如图,⊙O的直径AB=12cm,AM和BN是它的两条切线,DE切⊙O于E,交AM于D

如图,⊙O的直径AB=12cm,AM和BN是它的两条切线,DE切⊙O于E,交AM于D,交BN于C

 

 

(1)若AD=4cm,求BC的长;

(2)设AD=x,BC=y,求y与x的函数关系式;

(3)梯形ABCD的面积为78cm2,求AD的长

 

答案

解:(1)如图,过点D作DF⊥BC于点F,

        ∵AM,BN,CD都是⊙O的切线

∴MAO=∠NBO =90°,AD=DE,CB=CE

∴四边形ABFD是矩形

∴BF=AD=DE=4cm,DF=AB=12cm

设BC=CE=xcm,则CF=(x-4)cm,CD=(x+4)cm

在Rt△DCF中,CD2=DF2+CF2    即   (x+4)2=122+(x-4)2

解得X=9

∴BC的长为9cm                                     ………………3分

(2)由(1)可知DF=AB=12cm,当AD=x,BC=y时,CD=x+y

在Rt△DCF中,CD2=DF2+CF2   即  (x+y)2=122+(y-x)2

     化简得y=(x>0)                                     ………………6分

(3)  ∵梯形ABCD是直角梯形,则S梯形ABCD

      设AD=x,则(2)可知BC=

   

化简得  解得x=4或x=9

∴AD的长为4cm或9cm                             ………………9分

解析:略

 

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