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Dla tego przykladu bedzie P=3*6²√3\2=3*36√3\2=54√3
a²√3
________ - pole trojkata rownobocznego.
4
wiec pole szesciokata foremnego o boku a=6 wynosi
a²√3
_______ * 6 = pole szesciokata [6*6²√3 ]/4 co daje 54 √3(j²)
4
gotowe.