Geometrie-Stereometrie-Kreiszylinder

$V = r^{2} \cdot \pi \cdot h$
1 2
$r = \sqrt{\frac{ V}{\pi \cdot h}}$
1
$h = \frac{ V}{r^{2} \cdot \pi }$
1
$O = 2\cdot r\cdot \pi \cdot (r+h)$
1 2
$r = 0,5\cdot (-h+\sqrt{h^{2} +\frac{O}{\pi }})$
1
$h = \frac{0-2\cdot \pi \cdot r^{2} }{ 2\cdot r\cdot \pi }$
1
Beispiel Nr: 01
$\begin{array}{l} \text{Gegeben:}\\\text{Körperhöhe} \qquad h \qquad [m] \\ \text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Oberfläche} \qquad O \qquad [m^{2}] \\ \\ \text{Gesucht:} \\\text{Radius} \qquad r \qquad [m] \\ \\ r = 0,5\cdot (-h+\sqrt{h^{2} +\frac{O}{\pi }})\\ \textbf{Gegeben:} \\ h=4m \qquad \pi=3\frac{16}{113} \qquad O=30m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ r = 0,5\cdot (-h+\sqrt{h^{2} +\frac{O}{\pi }}) \\ h=4m\\ \pi=3\frac{16}{113}\\ O=30m^{2}\\ r = 0,5\cdot (-4m+\sqrt{h^{2} +\frac{30m^{2}}{3\frac{16}{113} }})\\\\r=0,962m \\\\\\ \small \begin{array}{|l|} \hline h=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline O=\\ \hline 30 m^2 \\ \hline 3\cdot 10^{3} dm^2 \\ \hline 3\cdot 10^{5} cm^2 \\ \hline 3\cdot 10^{7} mm^2 \\ \hline \frac{3}{10} a \\ \hline 0,003 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline r=\\ \hline 0,962 m \\ \hline 9,62 dm \\ \hline 96,2 cm \\ \hline 962 mm \\ \hline 9,62\cdot 10^{5} \mu m \\ \hline \end{array} \end{array}$