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圓周長C=2πR=4π(CM),所以R=2內接正三角形ABC,連結OA、OB,已求OA=OB=2作OD⊥AB於點D∠AOB=120°(A、B、C三分圓周)OA=OB且OD⊥AB於點D,∠AOD=∠BOD=60°RT△AOD中,∠OAD=30°,OD=½;OA=1,AO=√3,AB=2AO=2√3S…
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