【答案】
分析:(1)在直角△BCE中利用三角函數(shù)即可求得BE,EC的長度,則三角形的周長即可求得;
(2)取FC的中點P,連接E、P,易證EP是直角梯形ABCF的中位線,以及直角三角形的性質,以及梯形的中位線定理即可證得;
(3)取AB的中點Q,連接Q、P,則QP是直角梯形ABCF的中位線,QP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/0.png)
,EP是Rt△EFC斜邊上的中線,EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/1.png)
,要使得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/2.png)
,只需EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/3.png)
QP,即Rt△PQE是等腰直角三角形,即可表示出FA、AE的長度,然后根據Rt△EBC∽Rt△FAE,相似三角形的對應邊的比相等可以得到關于x的方程,從而求解.
解答:解:(1)如圖:∵∠A=∠B=90°,BC=6,BE=x,∠BCE=30°
∴Rt△EBC中,BE=BCtan30°=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/4.png)
,EC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/6.png)
∴△BCE的周長=BC+EB+EC=6+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/7.png)
(2)如圖:取FC的中點P,連接EP,
∵∠A=∠B=90°,AD=5,AB=10,BC=6,BE=x=5,EF⊥CE,
∴EP是直角梯形ABCF的中位線,EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/8.png)
EP也是Rt△EFC斜邊上的中線,EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/9.png)
∴EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/11.png)
,即CF=AF+BC
(3)如圖:取AB的中點Q,連接QP,
∵∠A=∠B=90°,AD=5,AB=10,BC=6,BE=x,EF⊥CE,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/images12.png)
∴AE=10-x,QE=|5-x|,∠AFE+∠AEF=90°,∠BEC+∠AEF=90°
QP是直角梯形ABCF的中位線,QP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/12.png)
,∠PQE=90°
EP是Rt△EFC斜邊上的中線,EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/13.png)
要使得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/14.png)
,只需EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/15.png)
QP,即Rt△PQE是等腰直角三角形,QP=QE=|5-x|
∴AF=2QP-BC=2|5-x|-6
∵∠A=∠B=90°,EF⊥CE,
∴∠AFE+∠AEF=90°,∠BEC+∠AEF=90°
∴∠AFE=∠BEC
∴Rt△EBC∽Rt△FAE
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/16.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/17.png)
當0≤x≤5時,|5-x|=5-x,2|5-x|-6=4-2x
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/19.png)
(舍),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/20.png)
當5<x≤10時,|5-x|=x-5,2|5-x|-6=2x-16
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/21.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/23.png)
(舍)
綜上所述:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193016938944327/SYS201311011930169389443023_DA/24.png)
時,
點評:本題是相似三角形的判定與性質,以及直角三角形的性質,梯形的中位線定理的綜合應用,正確作出輔助線是關鍵.