(2010•扬州二模)(1)如图1,已知∠AOB,OA=OB,点E在OB边上,四边形AEBF是平行四边形,请你只用无刻度的直尺在图中画出∠AOB的平分线;(保留作图痕迹,不要求写作法)
(2)如图2,在10×10的正方形网格中,点A(0,0)、B(5,0)、C(3,6)、D(-1,3),
①依次连接A、B、C、D四点得到四边形ABCD,四边形ABCD的形状是______;
②在x轴上找一点P,使得△PCD的周长最短(直接画出图形,不要求写作法),此时,点P的坐标为______,最短周长为______
【答案】
分析:(1)根据平行四边形的性质可知∠AOB的平分线必定经过平行四边形的中心即对角线的交点.所以先做平行四边形的对角线,再作∠AOB的平分线.
(2)在x轴上找一点P,使得△PCD的周长最短,即PD+PC最小,所以可先做点D关于x轴的对称点D′,连接CD′,与x轴相交于点P.所以P(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/0.png)
,0),最短周长为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/1.png)
.
解答:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/images2.png)
解:(1)如图所示;(2分)
(2)①等腰梯形;(4分)
②D关于x轴的对称点D′,连接CD′,则D′(-1,-3),
设过点CD′的直线解析式为:y=kx+b(k≠0),把C、D′两点坐标代入得,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/2.png)
,解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/3.png)
,
故直线CD′的解析式为:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/4.png)
x-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/5.png)
,
当y=0时,x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/6.png)
,
故P点坐标为:(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/7.png)
,0)
故答案为:P(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/8.png)
,0);
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230651759033008/SYS201310212306517590330023_DA/9.png)
(其中画图正确得2分)(10分)
点评:主要考查了复杂作图和轴对称作图.熟悉平行四边形的性质和轴对称的性质是解题的关键.