试题分析:(1)由BD是直径,根据圆周角定理,可得∠DAB=90°,又由FG⊥AB,可得FG∥AD,即可判定△FOE∽△ADE,根据相似三角形的对应边成比例,即可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024753319778.png)
,然后由O是BD的中点,DA∥OH,可得AD=2OH,则可证得OF•DE=OE•2OH;
(2)由⊙O的半径为12,且OE:OF:OD=2:3:6,即可求得OE,DE,OF的长,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024753319778.png)
,求得AD的长,又由在Rt△ABC中,OB=2OH,可求得∠BOH=60°,继而可求得BH的长,又由S
阴影=S
扇形GOB-S
△OHB,即可求得答案.
(1)∵BD是直径,
∴∠DAB=90°.
∵FG⊥AB,
∴DA∥FO.
∴△FOE∽△ADE.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024753319778.png)
,即OF•DE=OE•AD
∵O是BD的中点,DA∥OH,
∴AD=2OH
∴OF•DE=OE•2OH;
(2)∵⊙O的半径为12,且OE:OF:OD=2:3:6,
∴OE=4,ED=8,OF=6
代入(1)中OF•DE=OE•AD,得AD=12.
∴OH=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024753397338.png)
AD=6.
在Rt△OHB中,OB=2OH,
∴∠OBH=30°,
∴∠BOH=60°.
∴BH=BO•sin60°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024753413775.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230247534292265.png)
点评:此题综合性较强,难度适中,注意数形结合思想的应用,证得△FOE∽△ADE是解此题的关键.