如图,AB是⊙O的直径,弦CD⊥AB,垂足为E,若AE=2,CD=8,则⊙O的半径为(  )A.4B.5C.8D.1

2025-05-09 07:33:31
推荐回答(1个)
回答1:

解答:解:连接OC,
∵CD⊥AB,∴CE=

1
2
CD=4
OC=OA,OE=OA-AE,由勾股定理可得OC2=CE2+(OA-AE)2,解得OC=5
故选B.