[tex]h = 9\\oraz\\h = \frac{a\sqrt{3}}{2}\\\\\frac{a\sqrt{3}}{2} = 9 \ \ /\cdot2\\\\a\sqrt{3} = 18 \ \//:\sqrt{3}\\\\a = \frac{18}{\sqrt{3}}=\frac{18}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{18\sqrt{3}}{3} = 6\sqrt{3}\\\\\boxed{a = 6\sqrt{3}}[/tex]
Odp. Bok tego trójkąta wynosi 6√3 [j].
