[tex]a)\\\\a=9\ cm\\\\wzor\ na\ pole\ trojkata\ rownobocznego:\\\\P=\frac{a^2\sqrt{3}}{4}\\\\P=\frac{9^2\sqrt{3}}{4}=\frac{81\sqrt{3}}{4}=20,25\sqrt{3}\ cm^2[/tex]
[tex]b)\\\\a=22\ cm \\\\P=\frac{22^2\sqrt{3}}{4}=\frac{484\sqrt{3}}{4}=121\sqrt{3}\ cm^2[/tex]
