Odpowiedź:
[tex]h=\frac{a\sqrt{3}}{2}\\\\h=12\sqrt{3}\\\\12\sqrt{3}=\frac{a\sqrt{3}}{2}\ \ /\cdot2\\\\24\sqrt{3}=a\sqrt{3}\\\\a\sqrt{3}=24\sqrt{3}\ \ /:\sqrt{3}\\\\a=\frac{24\sqrt{3}}{\sqrt{3}}\\\\a=24\\\\\\P=\frac{a^2\sqrt{3}}{4}\\\\P=\frac{24^2\sqrt{3}}{4}=\frac{\not576^1^4^4\sqrt{3}}{\not4_{1}}=144\sqrt{3}\\\\\\Odp.Pole\ \ tr\'ojkata\ \ r\'ownobocznego\ \ wynosi\ \ 144\sqrt{3}[/tex]
