若正△ABC外接圆的半径为R，则△ABC的面积为______
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连接OB，OA，延长AO交BC于D，∵正△ABC外接圆是⊙O，
∴AD⊥BC，BD=CD=
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即OD=
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由勾股定理得：BD=
| OB2? OD2 |
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即BC=2BD=
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则△ABC的面积是
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连接OB，OA，延长AO交BC于D，| 1 |
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| OB2? OD2 |
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