[tex]\alpha = 60^{o}\\L = 16 \ cm\\oraz\\L = 4a\\\\4a = 16 \ cm \ \ /:4\\\\a = 4 \ cm\\\\\\P = a^{2} sin\alpha\\\\P = (4 \ cm)^{2}\cdot sin60^{o}\\\\P = 16 \ cm^{2}\cdot\frac{\sqrt{3}}{2}\\\\\boxed{P = 8\sqrt{3} \ cm^{2}}[/tex]
rozwiązanie w załączniku
