∵△ABC是等边三角形
∴∠BAC=∠ABC=∠ACB=60°
∴∠BOC=120°
∵OB=OC=R
OD⊥BC
∴垂经定理：BD=CD
∠COD=∠BOD=60°
∴∠OCD=∠OBD=30°
∴边心距OD=1/2OC=R/2
∴边长=2BD=2√(OB²-OD²)=2√[R²-(R/2)²]=√3R
面积=3S△BOC=3×1/2×√3R×R/2=3√3R/42√[R²-(R/2)²]为什么等于√3R边长=BD+CD=2BD=2√(OB²-OD²)=2√(R²-R²/4)=2√(4R²/4-R²/4)=2√(3R²/4)=2×R/2×√3=√3R