题目内容
如图,以Rt△ABO的直角顶点O为原点,OA所在的直线为x轴,OB所在的直线为y轴,建立平面直角坐标系.已知OA=4,OB=3,一动点P从O出发沿OA方向,以每秒1个单位长度的速度向A点匀速运动,到达A点后立即以原速沿AO返回;点Q从A点出发沿AB以每秒1个单位长度的速度向点B匀速运动.当Q到达B时,P、Q两点同时停止运动,设P、Q运动的时间为t秒(t>0).(1)试求出△APQ的面积S与运动时间t之间的函数关系式;
(2)在某一时刻将△APQ沿着PQ翻折,使得点A恰好落在AB边的点D处,如图①.求出此时△APQ的面积.
(3)在点P从O向A运动的过程中,在y轴上是否存在着点E使得四边形PQBE为等腰梯形?若存在,求出点E的坐标;若不存在,请说明理由.
(4)伴随着P、Q两点的运动,线段PQ的垂直平分线DF交PQ于点D,交折线QB-BO-OP于点F. 当DF经过原点O时,请直接写出t的值.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_ST/images0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_ST/images1.png)
【答案】分析:过Q作QH⊥AP于H点,构造直角三角形APQ.
(1)在Rt△AOB中,利用勾股定理求得AB;①P由O向A运动时,OP=AQ=t,AP=4-t.根据平行线截线段成比例的性质求得QH,然后求△APQ的面积;②P由A向O运动时,AP=t-4,AQ=t,由直角三角形ABO中的锐角的正弦求得QH=
,然后求△APQ的面积;
(2)根据翻折的性质知△APQ≌△DPQ,∠AQP=90°.在直角三角形AOB与直角三角形APQ中通过∠A的余弦值求得cosA=
=
=
.①当0<t<4时,求得t值;②当4<t≤5时,求得t值;然后将其代入(1)中的函数解析式;
(3)①若PE∥BQ,则梯形PQBE是等腰梯形.过E、P分分别作EM⊥AB于M,PN⊥AB于N.构造矩形PNME.则有BM=QN,由PE∥BQ,
得
,从而求得MB的值;在直角三角形APN中根据AP求得QN的值,然后由BM=QN,求得t,所以点E的坐标就迎刃而解了;②若PQ∥BE,则等腰梯形PQBE中BQ=EP且PQ⊥OA于P点.由OP+AP=OA求得t值;
(4)①当P由O向A运动时,OQ=OP=AQ=t.再有边角关系求得BQ=AQ=
AE,解得t值;②②当P由A向O运动时,OQ=OP=8-t.在Rt△OGQ中,利用勾股定理得OQ2=QG2+OG2,列出关于t的方程,解方程即可.
解答:
解:(1)在Rt△AOB中,OA=4,OB=3
∴AB=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/6.png)
①P由O向A运动时,OP=AQ=t,AP=4-t
过Q作QH⊥AP于H点.
由QH∥BO,得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/7.png)
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/8.png)
即
(0<t<4)
②当4<t≤5时,即P由A向O运动时,AP=t-4AQ=t
sin∠BAO=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/10.png)
QH=
,
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/12.png)
=
;
综上所述,S△APQ=
;
(2)由题意知,此时△APQ≌△DPQ,∠AQP=90°,
∴cosA=
=
=
,
当0<t<4∴
即![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/19.png)
当4<t≤5时,
=
,t=-16(舍去)
∴
;
(3)存在,有以下两种情况
①若PE∥BQ,则等腰梯形PQBE中PQ=BE
过E、P分分别作EM⊥AB于M,PN⊥AB于N.
则有BM=QN,由PE∥BQ,
得
,
∴
;
又∵AP=4-t,
∴AN=
,
∴
,
由BM=QN,得![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/27.png)
∴
,
∴
;
②若PQ∥BE,则等腰梯形PQBE中
BQ=EP且PQ⊥OA于P点
由题意知![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/30.png)
∵OP+AP=OA,
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/31.png)
∴t=
,
∴OE=
,
∴点E(0,-
)
由①②得E点坐标为(0,
)或(0,-
).
(4)①当P由O向A运动时,OQ=OP=AQ=t.
可得∠QOA=∠QAO∴∠QOB=∠QBO
∴OQ=BQ=t∴BQ=AQ=
AE
∴
;
②当P由A向O运动时,OQ=OP=8-t
BQ=5-t,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/39.png)
在Rt△OGQ中,OQ2=QG2+OG2
即(8-t)2=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/40.png)
∴t=5
点评:本题主要考查了解直角三角形的应用,相似三角形的性质以及二次函数等知识点的综合应用,弄清相关线段的大小和比例关系是解题的关键.
(1)在Rt△AOB中,利用勾股定理求得AB;①P由O向A运动时,OP=AQ=t,AP=4-t.根据平行线截线段成比例的性质求得QH,然后求△APQ的面积;②P由A向O运动时,AP=t-4,AQ=t,由直角三角形ABO中的锐角的正弦求得QH=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/0.png)
(2)根据翻折的性质知△APQ≌△DPQ,∠AQP=90°.在直角三角形AOB与直角三角形APQ中通过∠A的余弦值求得cosA=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/3.png)
(3)①若PE∥BQ,则梯形PQBE是等腰梯形.过E、P分分别作EM⊥AB于M,PN⊥AB于N.构造矩形PNME.则有BM=QN,由PE∥BQ,
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/4.png)
(4)①当P由O向A运动时,OQ=OP=AQ=t.再有边角关系求得BQ=AQ=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/5.png)
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/images6.png)
∴AB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/6.png)
①P由O向A运动时,OP=AQ=t,AP=4-t
过Q作QH⊥AP于H点.
由QH∥BO,得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/7.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/8.png)
即
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/9.png)
②当4<t≤5时,即P由A向O运动时,AP=t-4AQ=t
sin∠BAO=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/10.png)
QH=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/11.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/12.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/images15.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/14.png)
(2)由题意知,此时△APQ≌△DPQ,∠AQP=90°,
∴cosA=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/17.png)
当0<t<4∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/18.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/19.png)
当4<t≤5时,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/21.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/22.png)
(3)存在,有以下两种情况
①若PE∥BQ,则等腰梯形PQBE中PQ=BE
过E、P分分别作EM⊥AB于M,PN⊥AB于N.
则有BM=QN,由PE∥BQ,
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/23.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/24.png)
又∵AP=4-t,
∴AN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/25.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/26.png)
由BM=QN,得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/27.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/28.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/29.png)
②若PQ∥BE,则等腰梯形PQBE中
BQ=EP且PQ⊥OA于P点
由题意知
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/30.png)
∵OP+AP=OA,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/31.png)
∴t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/32.png)
∴OE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/33.png)
∴点E(0,-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/34.png)
由①②得E点坐标为(0,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/35.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/36.png)
(4)①当P由O向A运动时,OQ=OP=AQ=t.
可得∠QOA=∠QAO∴∠QOB=∠QBO
∴OQ=BQ=t∴BQ=AQ=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/37.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/38.png)
②当P由A向O运动时,OQ=OP=8-t
BQ=5-t,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/39.png)
在Rt△OGQ中,OQ2=QG2+OG2
即(8-t)2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022155844641666259/SYS201310221558446416662025_DA/40.png)
∴t=5
点评:本题主要考查了解直角三角形的应用,相似三角形的性质以及二次函数等知识点的综合应用,弄清相关线段的大小和比例关系是解题的关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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