Răspuns :
[tex]\it h_3=\dfrac{\ell\sqrt3}{2}\ \ \ \ \ (1)\\ \\ h_3=9\sqrt2\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow \dfrac{\ell\sqrt3}{2}=9\sqrt2 \Rightarrow \ell\sqrt3=2\cdot9\sqrt2\Big|_{\cdot\sqrt3} \Rightarrow \ell\cdot3=18\sqrt6|_{:3}\Rightarrow\\ \\ \Rightarrow \ell=6\sqrt6\\ \\ \\ R=\dfrac{\ell\sqrt3}{3}=\dfrac{6\sqrt6\sqrt3}{3}=2\sqrt{18}=2\sqrt{9\cdot2}=2\cdot3\sqrt2=6\sqrt2\\ \\ a_3=\dfrac{R}{2}=\dfrac{6\sqrt2}{2}=3\sqrt2[/tex]
[tex]\it \mathcal{A}=\dfrac{\ell\cdot h}{2}=\dfrac{6\sqrt6\cdot9\sqrt2}{2}=\dfrac{}{}3\cdot9\sqrt{12}=27\sqrt{4\cdot3}=27\cdot2\sqrt3=54\sqrt3[/tex]