Beispiel Nr: 05
$ \text{Gegeben:}\\\text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Radius} \qquad r \qquad [m] \\ \\ \text{Gesucht:} \\\text{Fläche} \qquad A \qquad [m^{2}] \\ \\ A = \frac{r^{2} \cdot \pi \cdot \alpha }{ 360}\\ \textbf{Gegeben:} \\ \alpha=30^{\circ} \qquad \pi=3,14 \qquad r=2\frac{1}{2}m \qquad \\ \\ \textbf{Rechnung:} \\ A = \frac{r^{2} \cdot \pi \cdot \alpha }{ 360} \\ \alpha=30^{\circ}\\ \pi=3,14\\ r=2\frac{1}{2}m\\ A = \frac{(2\frac{1}{2}m)^{2} \cdot 3,14 \cdot 30^{\circ} }{ 360}\\\\A=1,64m^{2} \\\\\\ \tiny \begin{array}{|l|} \hline alpha=\\ \hline 30 ° \\ \hline 1,8\cdot 10^{3} \text{'} \\ \hline 1,08\cdot 10^{5} \text{''} \\ \hline 33,3 gon \\ \hline 0,524 rad \\ \hline 0,000524 mrad \\ \hline \end{array} \tiny \begin{array}{|l|} \hline r=\\ \hline 2,5 m \\ \hline 25 dm \\ \hline 250 cm \\ \hline 2,5\cdot 10^{3} mm \\ \hline 2,5\cdot 10^{6} \mu m \\ \hline 0,0025 km \\ \hline \end{array} \tiny \begin{array}{|l|} \hline A=\\ \hline 1,64 m^2 \\ \hline 164 dm^2 \\ \hline 1,64\cdot 10^{4} cm^2 \\ \hline 1,64\cdot 10^{6} mm^2 \\ \hline 0,0164 a \\ \hline 0,000164 ha \\ \hline 1,64\cdot 10^{-6} km^2 \\ \hline \end{array}$