【文档说明】《山东中考真题数学》2014年山东省临沂市中考数学试题及答案.pdf，共(17)页，665.331 KB，由envi的店铺上传

转载请保留链接：https://www.doc5u.com/view-f02549ec259b5e6d40ce1552bb57b7ce.html

以下为本文档部分文字说明：

绝密★启用前试卷类型：A2014年临沂市初中学生学业考试试题数学注意事项：1．本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题），共8页，满分120分，考试时间120分钟．答卷前，考生务必用0.5毫米黑色签字笔将自己的姓名、准考证号、座号填写在试卷和答题卡规定的位置．考试结束

后，将本试卷和答题卡一并交回．2．答题注意事项见答题卡，答在本试卷上不得分．第Ⅰ卷（选择题共42分）一、选择题（本大题共14小题，每小题3分，共42分）在每小题所给出的四个选项中，只有一项是符合题目要求的．1．－3的相反数是（A）3．（B）－3．（C）13．（D）13．2．根据世界贸易组织(

WTO)秘书处初步统计数据，2013年中国货物进出口总额为4160000000000美元，超过美国成为世界第一货物贸易大国．将这个数据用科学记数法可以记为（A）124.1610美元．（B）134.1610美元．（C）120

.41610美元．（D）1041610美元．[来源:Zxxk.Com]3．如图，已知l1∥l2，∠A=40°，∠1=60°，则∠2的度数为（A）40°．（B）60°．（C）80°．（D）100°．4．下列计算正确的是2C（第3

题图）l1AB1l2（A）223aaa．（B）2363)abab（．（C）22()mmaa．（D）326aaa．5．不等式组-2≤11x的解集，在数轴上表示正确的是（A）（B）（C）（D）6．当2a时，22211(1)aaaa的结果是（A）32．（B）3

2．（C）12．（D）12．7．将一个n边形变成n+1边形，内角和将（A）减少180°．（B）增加90°．（C）增加180°．（D）增加360°．8．某校为了丰富学生的校园生活，准备购买一批陶笛，已知

A型陶笛比B型陶笛的单价低20元，用2700元购买A型陶笛与用4500元购买B型陶笛的数量相同，设A型陶笛的单价为x元，依题意，下面所列方程正确的是（A）2700450020xx．（B）2700450020xx．（C）2700450020xx．（D）270045

0020xx．01－1－2－301－1－2－301－1－2－301－1－2－39．如图，在⊙O中，AC∥OB，∠BAO=25°，则∠BOC的度数为（A）25°．（B）50°．（C）60°．（D）80°．10．从1，2，3，4中任

取两个不同的数，其乘积大于4的概率是（A）16．（B）13．（C）12．（D）23．11．一个几何体的三视图如图所示，这个几何体的侧面积为（A）2cm2．（B）4cm2．（C）8cm2．（D）16cm2．12．请你计算：(1)(1)xx，2(1)(1)xxx，…，猜想2(

1)(1xxx…)nx的结果是[来源:学科网]（A）11nx．（B）11nx．（C）1nx．（D）1nx．（第11题图）2cm主视图左视图俯视图CBAO（第9题图）B15°60°75°（第13题图）AC东北13．如图，

在某监测点B处望见一艘正在作业的渔船在南偏西15°方向的A处，若渔船沿北偏西75°方向以40海里/小时的速度航行，航行半小时后到达C处，在C处观测到B在C的北偏东60°方向上，则B，C之间的距离为（A）20海里．（B）103海里．（C）202海里．（D）30海里．14．在平面直

角坐标系中，函数22(yxxx≥0)的图象为1C，1C关于原点对称的图象为2C，则直线ya（a为常数）与1C，2C的交点共有（A）1个．（B）1个，或2个．（C）1个，或2个，或3个．（D）1个，或

2个，或3个，或4个．第Ⅱ卷（非选择题共78分）注意事项：1．第Ⅱ卷分填空题和解答题．2．第Ⅱ卷所有题目的答案，考生须用0.5毫米黑色签字笔答在答题卡规定的区域内，在试卷上答题不得分．二、填空题（本大题共5小题，每小题3分

，共15分）15．在实数范围内分解因式：36xx.16．某中学随机抽查了50名学生，了解他们一周的课外阅读时间，结果如下表所示：则这50名学生一周的平均课外阅读时间是小时.17．如图，在ABCD中，10BC，9sin10B，ACBC，则ABCD的面积是.1

8．如图，反比例函数4yx的图象经过直角三角形OAB的顶点A，D为斜边OA的中点，则过点D的反比例函数的解析式为.19．一般地，我们把研究对象统称为元素，把一些元素组成的总体称为集合．一个给定集合中的元素是互不相同．．．．的，也就是说，集合中的元素是不重复出现的．如一组数1，

1，2，3，4就可以构成一个集合，记为A={1，2，3，4}．类比实数有加法运算，集合也可以“相加”.定义：集合A与集合B中的所有元素组成的集合称为集合A与集合B的和，记为A+B.若A={－2，0，1，5，7}，B={－3，0，1，3，5}，则

A+B=.时间（小时）4567人数1020155（第18题图）ADBC（第17题图）yxOABD三、解答题（本大题共7小题，共63分）20．（本小题满分7分）计算：11sin6032831．21．（本小题满分7分）随着人民生活水平的提高，购买老年代步车的人越来越多．这些老

年代步车却成为交通安全的一大隐患．针对这种现象，某校数学兴趣小组在《老年代步车现象的调查报告》中就“你认为对老年代步车最有效的的管理措施”随机对某社区部分居民进行了问卷调查，其中调查问卷设置以下选项（只

选一项）：A：加强交通法规学习；B：实行牌照管理；C：加大交通违法处罚力度；D：纳入机动车管理；E：分时间分路段限行.调查数据的部分统计结果如下表：（第21题图）（1）根据上述统计表中的数据可得m=_______，n=______，a=________；（2）在答题卡中

，补全条形统计图；（3）该社区有居民2600人，根据上述调查结果，请你估计选择“D：纳入机动车管理”的居民约有多少人？管理措施回答人数百分比A255%B100mC7515%Dn35%E12525%合计a100%ABCDE管理措施人数200175150125100755025A22．（本小

题满分7分）如图，已知等腰三角形ABC的底角为30°，以BC为直径的⊙O与底边AB交于点D，过D作DEAC，垂足为E.（1）证明：DE为⊙O的切线；（2）连接OE，若BC=4，求△OEC的面积.23．（本小题满分9分）对一张矩形纸片ABCD进行折叠，具体操作如下：第一

步：先对折，使AD与BC重合，得到折痕MN，展开；第二步：再一次折叠，使点A落在MN上的点A处，并使折痕经过点B，得到折痕BE，同时，得到线段BA，EA，展开，如图1；第三步：再沿EA所在的直

线折叠，点B落在AD上的点B处，得到折痕EF，同时得到线段BF，展开，如图2.（1）证明：30ABE°；（2）证明：四边形BFBE为菱形.24．（本小题满分9分）某景区的三个景点A，B，C在同一线路上，甲、乙两名游客从景点A出

发，甲步行到景点C，乙乘景区观光车先到景点B，在B处停留一段时间后，再步行到景点C.甲、乙两人离开景点A后的路程S（米）关于时间t（分钟）的函数图象如图所示.根据以上信息回答下列问题：（1）乙出发后多长时间与甲相遇？（2）要使甲到达景点C时，乙与C的路程不超过400米，则乙从景点B步行

到景点C的速度至少为多少？（结果精确到0.1米/分钟）（第23题图）BCNA＇图1ABDCNA＇FB＇图2E（第24题图）t（分钟）MEDAM甲乙3020609030005400S（米）0（第22题图）BCODE25．（本小题满分11分）

问题情境：如图1，四边形ABCD是正方形，M是BC边上的一点，E是CD边的中点，AE平分DAM.探究展示：（1）证明：AMADMC；（2）AMDEBM是否成立？若成立，请给出证明；若不成立，请说明理由.拓展延伸：（3

）若四边形ABCD是长与宽不相等的矩形，其他条件不变，如图2，探究展示（1）、（2）中的结论是否成立？请分别作出判断，不需要证明.26．（本小题满分13分）如图，在平面直角坐标系中，抛物线与x轴交于点A(-1，0)和点B(1，0)，直线21

yx与y轴交于点C，与抛物线交于点C，D.（1）求抛物线的解析式；（2）求点A到直线CD的距离；（3）平移抛物线，使抛物线的顶点P在直线CD上，抛物线与直线CD的另一个交点为Q，点G在y轴正半轴上，当以G，P，Q三点为顶点的三角形为等腰直角三角形时，求出所有符合条件的G

点的坐标.ABMDEC图1ABM图2DEC（第25题图）（第26题图）xyABCDO绝密★启用前试卷类型：A2014年临沂市初中学生学业考试试题数学参考答案及评分标准一、选择题（每小题3分，共42分）题号1234567891011121314答案AADBBDCDBCBACC二、填空题（每小题

3分，共15分）15．(6)(6)xxx；16．5.3；17．1819；18．1yx；19．｛－3，－2，0，1，3，5，7｝．（注：各元素的排列顺序可以不同）20．解：原式=31313228(31)(31)=313222

·······························································（6分）=122=32．·································

···································（7分）（注：本题有3项化简，每项化简正确得2分）21．（1）20%，175，500．························

·········································（3分）（2）（注：画对一个得1分，共2分）……………（2分）管理措施人数200175150125100755025ABCDEBCODEGFA（3）∵2600×35%=910（人），∴选择D选项的居民约有910人.·

···················································（2分）22．（1）（本小问3分）证明：连接OD.∵OB=OD,∴∠OBD=∠ODB.又∵∠A=∠B=30°

,∴∠A=∠ODB,∴DO∥AC．································（2分）∵DE⊥AC，∴OD⊥DE．∴DE为⊙O的切线．··························································

·················（3分）（2）（本小问4分）连接DC．∵∠OBD=∠ODB=30°，∴∠DOC=60°．∴△ODC为等边三角形．∴∠ODC=60°，∴∠CDE=30°．又∵BC=4，∴DC=2,∴CE=1．·········

··················································································（2分）方法一：过点E作EF⊥BC，交BC的延长线于点F．∵∠ECF=∠A+∠B=60°,∴EF=C

E·sin60°=1×32=32．···························································（3分）∴S△OEC11332.2222OCEF·······································

··············（4分）方法二：[来源:Z.xx.k.Com]过点O作OG⊥AC，交AC的延长线于点G．∵∠OCG=∠A+∠B=60°，∴OG=OC·sin60°=2×32=3．·········································

·················（3分）∴S△OEC11313.222CEOG······················································（4分）方法三：∵OD∥CE，∴S△OEC=S△DEC．又∵DE=DC·cos30°=2×

32=3,························································（3分）∴S△OEC11313.222CEDE·······················

·······························（4分）23．证明：（1）（本小问5分）由题意知，M是AB的中点，△ABE与△A'BE关于BE所在的直线对称.[来源:学&科&网Z&X&X&K]∴AB=A'B，∠ABE=

∠A'BE.················（2分）在Rt△A'MB中，12MBA'B，∴∠BA'M=30°,···········································································

·········（4分）∴∠A'BM=60°，∴∠ABE=30°.·················································································

····（5分）（2）（本小问4分）∵∠ABE=30°，∴∠EBF=60°，∠BEF=∠AEB=60°，∴△BEF为等边三角形.················（2分）由题意知，△BEF与△B'EF关于EF所在的直线对称.∴BE=B'E=B'F=BF,∴四边形BF

'BE为菱形.·······································································（4分）24．解：（1）（本小问5分）当0≤t≤90时，设甲步行路程与时间的函数解析式为S=at.∵点(90，5400)在S=at的图象上

，∴a=60.∴函数解析式为S=60t.·····································································（1分）[来源:学§科§网]

当20≤t≤30时，设乙乘观光车由景点A到B时的路程与时间的函数解析式为S=mt+n.∵点(20，0)，(30，3000)在S=mt+n的图象上，∴200,303000.mnmn解得300,

6000.mn·················································（2分）∴函数解析式为S=300t-6000(20≤t≤30).·······································

··········（3分）根据题意，得60,3006000,StStCNBA＇图1EDAMB＇图2ABDCNA＇FME解得25,1500.ts····························

························································（4分）∴乙出发5分钟后与甲相遇.·······························································

···（5分）（2）（本小问4分）设当60≤t≤90时，乙步行由景点B到C的速度为v米／分钟，根据题意，得5400-3000-(90-60)v≤400，··································

·············（2分）解不等式，得v≥20066.73.·······························································（3分）∴乙步行由B到C的速度至少为66.7米／分钟.····················

·····················（4分）25.证明：（1）（本小问4分）方法一：过点E作EF⊥AM，垂足为F.∵AE平分∠DAM，ED⊥AD，∴ED=EF.·················

··················（1分）由勾股定理可得,AD=AF.······································（2分）又∵E是CD边的中点，∴EC=ED=EF.又∵EM=EM，

∴Rt△EFM≌Rt△ECM.∴MC=MF.·························································································（3分）∵

AM=AF+FM，∴AM=AD+MC.···················································································（4分）方法二：连

接FC.由方法一知，∠EFM=90°,AD=AF，EC=EF.·······························（2分）则∠EFC=∠ECF，∴∠MFC=∠MCF.∴MF=MC.·····················

····································································（3分）∵AM=AF+FM，∴AM=AD+MC.···········································

········································（4分）方法三：延长AE，BC交于点G.∵∠AED=∠GEC，∠ADE=∠GCE=90°，DE=EC，∴△ADE≌△GCE.∴AD=GC,∠DAE=∠G.···························

·············································（2分）又∵AE平分∠DAM，∴∠DAE=∠MAE，∴∠G=∠MAE，∴AM=GM，·······························

·······················································（3分）CGABMDEFN∵GM=GC+MC=AD+MC，∴AM=AD+MC.····························

·······················································（4分）方法四：连接ME并延长交AD的延长线于点N，∵∠MEC＝∠NED，EC＝ED，∠MCE＝∠NDE=90°，∴△MCE≌△NDE.∴MC=ND，∠CME=∠DNE.··

·······························································（2分）由方法一知△EFM≌△ECM，∴∠FME=∠CME，∴∠AMN=∠ANM.·······················

························································（3分）∴AM=AN=AD+DN=AD+MC.································

································（4分）（2）（本小问5分）成立.··········································（1分）方法一：延长CB

使BF=DE，连接AF，∵AB=AD，∠ABF=∠ADE=90°，∴△ABF≌△ADE，∴∠FAB=∠EAD，∠F=∠AED.·······（2分）∵AE平分∠DAM，∴∠DAE=∠MAE.∴∠FAB=∠MAE，∴∠FAM=∠FAB+∠BAM=∠BAM+∠MAE=∠B

AE.····································（3分）∵AB∥DC，∴∠BAE=∠DEA，∴∠F=∠FAM，∴AM=FM.······················································

···································（4分）又∵FM=BM+BF=BM+DE，∴AM=BM+DE.·················································································

··（5分）方法二：设MC=x，AD=a.由（1）知AM=AD+MC=a+x.在Rt△ABM中，∵222AMABBM，∴222()()axaax，······························································

·······（3分）∴14xa.··························································································（4分）ABMDECF∴34BMa

，54AMa，∵BM+DE=315424aaa，∴AMBMDE.··············································································（5分）（3）（本小问2分）

AM=AD+MC成立，·············································································（1分）AM=DE+BM不成立.···························

················································（2分）26.（1）（本小问3分）解：在21yx中，令0x，得1y.∴C(0，-1)···························

·······（1分）∵抛物线与x轴交于A(-1，0),B(1，0),∴C为抛物线的顶点.设抛物线的解析式为21yax，将A(-1，0)代入，得0=a-1.∴a=1.∴抛物线的解析式为21yx.········（3分）（2）（本小问5分）方法一：设直线21y

x与x轴交于E，则1(2E，0).························································································（1分）∴2151()22CE,13122AE

.····················································································（2分）连接AC，过A作AF⊥CD，垂足为F，S△CAE1122AEOCCEAF

，·····························································（4分）即131512222AF，∴355AF.································

······················································（5分）方法二：由方法一知，∠AFE=90°，32AE，52CE.··························································

·（2分）在△COE与△AFE中，图1xyABCDOFEM∠COE=∠AFE=90°，∠CEO=∠AEF，∴△COE∽△AFE.∴AFAECOCE，·········································

···········································（4分）即32152AF.∴355AF.··········································································

············（5分）（3）（本小问5分）由2211xx，得10x，22x.∴D(2，3).···········································································

···············（1分）如图1，过D作y轴的垂线，垂足为M，由勾股定理，得222425CD.·············································

······························（2分）在抛物线的平移过程中，PQ=CD.（i）当PQ为斜边时，设PQ中点为N，G(0，b)，则GN=5.∵∠GNC=∠EOC=90°，∠GCN=∠ECO，∴△GNC∽△EOC．∴GNCGOECE

，∴511522b，∴b=4.∴G(0，4).·································（3分）（ii）当P为直角顶点时，设G(0，b)，则25PG，同（i）可得b=9，则G(0，9).···································

·····················································（4分）（iii）当Q为直角顶点时，同（ii）可得G(0，9).xyECOGQPN图2综上所述，符合条件的点G有两个，分别是1G(0，4)，2G(0，9).·

················（5分）xyECDOGQP图3xyECOGQP图4获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com