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⅓h=⅓a√3:2=⅙a√3
pole okregu wpisanego=πr²=π×(⅙a√3)²=¹/₁₂a²π
r okregu opisanego=⅔hΔ=⅔×a√3:2=⅓a√3
pole okregu opisanego=π×(⅓a√3)²=⅓a²π
pole pierścienia=⅓a²π-¹/₁₂a²π=75πcm²/:π
¼a²=75
a²=75×4
a²=300
a=10√3
poleΔ=a²√3:4=(10√3)²√3:4=75√3cm²