题目内容
如图(1)已知,矩形ABDC的边AC=3,对角线长为5,将矩形ABDC置于直角坐系内,点D与原点O重合.且反比例函数y=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_ST/0.png)
(1)求点A的坐标;
(2)若矩形ABDC从图(1)的位置开始沿x轴的正方向移动,每秒移动1个单位,1秒后点A刚好落在反比例函数y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_ST/1.png)
(3)矩形ABCD继续向x轴的正方向移动,AB、AC与反比例函数图象分别交于P、Q如图(2),设移动的总时间为t(1<t<5),分别写出△BPD的面积S1、△DCQ的面积S2与t的函数关系式;
(4)在(3)的情况下,当t为何值时,S2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_ST/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_ST/images3.png)
【答案】分析:(1)连接OA,根据勾股定理求出OC,即可得出答案;
(2)求出A的坐标,把A的坐标代入反比例函数的解析式,求出k即可;
(3)求出BP,根据三角形的面积公式求出S1即可;求出t秒后A的坐标,得出Q的横坐标,代入解析式求出Q的纵坐标,求出CQ,根据三角形的面积公式求出S2即可;
(4)把S1、S2代入已知,得出关于t的方程,求出t的值即可.
解答:解:(1)连接OA,OA=5,AC=3,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/images0.png)
由勾股定理得:OC=
=
=4,
∴点A的坐标是(4,3).
(2)4+1=5,
∴1秒后点A的坐标是(5,3),
代入y=
得:3=
,
∴k=15.
(3)∵A在双曲线上时t=1,
∴AP=t-1,
BP=BA-AP=4-(t-1)=5-t,
∴S1=
BP×BD=
×(5-t)×3=-
t+
,
t秒后A的坐标是(4+t,3),
把x=4+t代入y=
得:y=
,
∴Q的坐标是(4+t,
),
∴S2=
×DC×CQ=
×4×
=
,
即S1=-
t+
,S2=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/17.png)
(4)∵S2=
S1,
∴
=
×(-
t+
),
解得:t=3,t=-2(舍去),
当t=3时,S2=
S1.
点评:本题考查了用待定系数法求反比例函数的解析式,点的坐标,三角形的面积,矩形的性质等知识点的应用,熟练的运用性质进行计算是解此题的关键,主要考查了学生的计算能力和运用性质进行推理的能力,题目较好,难度适中.
(2)求出A的坐标,把A的坐标代入反比例函数的解析式,求出k即可;
(3)求出BP,根据三角形的面积公式求出S1即可;求出t秒后A的坐标,得出Q的横坐标,代入解析式求出Q的纵坐标,求出CQ,根据三角形的面积公式求出S2即可;
(4)把S1、S2代入已知,得出关于t的方程,求出t的值即可.
解答:解:(1)连接OA,OA=5,AC=3,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/images0.png)
由勾股定理得:OC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/1.png)
∴点A的坐标是(4,3).
(2)4+1=5,
∴1秒后点A的坐标是(5,3),
代入y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/3.png)
∴k=15.
(3)∵A在双曲线上时t=1,
∴AP=t-1,
BP=BA-AP=4-(t-1)=5-t,
∴S1=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/7.png)
t秒后A的坐标是(4+t,3),
把x=4+t代入y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/9.png)
∴Q的坐标是(4+t,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/10.png)
∴S2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/14.png)
即S1=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/17.png)
(4)∵S2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/18.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/19.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/21.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/22.png)
解得:t=3,t=-2(舍去),
当t=3时,S2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200953515639753/SYS201311032009535156397027_DA/23.png)
点评:本题考查了用待定系数法求反比例函数的解析式,点的坐标,三角形的面积,矩形的性质等知识点的应用,熟练的运用性质进行计算是解此题的关键,主要考查了学生的计算能力和运用性质进行推理的能力,题目较好,难度适中.
![](http://thumb2018.1010pic.com/images/loading.gif)
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