【答案】
分析:(1)由已知和勾股定理先求出BC,再由D,E分別是AC,BC的中點,求出AD、DE、BE,從而求出t;
(2)先求出當點P運動到點D時所用時間,得出AQ的長,即可求出BQ的長,再根據(jù)△BPQ的面積=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/0.png)
BQ•AP進行計算即可;
(3)由已知用t表示出AQ、AP、BQ,再由∠A=90°,通過面積公式求出S與t的函數(shù)關(guān)系式;
(4)通過假設(shè),分兩種情況討論即可求解.
解答:解:(1)已知Rt△ABC中,∠A=90°,AB=6,AC=8,
由勾股定理得:BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/2.png)
=10,
又由D,E分別是AC,BC的中點,
∴AD=4,DE=3,BE=5,
∴當點P到達終點B時所用時間t=(4+3+5)÷3=4(秒),
答t的值為4秒.
(2)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/images3.png)
當點P運動到點D時,所用時間為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/3.png)
秒,
所以AQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/4.png)
×2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/5.png)
,
∴BQ=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/7.png)
,
∴△BPQ的面積=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/8.png)
BQ•AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/9.png)
×4=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/10.png)
;
(3)①如圖,當點P在AD上(不包含D點),
由已知得:AQ=2t,AP=3t,
∴BQ=AB-AQ=6-2t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/images12.png)
已知∠A=90°,
∴△BPQ的面積S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/11.png)
BQ•AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/12.png)
(6-2t)•3t=-3t
2+9t,
所以Q在線段AB上運動時,S與t的函數(shù)關(guān)系式為S=-3t
2+9t.
②如圖當點P在DE(包括點D、E)上,
過點P作PF⊥AB于F,
則PF=AD=4,
∴△BPQ的面積S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/13.png)
BQ•PF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/14.png)
(6-2t)•4=12-4t,
所以此時Q在線段AB上運動時,S與t的函數(shù)關(guān)系式為S=12-4t.
③當點P在BE上(不包括E點),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/images17.png)
由已知得:BP=3+4+5-3t=12-3t,
過點P作PF⊥AB于F,
∴PF∥AC,
∴△BPF∽△BCA,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/16.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/18.png)
,
∴PF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/19.png)
,
∴△BPQ的面積S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/20.png)
BQ•PF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/21.png)
(6-2t)•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/23.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/24.png)
t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/25.png)
,
所以Q在線段AB上運動時,S與t的函數(shù)關(guān)系式為S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/26.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/27.png)
t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/28.png)
,
(4)若PQ∥DB,則點P、Q必在DB同側(cè).分兩種情況:
①當點Q在AB上,點P在AD上時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/images32.png)
假設(shè)PQ∥DB成立,
則△AQP∽△ABD,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/29.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/30.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/32.png)
,
此時方程的解是t=0,但此解不符合題意,
則PQ∥DB不成立,
②當3<t<4時,點Q在AB延長線上,點P在EB上,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/images37.png)
此時PB=12-3t,PC=3t-7,BQ=2t-6.
若PQ∥DB,設(shè)直線PQ交DE與N,
∵DE∥AB,
∴△PEN∽△PBQ,
∴EN:BQ=PE:PB,
則EN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/33.png)
;
又∵NQ∥DB,
∴EN:ED=EP:EB,
則EN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/34.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/36.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/37.png)
符合題意.
綜上所述,當t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163221301331679/SYS201310221632213013316025_DA/38.png)
時,PQ∥DB.
點評:此題考查的知識點是勾股定理、三角形中位線定理及相似三角形的判定與性質(zhì),關(guān)鍵是通過勾股定理三角形中位線定理求解,以及通過假設(shè)推出錯誤結(jié)論論證.