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r okregu wpisanego=hΔ
4π=a√3:2
a√3=8π
a=8π√3:2
pole=6a²√3:4=6(8π√3:2)²√3:4=
72√3 π²j.²
2]
r okregu wpisanego=⅓h
⅓a√3:2=3
a√3=18
a=18√3:'3
a=6√3= bok trójkąta