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?12msin?． ··········································· 7分 2
（3）存在．分两种情况说明如下： ···················································································· 8分 ①当AB与CD相交时，
由（2）及AB?CD?2R知S四边形ACBD?②当AB与CD不相交时，如图④ 1AB·CD·sin??R2sin?． ······················· 9分 2A D 2 ∵AB?CD?2R，OC?OD?OA?OB?R， ∴?AOB??COD?90°，
而S四边形ABCD?SRt△AOB?SRt△OCD?S△AOD?S△BOC B O 3 1 H E C ?R?S△AOD?S△BOC2（第25题答案图④）
． ····································································································· 10分
延长BO交?O于点E，连接EC，则?1??3??2??3?90°． ∴?1??2． ∴△AOD≌△COE． ∴S△AOD?S△OCE． ∴S△AOD?S△BOC?S△OCE?S△BOC?S△BCE． 过点C作CH?BE，垂足为H， 则S△BCE?1BE·CH?R·CH． 2∴当CH?R时，S△BCE取最大值R2． ············································································ 11分
综合①、②可知，当?1??2?90°，即四边形ABCD是边长为2R的正方形时， ············································································ 12分 S四边形ABCD?R2?R2?2R2为最大值． ·
11