(2009•威海)已知⊙O是△ABC的外接圆,若AB=AC=5,BC=6,则⊙O的半径为(  ) A.4 B.3.25 C.3.125 D.2.25

问题描述:

(2009•威海)已知⊙O是△ABC的外接圆,若AB=AC=5,BC=6,则⊙O的半径为(  )
A. 4
B. 3.25
C. 3.125
D. 2.25

过A作AD⊥BC于D,
△ABC中,AB=AC,AD⊥BC,
则AD必过圆心O,
Rt△ABD中,AB=5,BD=3
∴AD=4
设⊙O的半径为x,
Rt△OBD中,OB=x,OD=4-x
根据勾股定理,得:OB2=OD2+BD2,即:
x2=(4-x)2+32,解得:x=

25
8
=3.125.
故选C.