江苏省苏州市2017届中考数学二模试题(共13页)

1、精选优质文档-倾情为你奉上江苏省苏州市2017届中考数学二模试题9如图ABC中，C=90°，AC=4，BC=3，将ABC绕点A逆时针旋转，使点C落在线段AB上的点E处，点B落在点D处，则B、D两点间的距离为（）A2 B C3 D210如图点A、B在反比例函数y=（k0，x0）图象上，BCx轴，交y轴于点C，动点P从坐标原点O出发，沿OABC（图中“”所示路线）匀速运动，终点为C，过P作PMx轴，垂足为M设三角形OMP的面积为S，P点运动时间为t，则S关于x的函数图象大致为 A B C D三、解答题：本大题共11小题，共76分把解答过程写在答题卡相应位置上，解答时应写出必要的计算过程、

2、推演步骤或文字说明，作图时用2B铅笔或黑色墨水签字笔19（本题满分5分）计算：20（本题满分5分）解不等式组：参考答案一、选择题（每小题3分，共30分）题号12345678910答案CBCAADBDBA二、选择题（每小题3分，共24分）11(a + 3)(a - 3) 124.51×107138 141560016 17一4 18、 三、解答题（共11大题，共76分）19（本题共5分）解：原式= 3-2 + 13分=25分20(本题共5分)解：由式得：x>32分由式得：x4分不等式组的解集为： 5分21 (本题共6分)解：原式=1分=2分=4分当x=时，原式=5分=6分22（本

3、题满分6分） 解：设骑电瓶车学生的速度为x km/h，汽车的速度为2x km/h，可得：··········1分，································&

4、#183;··············································3分解得x15，··

5、;··················································

6、;··································4分经检验，x15是原方程的解，·············&

7、#183;················································5分2x2&#

8、215;1530.答：骑车学生的速度和汽车的速度分别是15 km/h，30 km/h.·························6分23（本题共8分）(1)证明：四边形ABCD为平行四边形ADBC，ABCD，AB=CD，··········&

9、#183;·················································&

10、#183;········1分B+ C=180°,AEB=DAE，·····································

11、·························2分AE是BAD的角平分线BAE=DAE， BAE=DAE，··················3分AB=BE，BE=C

12、D ·················································

13、;·······························4分（2）解：AB=BE，BEA=60°， ABE是等边三角形， AE=AB=4， ········

14、83;·················································

15、83;·································5分BFAE， AF=EF=2， BF=，········· ··&

16、#183;···············································6分 ADBC，

17、 D=ECF，DAF=E， 在ADF和ECF中， ， ADFECF（AAS），  ········· ······························

18、;······7分ADF的面积=ECF的面积， 平行四边形ABCD的面积=ABE的面积=AEBF的面积=×4×2=4 ·······8分24 （本题共8分）1）由题意可得，本次抽样调查中，选择曲目代号为A的学生占抽样总数的百分比为：×100%=20%··············&

19、#183;···································2分（2）由题意可得，选择C的人数有：30÷363044=70（人）补全的图柱状图正确 ······

20、···································5分（3）由题意可得，全校选择此必唱歌曲共有：1260×=490（人），答：全校共有490名学生选择此必唱歌曲·····

21、···································8分25（本题共8分）解：（1）设点D的坐标为（4，m）（m0），则点A的坐标为（4，3+m），点C为线段AO的中点，点C的坐标为（2，）点C、点D均在反比例函数y=的函数图象上，

22、···························1分 解得：·········2分反比例函数的解析式为y=··········

23、3;·····························3分（2）m=1，点A的坐标为（4，4），·················&

24、#183;······················4分OB=4，AB=4在RtABO中，OB=4，AB=4，ABO=90°，OA=4，cosOAB=··················&

25、#183;·····················5分（3）m=1，点C的坐标为（2，2），点D的坐标为（4，1）设经过点C、D的一次函数的解析式为y=ax+b，则有，解得：·····7分经过C、D两点的一次函数解析式为y=x+3 ·········

26、;·······························8分26（本题共10分）证明： 如图，连接OC，PA切O于APAO=90º 1分OPBC，AOP=OBC，COP=OCBOC=OB，OBC=OCB，AOP=COP 2分 又OA=OC，OP=OP， PAOPCO (S

27、AS)PAO=PCO=90 º，又OC是O的半径， PC是O的切线 3分解法不唯一. 解：由（1）得PA，PC都为圆的切线，PA=PC，OP平分APC，ADO=PAO=90 º，PAD+DAO=DAO+AOD，PAD =AOD，第23题答图BAEPODCMADOPDA 4分，AC=8， PD=，AD=AC=4，OD=3，AO=5， 5分由题意知OD为ABC的中位线，BC=2OD=6，AB=10 S阴=S半OSACB=答：阴影部分的面积为6分（3）如图，连接AE，BE，过点B作BMCE于点M7分CMB=EMB=AEB=90º，又点E是的中点，ECB=CBM=ABE=

28、45º，CM=MB =，BE=ABcos450 =，8分 EM=，CE=CM+EM= ·······················9分答：CE的长为cm 10分27（本题共10分）解：（1）垂直；1分BC=CF+CD；···········

29、3;···············2分（2）成立，正方形ADEF中，AD=AF，BAC=DAF=90°，BAD=CAF，在DAB与FAC中，DABFAC，························

30、;···4分B=ACF，CF=BDACB+ACF=90°，即CFBD；BC=BD+CD，BC=CF+CD；···························6分（3）解：过A作AHBC于H，过E作EMBD于M，ENCF于N，BAC=90°，AB=AC，BC=AB=4，AH=B

31、C=2，CD=BC=1，CH=BC=2，DH=3，···························7分由（2）证得BCCF，CF=BD=5，四边形ADEF是正方形，AD=DE，ADE=90°，BCCF，EMBD，ENCF，四边形CMEN是矩形，······

32、;·····················8分NE=CM，EM=CN，AHD=ADC=EMD=90°，ADH+EDM=EDM+DEM=90°，ADH=DEM，在ADH与DEM中，ADHDEM，EM=DH=3，DM=AH=2，CN=EM=3，EN=CM=3，·········&#

33、183;·················9分ABC=45°，BGC=45°，BCG是等腰直角三角形，CG=BC=4，GN=1，EG=·······················

34、···10分28 （本题共10分）解：（1）点B是点A关于y轴的对称点，抛物线的对称轴为y轴，抛物线的顶点为（0，），故抛物线的解析式可设为y=ax2+A（1，2）在抛物线y=ax2+上，a+=2，解得a=，抛物线的函数关系表达式为y=x2+；··························2分（2）当点

35、F在第一象限时，如图1，令y=0得，x2+=0，解得：x1=3，x2=3，点C的坐标为（3，0）设直线AC的解析式为y=mx+n，则有，解得，直线AC的解析式为y=x+·········3分设正方形OEFG边长为p，则F（p，p）点F（p，p）在直线y=x+上，p+=p，解得p=1，点F的坐标为（1，1）·························4分当点F在第二象限时，同理可得：点F的坐标为（3，3），此时点F不在线段AC上，故舍去···········