot \pi } $
Geometrie-Kreis-Kreis
$d= 2\cdot r$
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$r= \frac{d}{2}$
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$A = r^{2} \cdot \pi$
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$r = \sqrt{\frac{A}{\pi }}$
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$U = 2\cdot r\cdot \pi$
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$r = \frac{ U}{2\cdot \pi }$
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Beispiel Nr: 06
$\begin{array}{l}
\text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\
\text{Radius} \qquad r \qquad [m] \\
\\ \text{Gesucht:} \\\text{Umfang} \qquad U \qquad [m] \\
\\ U = 2\cdot r\cdot \pi\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad r=\frac{1}{20}m \qquad \\ \\ \textbf{Rechnung:} \\
U = 2\cdot r\cdot \pi \\
\pi=3\frac{16}{113}\\
r=\frac{1}{20}m\\
U = 2\cdot \frac{1}{20}m\cdot 3\frac{16}{113}\\\\U=0,314m
\\\\ \small \begin{array}{|l|} \hline r=\\ \hline \frac{1}{20} m \\ \hline \frac{1}{2} dm \\ \hline 5 cm \\ \hline 50 mm \\ \hline 5\cdot 10^{4} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline U=\\ \hline 0,314 m \\ \hline 3\frac{16}{113} dm \\ \hline 31,4 cm \\ \hline 314 mm \\ \hline 314159\frac{27}{100} \mu m \\ \hline \end{array} \end{array}$