题目如图,正八边形ABCDEFGHABCDEFGHABCDEFGH内接于⊙O\odot O⊙O,连接AFAFAF、BFBFBF,若S△ABF=42S_{△ABF}=4\sqrt{2}S△ABF=42,则⊙O\odot O⊙O的半径为( )A.111B.2\sqrt{2}2C.222\sqrt{2}22D.222知识点:正多边形和圆章节:图形的性质 / 圆