👤

oblicz wysokosc i obwód teojkata rownobocznegi o polu rownym 16 pierwiastkow z 3​

Odpowiedź :

[tex]P = \frac{a^{2}\sqrt{3}}{4}\\oraz\\P = 16\sqrt{3}\\\\\frac{a^{2}\sqrt{3}}{4} = 16\sqrt{3} \ \ /\cdot\frac{4}{\sqrt{3}}\\\\a^{2} = 64\\\\a = \sqrt{64}\\\\\underline{a = 8}\\\\Ob = 3a\\\\Ob = 3\cdot8\\\\\boxed{Obw = 24}[/tex]

[tex]h = \frac{a\sqrt{3}}{2}\\\\h = \frac{8\sqrt{3}}{2}\\\\\boxed{h = 4\sqrt{3}}[/tex]