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
的中点,.································ 1分
.·································································· 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.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网