[tex]V= \pi R^{2}*H \\ \pi R^{2} H=6 \pi ; R^{2}H=6; R^{2}= \frac{6}{H} (1) [/tex]
Din conditia problemei
[tex]tg \alpha = \frac{H}{2R} \\ tg30= \frac{H}{2R} ; \frac{1}{ \sqrt{3} }= \frac{H}{2R} ;R= \frac{ \sqrt{3}H }{2 } \\ [/tex]
inlocuind in relatia de mai sus (1) obtinem
[tex] \frac{3H^2}{4}= \frac{6}{H} ;3H^3=24;H^3=8;H=2 \\ [/tex]
Atunci
[tex]R= \frac{ \sqrt{3}*2 }{2}= \sqrt{3} \\ A_{sec}=H*2R =2*2 \sqrt{3}=4 \sqrt{3}cm^2 [/tex]
