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a=3×2=6= bok małego Δ
pole =a²√3:4=6²√3:4=9√3= pole małego Δ
2√3=⅓ a√3:2
2√3=a√3:6
a=2×6=12
bok duzego Δ
pole=12²√3:4=36√3
pole zacieniowane=36√3-9√3=27√3j.²