Odpowiedź:
[tex]P = \frac{a^{2} \sqrt{3} }{4} \\Pa = \frac{3^{2} \sqrt{3} }{4} = \frac{9 \sqrt{3} }{4} \\Pb = \frac{11^{2} \sqrt{3} }{4} = \frac{121 \sqrt{3} }{4} \\Pc = \frac{15^{2} \sqrt{3} }{4} = \frac{225 \sqrt{3} }{4} \\[/tex]
[tex]\frac{9\sqrt{3} }{4} : \frac{121\sqrt{3} }{4} : \frac{225\sqrt{3} }{4}\\9:121:225[/tex]