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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: 07
$\begin{array}{l}
\text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\
\text{Radius} \qquad r \qquad [m] \\
\\ \text{Gesucht:} \\\text{Umfang} \qquad U \qquad [m] \\
\\ U = 2\cdot r\cdot \pi\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad r=2\frac{1}{4}m \qquad \\ \\ \textbf{Rechnung:} \\
U = 2\cdot r\cdot \pi \\
\pi=3\frac{16}{113}\\
r=2\frac{1}{4}m\\
U = 2\cdot 2\frac{1}{4}m\cdot 3\frac{16}{113}\\\\U=14,1m
\\\\ \small \begin{array}{|l|} \hline r=\\ \hline 2\frac{1}{4} m \\ \hline 22\frac{1}{2} dm \\ \hline 225 cm \\ \hline 2,25\cdot 10^{3} mm \\ \hline 2,25\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline U=\\ \hline 14,1 m \\ \hline 141 dm \\ \hline 1,41\cdot 10^{3} cm \\ \hline 1,41\cdot 10^{4} mm \\ \hline 14137167\frac{3}{20} \mu m \\ \hline \end{array} \end{array}$