[tex]przekatna\ kwadratu:\ a=3\ cm\\\\ d=a\sqrt{2}\\\\ a\sqrt{2}=3\ \ |:\sqrt{2}\\\\ a=\frac{3}{\sqrt{2}} *\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\ cm \\\\\\pole\ kwadratu:\\\\P_{k}=a^2\\\\P_{k}=(\frac{3\sqrt{2}}{2})^2=\frac{9*2}{4}= \frac{18}{4}=4,5\ cm^2[/tex]
[tex]wysokosc\ trojkata\ rownobocznego:\ \ h=3\ cm\\\\h=\frac{a\sqrt{3}}{2}\\\\\frac{a\sqrt{3}}{2}=3\ \ |*2\\\\a\sqrt{3}=6\ \ |:\sqrt{3}\\\\a=\frac{6}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} =\frac{6\sqrt{3}}{3}=2\sqrt{3}\ cm\\\\\\P_{t}=\frac{a^2\sqrt{3}}{4}\\\\P_{t}=\frac{(2\sqrt{3})^2\sqrt{3}}{4}=\frac{4*3\sqrt{3}}{4}= 3\sqrt{3}\approx 3*1,73\approx 5,19\ cm^2[/tex]
[tex]4,5<3\sqrt{3}\\\\P_{k} < P_{t}\\\\odp.\ Pole\ kwadratu\ jest\ mniejsze\ od\ pola\ trojkata .[/tex]
