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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: 09
$\begin{array}{l}
\text{Gegeben:}\\\text{Kreiszahl} \qquad \pi \qquad [] \\
\text{Umfang} \qquad U \qquad [m] \\
\\ \text{Gesucht:} \\\text{Radius} \qquad r \qquad [m] \\
\\ r = \frac{ U}{2\cdot \pi }\\ \textbf{Gegeben:} \\ \pi=3\frac{16}{113} \qquad U=4m \qquad \\ \\ \textbf{Rechnung:} \\
r = \frac{ U}{2\cdot \pi } \\
\pi=3\frac{16}{113}\\
U=4m\\
r = \frac{ 4m}{2\cdot 3\frac{16}{113} }\\\\r=0,637m
\\\\ \small \begin{array}{|l|} \hline U=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline r=\\ \hline 0,637 m \\ \hline 6\frac{26}{71} dm \\ \hline 63,7 cm \\ \hline 637 mm \\ \hline 6,37\cdot 10^{5} \mu m \\ \hline \end{array} \end{array}$