Răspuns :
Notăm secțiunea axială ABA'B'. AB = 2R, AA'=h (G)
[tex]\it \mathcal{A}_{sec.ax.}=AA'\cdot AB=h\cdot2R=180\bigg|_{:2} \Rightarrow h\cdot R=90\ \ \ \ \ \ (1)\\ \\ \\ \mathcal{A}_b=\pi R^2=25\pi\bigg|_{:\pi} \Rightarrow R^2=25 \Rightarrow R=5\ cm\ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow h\cdot5=90\bigg|_{:5} \Rightarrow h=18\ cm\ =\ G\\ \\ \\ \mathcal{A}_\ell=2\pi R G=2\pi\cdot5\cdot18=180\pi\ cm^2[/tex]