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a=bok Δ
⅔h Δ=⅔a√3:2=⅓a√3
r koła=⅓a√3
pole koła=πr²=3,14×(⅓a√3)²=⅓a²≈0,33a²
poleΔ=a²√3:4
poleΔ=¼a²√3≈0,43a²
0,43a²=100%
0,33 a²=x%
x=0,33×100:0,43≈76.74%
odp. pole trójkąta stanowi około 76,74% pola koła