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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
$\begin{array}{l}
\text{Gegeben:}\\\text{Radius} \qquad r \qquad [m] \\
\\ \text{Gesucht:} \\\text{Durchmesser} \qquad d \qquad [m] \\
\\ d= 2\cdot r\\ \textbf{Gegeben:} \\ r=\frac{1}{5}m \qquad \\ \\ \textbf{Rechnung:} \\
d= 2\cdot r \\
r=\frac{1}{5}m\\
d= 2\cdot \frac{1}{5}m\\\\d=\frac{2}{5}m
\\\\\\ \small \begin{array}{|l|} \hline r=\\ \hline \frac{1}{5} m \\ \hline 2 dm \\ \hline 20 cm \\ \hline 200 mm \\ \hline 2\cdot 10^{5} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline d=\\ \hline \frac{2}{5} m \\ \hline 4 dm \\ \hline 40 cm \\ \hline 400 mm \\ \hline 4\cdot 10^{5} \mu m \\ \hline \end{array} \end{array}$