题目内容
(9分)已知,,(如图).是射线上的动点(点与点不重合),是线段的中点.
(1)设,的面积为,求关于的函数关系式,并写出自变量的取值范围;
(2)如果以线段为直径的圆与以线段为直径的圆外切,求线段的长;
(3)连结,交线段于点,如果以为顶点的三角形与相似,求线段的长.
(1)设,的面积为,求关于的函数关系式,并写出自变量的取值范围;
(2)如果以线段为直径的圆与以线段为直径的圆外切,求线段的长;
(3)连结,交线段于点,如果以为顶点的三角形与相似,求线段的长.
解:(1)取中点,连结,
为的中点,,.································ 1分
又,.·································································· 2分
,得;··············································· 3分
(2)过D作DP⊥BC,垂足为P,∠DAB=∠ABC=∠BPD=90°,
∴四边形ABPD是矩形.
以线段为直径的圆与以线段为直径的圆外切,
,又,∴DE=BE+AD-AB=x+4-2=x+2……4分
PD=AB=2,PE= x-4,DE2= PD2+ PE2,…………………………………………………5分
∴(x+2)2=22+(x-4)2,解得:.
∴线段的长为.…………………………………………………………………………6分
(3)由已知,以为顶点的三角形与相似,
又易证得.··································································· 7分
由此可知,另一对对应角相等有两种情况:①;②.
①当时,,..
,易得.得;················································ 8分
②当时,,.
.又,.
,即=,得x2=[22+(x-4)2].
解得,(舍去).即线段的长为2.······································· 9分
综上所述,所求线段的长为8或2.
为的中点,,.································ 1分
又,.·································································· 2分
,得;··············································· 3分
(2)过D作DP⊥BC,垂足为P,∠DAB=∠ABC=∠BPD=90°,
∴四边形ABPD是矩形.
以线段为直径的圆与以线段为直径的圆外切,
,又,∴DE=BE+AD-AB=x+4-2=x+2……4分
PD=AB=2,PE= x-4,DE2= PD2+ PE2,…………………………………………………5分
∴(x+2)2=22+(x-4)2,解得:.
∴线段的长为.…………………………………………………………………………6分
(3)由已知,以为顶点的三角形与相似,
又易证得.··································································· 7分
由此可知,另一对对应角相等有两种情况:①;②.
①当时,,..
,易得.得;················································ 8分
②当时,,.
.又,.
,即=,得x2=[22+(x-4)2].
解得,(舍去).即线段的长为2.······································· 9分
综上所述,所求线段的长为8或2.
略
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