Odpowiedź:
P = 27√3 [j²]
Szczegółowe wyjaśnienie:
[tex]h = \frac{a\sqrt{3}}{2} \ - \ wysokosc \ trojkata \ rownobocznego\\\\R = \frac{2}{3}h = \frac{a\sqrt{3}}{3} \ - \ promien \ okregu \ opisanego \ na \ trojkacie \ rownobocznym\\\\r = \frac{1}{3}h = \frac{a\sqrt{3}}{6} \ - \ promien \ okregu \ wpisanego \ w \ trojkat \ rownoboczny\\\\\\R - r = 3\\\\\frac{a\sqrt{3}}{3}-\frac{a\sqrt{3}}{6}=3\\\\\frac{2a\sqrt{3}}{6}-\frac{a\sqrt{3}}{6} = 3\\\\\frac{a\sqrt{3}}{6} = 3 \ \ \ |\cdot\frac{6}{\sqrt{3}}[/tex]
[tex]a = \frac{3\cdot6}{\sqrt{3}} = \frac{3\cdot6}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{3\cdot6\sqrt{3}}{3} = 6\sqrt{3}[/tex]
[tex]P = \frac{a^{2}\sqrt{3}}{4}\\\\P = \frac{(6\sqrt{3})^{2}\cdot\sqrt{3}}{4} = \frac{36\cdot3\sqrt{3}}{4}=9\cdot3\sqrt{3}\\\\\boxed{P = 27\sqrt{3}}[/tex]
