Question

如图24,已知ABCD是矩形纸片,E是AB上一点,BE∶EA =5∶3,EC=,把△BCE沿折痕EC翻折,若B点恰好落在AD边上,设这个点为F,

图24

(1)求AB、BC的长度各是多少;

(2)若⊙O内切于以F、E、B、C为顶点的四边形,求⊙O的面积.

Answer 1

思路分析:考察所给的条件,翻折△BCE,则△CBE≌△CFE,这样图形中提供了很多的线段相等、角相等.

解:(1)连结CE、CF、EF,设BE =5x,EA =3x.?

∵四边形ABCD是矩形,?

∴AB =CD =8x,AD =BC,∠B =∠A =∠D =90°.?

∵△CBE≌△CFE,?

∴EF =5x,FC=BC,?∠CFE =90°.??

∵∠AEF +∠EFC+∠DFC=180°,?

∴∠AFE +∠DFC=90°.?

又∵∠AEF +∠AFE =90°,∠AEF =∠DFC,?

∴sin∠AEF =sin∠DFC,即=.?

∴=,则FC =10x.?

∴ ==.?

∴x =3.∴AB =24,BC =30.?

(2)∵CE平分∠FCB和∠FEB,∴O在EC上.?

设⊙O和BC切于M,和AB切于N,连结OM、ON,设⊙O的半径为r,?

∴OM⊥BC,ON⊥AB.∴OM∥AB,ON∥BC.?

∴OM =BN =ON =BM =r.?

∴=,即=.∴r =10.?

∴⊙O的面积为100π.