辽宁省大连市第55中学旅顺实验中学2009－2010九年级第一学期期中联考
数学试题
（本试卷满分150分，考试时间120分钟．）
一、选择题（本大题共8小题，每小题3分，共24分）
1．下列根式中不是最简二次根式的是 （ ）．
A． B． C． D．
2．一元二次方程的解是 （ ）
A．x1 = 0 ，x2 = B． x1 = 0 ，x2 =
C．x1 = 0 ，x2 = D． x1= 0 ，x2 =
3．函数中，自变量x的取值范围是 （ ）
A．x < 2 B．x ≤．x > 2 D．x≥2
4．下列图形中，既是轴对称图形，又是中心对称图形的是 ( )
A B C D
5．如图1，∠AOB是⊙O的圆心角，∠AOB=80°，则弧所对圆周角∠ACB的度数是( )
A．40° B．45° C．50° D．80°
6．抛物线的顶点坐标是（ ）
A．（2，3） B．（－2，3） C．（2，－3） D．（－2，－3）
7．如图2，某公园的一座石拱桥是圆弧形（劣弧），其跨度为，
拱的半径为，则拱高为( )
A． B． C． D．米
8．图3（1）是一个横断面为抛物线形状的拱桥，
当水面在l时，拱顶（拱桥洞的最高点）离水
面，水面宽．如图3（2）建立平面直
角坐标系，则抛物线的关系式是（ ）
A． B．
C． D．
二、填空题（本题共9小题，每小题3分，共27分）
9．计算：= ．
10．如图4，点在以为直径的上，，则的长为 ．
11．若关于x的方程的一个根是0，则 ．
12．如图5，在平面直角坐标系中，点A的坐标为(1，4)，将线段OA绕点O顺时针旋转90°得到线段OA′，则点A′的坐标是 ．
13．在一幅长，宽的风景画的四周镶一条金色纸边，制成一幅矩形挂图，如图6，如果要使整个挂图的面积是，设金色纸边的宽为cm，那么满足的方程为 ．
14．若把代数式化为的形式，其中为常数，则= ．
15．阅读材料：设一元二次方程ax2+bx+c＝0(a≠0)的两根为x1，x2，则两根与方程系数之间有如下关系：x1+x2＝－，x1·x2＝.根据该材料填空：已知x1、x2是方程x2+6x+3＝0的两实数根，则+的值为 ．
16．已知抛物线（＞0）的对称轴为直线，且经过点试比较和的大小： _（填“>”，“<”或“=”）
17．如图7，四边形EFGH是由四边形经过旋转得到的．如果
用有序数对（2，1）表示方格纸上A点的位置，用（1，2）表示
B点的位置，那么四边形旋转得到四边形EFGH时
的旋转中心用有序数对表示是 ．
三、解答题（本题共3小题，每小题12分，共36分）
18．（1）计算 ． （2）解方程：．
19．某药品经过两次降价，每瓶零售价由100元降为81元．已知两次降价的百分率相同，求两次降价的百分率．
20．如图8，在⊙O中，D、E分别为半径OA、OB上的点，且AD＝BE．点C为中点，连接CD、CE．
求证：CD＝CE．
四、解答题（本题共3小题，其中21、22题各9分，23题10分，共28分）
21．如图9，在正方形网格中，每个小正方形的边长均为1个单位．
（1）作关于点P的对称图形．
（2）再把绕着逆顺时针旋转，得到，请你画出和（不要求写画法）．
22．如图10，直线交x轴于点A，交y轴于点B，抛物线的顶点为A，且经过点B．
（1）求该抛物线的解析式；
（2）求当时的值．
23．跳绳时，绳甩到最高处时的形状是抛物线.正在甩绳的甲、乙两名同学拿绳的手间距AB为，到地面的距离AO和BD均为，身高为的小丽站在距点O的水平距离为的点F处，绳子甩到最高处时刚好通过她的头顶点E.以点O为原点建立如图所示的平面直角坐标系, 设此抛物线的解析式为y=ax2＋bx＋0.9.
（1）求该抛物线的解析式；
（2）如果小华站在OD之间,且离点O的距离为，当绳子甩到最高处时刚好通过他的头顶，请你算出小华的身高；
（3）如果身高为的小丽站在OD之间,且离点O的距离为t米, 绳子甩到最高处时超过她的头顶,请结合图像,写出t的取值范围 .
五、解答题（本题共3小题，其中24题11分，25、26题各12分，共35分）
24．如图12，在中，，，，另有一等腰梯形（）的底边与重合，两腰分别落在上，且分别是的中点．
（1）求等腰梯形的面积；
（2）操作：固定，将等腰梯形以每秒1个单位的速度沿方向向右运动，直到点与点重合时停止．设运动时间为秒，运动后的等腰梯形为（如图）．
探究1：在运动过程中，四边形能否是菱形？若能，请求出此时的值；若不能，请说明理由．
探究2：设在运动过程中与等腰梯形重叠部分的面积为，求与的函数关系式．
25．已知正方形ABCD中，E为对角线BD上一点，过E点作EF⊥BD交BC于F，连接DF，G为DF中点，连接EG，CG．
（1）探究EG与CG的关系，并说明理由；
（2）将图14中△BEF绕B点逆时针旋转45º，如图15所示，取DF中点G，连接EG，CG．问（1）中的结论是否仍然成立？若成立，请给出证明；若不成立，请说明理由．
（3）将图14中△BEF绕B点旋转任意角度，如图16所示，再连接相应的线段，问（1）中的结论是否仍然成立？并说明理由．
26．定义一种变换：平移抛物线得到抛物线，使经过的顶点．设的对称轴分别交、于点，点是点关于直线的对称点．
（1）如图17，若：，经过变换后，得到：，点的坐标为，则
①的值等于______________；
②四边形为（ ）
A．平行四边形 B．矩形 C．菱形 D．正方形
（2）如图18，若：，经过变换后，点的坐标为，求的面积；
（3）如图19，若：，经过变换后，，点是直线上的动点，求点到点的距离和到直线的距离之和的最小值．
大连市第55中学初三第二次月考
数 学
参考答案及评分标准（仅供参考）
一、选择题
1．C，2．A， 3．D，4．B，5．A，6．A，7．B， 8．C
二、填空题
9．．
10．5．
11．1．
12．(4，－1)．
13．13 x2+40x－75=0．
14．．
15．11．
16．＞
17．（5，2）
三、解答题
18．计算：
解：（1）原式 ···················································································3分
． ···················································································6分
（2）解： ···················································································1分
···················································································2分
···················································································3分
···················································································4分
···················································································5分
∴ ···················································································6分
（其它解法参照给分）
19．解：设每次降价的百分率为x，根据题意得：···········································································1分
100=81 ···················································································6分
解得：=0.1，=1.9 ···················································································10分
经检验=1.9不符合题意，∴x=0.1=10%···················································································11分
答：每次降价百分率为10%． ···················································································12分
20． 证明（如图）：连接OC
∵
∴∠AOC=∠BOC ······························································4分
∵OA=OB，AD=BE
∴OD=OE ··········································································8分
∵OC=OC
∴△DOC≌EOC ······································································10分
∴CD=CE ···················································································12分
（注：其它方法参照给分）
四、解答题（本题共3小题，其中21、22题各9分，23题10分，共28分）
21．解：图略（1）4分，（2）5分．
22．解：（1）∵直线y=－x－2交x轴于点A，交y轴于点B，
∴点A的坐标为（－2，0），点B的坐标为（0，－2）···················································2分
∵抛物线的顶点为A，
设抛物线为， ···················································································4分
∵抛物线过点B（0，－2）
∴－2＝，a＝， ···················································································6分
∴． ············································································7分
（2）x<－2或x>0 （注：可直接写答案） ··················································································9分
23． 解：（1）由题意得点E（1,1.4）, B(6,0.9), 代入y=ax2+bx+0.9得
···················································································2分
解得 ···················································································4分
∴所求的抛物线的解析式是y=－0.1x2＋0.6x+0.9. ····················································5分
（2）把x=3代入y=－0.1x2＋0.6x+0.9得
y=－0.1×32＋0.6×3+0.9=1.8
∴小华的身高是 ···················································································7分
（3）1＜t＜5 ···················································································10分
五、解答题（本题共3小题，其中24题11分，25、26题各12分，共35分）
24．解：如图，（1）过点作于．
，，，为中点
． ··························································1分
又分别为的中点
·················································2分
等腰梯形的面积为6．···························································3分
（2）能为菱形
如图，由，
四边形是平行四边形························································4分
当时，四边形为菱形，··············5分
此时可求得··················································································6分
当秒时，四边形为菱形．·········································7分
（3）分两种情况：
①当时，
方法一：，
重叠部分的面积为：
当时，与的函数关系式为 ···················································9分
（方法二：当时，
，，
重叠部分的面积为：
当时，与的函数关系式为）
（注：方法二参照给分）
②当时，
设与交于点，则
，
作于，则
重叠部分的面积为：
·················································11分
25． 解：（1） CG=EG ，CG⊥EG． ··················································································2分
（2）（1）中结论仍然成立，即EG=CG，CG⊥EG．
证明：延长CG至M,使MG=CG，
连接MF，ME，EC， ··················································································3分
在△DCG 与△FMG中，
∵FG=DG，∠MGF=∠CGD，MG=CG，
∴△DCG ≌△FMG． ·········································································5分
∴MF=CD，∠FMG＝∠DCG．
∴MF∥CD∥AB．
∴．
在Rt△MFE 与Rt△CBE中，
∵ MF=CB，EF=BE，
∴△MFE ≌△CBE．
∴，EM=EC．
∴∠MEC＝∠MEF＋∠FEC＝∠CEB＋∠CEF＝90°．
∴ △MEC为等腰直角三角形．
∵ MG = CG，
∴ EG=MC．
∴ ，CG⊥EG． ··················································································7分
（3）（1）中的结论仍然成立，
即EG=CG， EG⊥CG．
证明：延长CG至M,使MG=CG，
连接MF，ME，EC，··················································································8分
在△DCG 与△FMG中，
∵FG=DG，∠MGF=∠CGD，MG=CG，
∴△DCG ≌△FMG．··················································································10分
∴MF=CD，∠FMG＝∠DCG．
∴MF∥CD∥AB．
∵∠ABF+∠EBC=∠ABC+∠EBF=135°
∠ABF+∠EFB+MFE=180°．
∴∠EBC=∠MFE
在△MFE 与△CBE中，
∵ MF=CB，∠EBC=∠MFE， EF=BE，
∴△MFE ≌△CBE．
∴，EM=EC．
∴∠MEC＝∠MEF＋∠FEC＝∠CEB＋∠CEF＝90°．
∴ △MEC为等腰直角三角形．
∵ MG = CG，
∴ EG=MC．
∴ ，CG⊥EG． ··················································································12分
26．解：(1) －2；D； ··················································································2分
(2) ∵ ： y＝a(x－2)2＋c－1，而（0，c）在上，可得a＝． ····················4分
∴ DB＝（＋c）－（c－1）＝2， ∴ ＝2． ············································ 6分
（3）当点在点的右侧时（如图1），
设AC与BD交于点N，抛物线，配方得，
其顶点坐标是（1，2）， ∵ AC＝2，∴ 点C的坐标为．
∵过点， ∴解析式为， ∴ B（，
∴ D（，
∴ ，∵ 点与点关于直线对称，
∴，且
∴ 四边形ABCD是菱形． ∴ PD＝PB．
作交于点， 则PD＋PH＝PB＋PH．
要使PD＋PH最小， 即要使PB＋PH最小，
此最小值是点B到AD的距离， 即△ABD边AD上的高．
∵＝1，＝，，∴＝，
故是等边三角形．
∴ ∴ 最小值为．·········································9分
当点在点的左侧时（如图2），同理， 最小值为．
综上，点到点的距离和到直线的距离之和
的最小值为． ··········································································12分