Pole trójkąta równobocznego obliczamy korzystając ze wzoru
[tex]P=\frac{a^2\sqrt{3}}{4}\ \ \ \ gdzie:\ \ a - dlugo\'s\'c\ \ boku\ \ tr\'ojkata[/tex]
[tex]a)\ \ a=0,6\\\\P=\frac{0,6^2\cdot\sqrt{3}}{4}=\frac{\not0,36^0^,^0^9\cdot\sqrt{3}}{\not4_{1}}=0,09\sqrt{3}\\\\\\b)\ \ a=1\frac{3}{5}\\\\P=\frac{(1\frac{3}{5})^2\cdot\sqrt{3}}{4}=\frac{(\frac{8}{5})^2\cdot\sqrt{3}}{4}=\frac{\frac{64}{25}\sqrt{3}}{4}=\frac{64}{25}\sqrt{3}:4=\frac{64}{25}\sqrt{3}\cdot\frac{1}{4}=\frac{64}{100}\sqrt{3}=0,64\sqrt{3}[/tex]
[tex]c)\ \ a=2\sqrt{2}\\\\P=\frac{(2\sqrt{2})^2\cdot\sqrt{3}}{4}=\frac{2^2\cdot(\sqrt{2})^2\cdot\sqrt{3}}{4}=\frac{4\cdot2\sqrt{3}}{4}=\frac{\not8^2\sqrt{3}}{\not4_{1}}=2\sqrt{3}\\\\\\d)\ \ a=3\sqrt{3}\\\\P=\frac{(3\sqrt{3})^2\cdot\sqrt{3}}{4}=\frac{3^2\cdot(\sqrt{3})^2\cdot\sqrt{3}}{4}=\frac{9\cdot3\sqrt{3}}{4}=\frac{27\sqrt{3}}{4}=\frac{27}{4}\sqrt{3}=6\frac{3}{4}\sqrt{3}=6,75\sqrt{3}[/tex]
[tex]e)\ \ a=4\sqrt{5}\\\\P=\frac{(4\sqrt{5})^2\cdot\sqrt{3}}{4}=\frac{4^2\cdot(\sqrt{5})^2\cdot\sqrt{3}}{4}=\frac{16\cdot5\sqrt{3}}{4}=\frac{\not80^2^0\sqrt{3}}{\not4_{1}}=20\sqrt{3}[/tex]
