南京市联合体中考二模数学试题及答案Word版

南京市联合体中考二模数学试题及答案Word版

‎2015年中考模拟试卷（二）‎ 数 学 化工园　雨花　栖霞　浦口四区联合体 注意事项：‎ ‎1．本试卷共6页．全卷满分120分．考试时间为120分钟．考生答题全部答在答题卡上，答在本试卷上无效．‎ ‎2．请认真核对监考教师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符合，再将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在答题卡及本试卷上．‎ ‎3．答选择题必须用2B铅笔将答题卡上对应的答案标号涂黑．如需改动，请用橡皮擦干净后，再选涂其他答案．答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡上的指定位置，在其他位置答题一律无效．‎ ‎4．作图必须用2B铅笔作答，并请加黑加粗，描写清楚．‎ 一、选择题（本大题共6小题，每小题2分，共12分．在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在答题卡相应位置上）‎ ‎1．﹣的相反数是 （ ▲ ） ‎ A. -2 B.2 C. - D. ‎2.下列计算正确的是（ ▲ ）‎ A. a3+a4=a7 B．2a3•a4=2a7 C．（2a4）3=8a7 D．a8÷a2=a4‎ ‎3.为调查某班学生每天使用零花钱的情况，张华随机调查了20名同学，结果如下表：‎ 每天使用零花钱（单位：元）‎ ‎1‎ ‎2‎ ‎3‎ ‎4‎ ‎5‎ 人数 ‎1‎ ‎3‎ ‎6‎ ‎5‎ ‎5‎ 则这20名同学每天使用的零花钱的众数和中位数分别是（ ▲ ）‎ A. 3，3‎ B. 3，3.5 ‎ C. 3.5，3.5‎ D. 3.5，3‎ ‎4．小张同学的座右铭是“态度决定一切”，他将这几个字写在一个正方体纸盒的每个面上，其平面展开图如图所示，那么在该正方体中， 和“一”相对的字是（ ▲ ）‎ A. 态 ‎ B. 度 ‎ C. 决 D. 切 ‎（第6题）‎ B A D C E F ‎（第5题）‎ A B C O ‎（第4题）‎ ‎5. 如图，⊙O是△ ABC的外接圆，∠OBC=42°，则∠A的度数是（ ▲ ）‎ A. 42° ‎ B. 48° ‎ C. 52° ‎ D. 58°‎ ‎6．如图，在矩形ABCD中，AB＝3，BC＝5，以B为圆心BC为半径画弧交AD于点E，连接CE，作BF⊥CE，垂足为F，则tan∠FBC的值为（ ▲ ）‎ A. B. C. D. 二、填空题（本大题共10小题，每小题2分，共20分，请在答题卡指定区域内作答．）‎ ‎7．代数式有意义，则 x的取值范围是 ▲ ．‎ ‎8. 分解因式：a3－4a＝ ▲ ．‎ ‎9. 计算－2cos30°－|1－|＝ ▲ .‎ ‎10. 反比例函数y＝的图象经过点（1，6）和（m，-3），则m＝ ▲ .‎ ‎11. 如图，在菱形ABCD中，AC＝2，∠ABC＝60°，则BD＝ ▲ .‎ B O A ‎1‎ C D ‎（第12题）‎ ‎12. 如图，在⊙O中， AO∥CD, ∠1＝30°，劣弧AB的长为3300千米，则⊙O的周长用科学计数法表示为 ▲ 千米．‎ A B C D ‎（第11题）‎ ‎13.某商品原价100元，连续两次涨价后，售价为144元，若平均增长率为x，则x＝ ▲ .‎ ‎14．直角坐标系中点A坐标为（5，3），B坐标为（1，0），将点A绕点B逆时针旋转90°得到点C，则点C的坐标为 ▲ .‎ ‎(第15题)‎ ‎15．二次函数y＝ax2+bx+c（a≠0）的图象如图所示，根据图象可知：方程ax2+bx+c＝k有两个不相等的实数根，则k的取值范围为 ▲ . ‎ O ‎(第16题)‎ ‎16.如图，在半径为2的⊙O中，两个顶点重合的内接正四边形与正六边形，则阴影部分的面积为 ▲ .‎ 三、解答题（本大题共11小题，共88分．请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或演算步骤）‎ ‎17．（6分）解方程组 ‎ ‎18.（6分）化简：（－x）÷．‎ ‎19.（8分）为了备战初三物理、化学实验操作考试，某校对初三学生进行了模拟训练．物理、化学各有3个不同的操作实验题目，物理用番号①、②、③代表，化学用字母a、b、c表示．测试时每名学生每科只操作一个实验，实验的题目由学生抽签确定．‎ ‎（1）小张同学对物理的①、②和化学的b、c实验准备得较好.请用树形图或列表法求他两科都抽到准备得较好的实验题目的概率；‎ ‎（2）小明同学对物理的①、②、③和化学的a实验准备得较好.他两科都抽到准备得较好的实验题目的概率为 ▲ .‎ ‎20. （8分）据报道，历经一百天的调查研究，南京PM 2.5源解析已经通过专家论证．各种调查显示，机动车成为PM 2.5的最大来源，一辆车每行驶20千米平均向大气里排放0.035千克污染物．校环保志愿小分队从环保局了解到南京100天的空气质量等级情况，并制成统计图和表：‎ ‎2014年南京市100天空气质量等级天数统计图 ‎ ‎ 空气质量等级 优 良 轻度污染 中度污染 重度污染 严重污染 天数（天）‎ ‎10‎ a ‎12‎ ‎8‎ ‎25‎ b ‎2014年南京市100天空气质量等级天数统计表 优 良 轻度 重度 严重 ‎10%‎ ‎25%‎ ‎12%‎ ‎8%‎ n°‎ ‎25%‎ 中度 （1） 表中a＝ ▲ ，b＝ ▲ ，图中严重污染部分对应的圆心角n＝ ▲ °.‎ ‎（2）请你根据“2014年南京市100天空气质量等级天数统计表”计算100天内重度污染和严重污染出现的频率共是多少？‎ ‎（3）小明是社区环保志愿者，他和同学们调查了机动车每天的行驶路程，了解到每辆车每天平均出行25千米．已知南京市2014年机动车保有量已突破200万辆，请你通过计算，估计2014年南京市一天中出行的机动车至少要向大气里排放多少千克污染物？‎ ‎21.（8分）如图， 在□ABCD中，E、F、G、H分别为AB、BC、CD、AD的中点，AF与EH交于点M，FG与CH交于点N.‎ ‎（1）求证：四边形MFNH为平行四边形；‎ ‎（2）求证：△AMH≌△CNF.‎ A B C D F G E H M N ‎22. （8分）端午节期间，某食堂根据职工食用习惯，购进甲、乙两种粽子260个，其中甲种粽子花费300圆，乙种粽子花费400元，已知甲种粽子单价比乙种粽子单价高20%，乙种粽子的单价是多少元？甲、乙两种粽子各购买了多少个？‎ ‎23.（8分）如图，为了测出某塔CD的高度，在塔前的平地上选择一点A，用测角仪测得塔顶D的仰角为30º，在A、C之间选择一点B （A、B、C三点在同一直线上），用测角仪测得塔顶D的仰角为75º，且AB间距离为‎40m．‎ ‎（1）求点B到AD的距离； ‎ ‎（2）求塔高CD（结果用根号表示）．‎ A B C D ‎（第23题）‎ ‎30°‎ ‎75°‎ ‎24．（8分）小林家、小华家、图书馆依次在一条直线上．小林、小华两人同时各自从家沿直线匀速步行到图书馆借阅图书，已知小林到达图书馆花了20分钟.设两人出发x（分钟）后，小林离小华家的距离为y（米），y与x的函数关系如图所示.‎ ‎（1）小林的速度为 ▲ 米/分钟 ，a＝ ▲ ，小林家离图书馆的距离为 ▲ 米；‎ ‎（2）已知小华的步行速度是‎40米/分钟，设小华步行时与自己家的距离为y1（米），请在图中画出y1（米）与x（分钟 ）的函数图象；‎ ‎（3）小华出发几分钟后两人在途中相遇？‎ 第题 ‎（第24题）‎ x（分钟）‎ y（米）‎ ‎4‎ ‎20‎ ‎240‎ O a ‎25．(8分)施工队要修建一个横断面为抛物线的公路隧道，其高度为‎6米，宽度OM为12米．现以O点为原点，OM所在直线为x轴建立直角坐标系（如图①所示）．‎ ‎（1）求出这条抛物线的函数表达式，并写出自变量x的取值范围；‎ ‎（2）隧道下的公路是双向行车道(正中间有一条宽1米的隔离带)，其中的一条行车道能否行驶宽2.5米、高5米的特种车辆？请通过计算说明；‎ ‎（第25题）‎ ‎26. （10分）如图，已知△ABC，AB=6、AC＝8，点D是BC边上一动点，以AD为直径的⊙O分别交AB、AC于点E、F.‎ ‎（1）如图①若∠AEF＝∠C，求证：BC与⊙O相切；‎ ‎（2）如图②，若∠BAC＝90°，BD长为多少时，△AEF与△ABC相似.‎ 图①‎ A D B C E F O D B O C F A E A B C 备用 ‎ ‎ 图②‎ ‎27. (10分)已知直角△ABC，∠ACB＝90°，AC＝3，BC＝4，D为AB边上一动点，沿EF折叠，点C与点D重合，设BD的长度为m.‎ ‎（1）如图①，若折痕EF的两个端点E、F在直角边上，则m的范围为 ▲ ;‎ ‎（2）如图②，若m等于2.5，求折痕EF的长度；‎ ‎（3）如图③，若m等于，求折痕EF的长度.‎ D F A C D B E F A C B A C B 图②‎ 图③‎ 图①‎ D E F E ‎ ‎ ‎2015中考数学模拟试卷（二）答案 一、选择题（本大题共6小题，每小题2分，共12分．）‎ 题号 ‎1‎ ‎2‎ ‎3‎ ‎4‎ ‎5‎ ‎6‎ 答案 D B B A B D 二、填空题（本大题共10小题，每小题2分，共20分．）‎ ‎7．x＞1 8. a（a－2）（a+2） 9. +1 10. ﹣2 11. 2 ‎ ‎12.3.96×104 13. （﹣2，4） 14.0.2 15. k＜2 16. 6－2 三、解答题（本大题共11小题，共88分．）‎ ‎17．‎ 解： ①×2得：4x+6y＝﹣10③‎ ‎②×3得：9x－6y＝36 ④‎ ‎③+④得：13x＝26‎ 解得： x＝2········································································································3分 把x＝2代入①得y＝﹣3····················································································5分 所以方程组的解为·················································································6分 ‎18.解原式＝[－]÷·············································································1分 ‎＝×·····························································································2分 ‎＝×··································································································3分 ‎＝×···················································································4分 ‎＝-x(x－1) ··············································································································5分 ‎＝﹣x2+x ················································································································6分 ‎19. （1）画图或列表正确·····································································································4分 共有9种等可能结果，期中两科都满意的结果有4种··································································5分 P（两科都满意）＝.·········································································································6分 ‎（2）···························································································································8分 ‎20. （1）25；20；72°······································································································3分 ‎（2）45% ···············································································································5分 ‎（3）200×0.035×10000×＝87500（千克）···········································································8分 ‎21. （1）证明：连接BD，∵E、F、G、H分别为AB、BC、CD、AD的中点，‎ ‎∴EH为△ABD的中位线，∴EH∥BD.‎ 同理FG∥BD.‎ ‎∴EH∥FG·······················································································································2分 在□ABCD中 ‎∴AD∥＝BC，‎ ‎∵H为AD的中点AH＝AD，‎ ‎∵F为BC的中点FC＝BC，‎ ‎∴AH∥＝FC ‎∴四边形AFCH为平行四边形，‎ ‎∴AF∥CH·······················································································································4分 又∵EH∥FG ‎∴四边形MFNH为平行四边形···························································································5分 ‎（2）∵四边形AFCH为平行四边形 ‎∴∠FAD＝∠HCB ···········································································································6分 ‎∵EH∥FG,∴∠AMH＝∠AFN ‎∵AF∥CH ‎∴∠AFN＝∠CNF ‎∴∠AMH＝∠CNF············································································································7分 又∵AH＝CF ‎∴△AMH≌△CNF·············································································································8分 ‎22.解：设乙种粽子的单价是x元，则甲种粽子的单价为（1+20%）x元，‎ 由题意得， +＝260，···················································································4分 解得：x=2.5，·················································································5分 经检验：x=2.5是原分式方程的解，························································································6分 ‎（1+20%）x=3，‎ 则买甲粽子为： ＝100个，乙粽子为：＝160个．················································7分 答：乙种粽子的单价是2.5元，甲、乙两种粽子各购买100个、160个．········································8分 ‎23. （1）作BE⊥AD，垂足为E，‎ A B C D ‎（第23题）‎ ‎30°‎ ‎75°‎ E 在Rt△AEB中，sinA＝，‎ ‎ ＝，BE＝20················3分 ‎（2）∠DBC是△ABD的外角 ‎∠ADB＝∠DBC－∠A＝45°，···············4分 在Rt△DEB中，tan∠EDB＝ ，1＝，‎ ED＝20·············································5分 在Rt△AEB中，cos∠EAB＝ ， EA＝20······························6分 AD＝ED+ EA＝20+20························································································7分 在Rt△ACD中，sin∠DAC＝ ， EA＝10+10·····················································8分 ‎24.（1）60；960；1200；····························3分 ‎（2）如图略（以（0,0）、（24,960）为端点的线段），····························5分 ‎（3）解法一：由题意得60x－240=40x，x=12，小华出发12分钟后两人在途中相遇．························8分 解法二：设小林在4~20分钟的函数表达式为y=kx+b，‎ 则，∴k=60，b=－240，下同解法一··········8分 ‎25.解：（1）设抛物线的函数表达式为y=a(x－6)2+6，∵图像过点（0,0）∴a =－，…………………2分 ‎∴y=－ (x－6)2+6＝－x2+2x，…………………3分 ‎0≤x≤12．…………………4分 ‎（2）当x＝3时，y＝－×9+2×3＝4.5．…………………6分 ‎∵4.5＜5，∴不能通过．…………………8分 ‎26.(1)证明：连接DF，在⊙O中∠AEF＝∠ADF····························1分 又∵∠AEF＝∠C∴∠ADF＝∠C····························2分 ‎∵AD为直径，∴∠AFD＝90°∴∠CFD＝90°∴∠C+∠CDF＝90°‎ ‎∴∠ADF+∠CDF＝90°∴∠ADC＝90°····························3分 又∵AD为直径∴BC与⊙O相切. ····························4分 ‎(2)情况一：若△AEF∽△ACB，则∠AEF＝∠C，由（1）知BC与⊙O相切. ∴BD＝3.6···············7分 情况二：若△AEF∽△ABC ∴∠AEF＝∠B，∴EF∥BC，‎ ‎∵∠EAF为直角，∴EF为直径，∴△AEO∽△ABD，‎ ‎∴＝＝＝，∴BD＝2EO＝EF ‎∵EF∥BC∴△AEF∽△ABC∴＝＝,即BD＝2EO＝EF＝BC＝5……………………10分 ‎27.解：（1）2≤m≤4；…………………2分 ‎（2）方法一、∵∠ACB＝90°，AC＝3，BC＝4，∴AB＝5，∵BD＝2.5，∴AD＝DB＝CD＝2.5，‎ ‎∵点C与点D关于对称，∴DE＝CE，CF＝DF，∴∠CAD=∠ECD=∠EDC，‎ ‎∴△ACD∽△CDE， ‎ ‎∴=，即＝，‎ ‎∴CE=；同理CF= ；∴EF=．…………………6分 方法二、作DG⊥BC，垂足为G，连接DF，△BGD∽△BCA，∴== ‎∴DG＝，CG＝GB＝2‎ 在Rt△FDG中，FG2+DG2＝DF2，（2－DF）2+1.52＝DF2，解得DF＝，CF＝DF＝…………………4分 ‎∵∠CEF+∠ECD=90°，∠DCF+∠ECD=90°，∴∠CEF=∠DCF，又∵∠ECF＝∠CGD＝90°‎ ‎∴△ECF∽△CGD∴=∴EF=.…………………6分 ‎（3）作DG⊥BC，垂足为G，作EH⊥BC,垂足为H，连接DF，△BGD∽△BCA，∴＝＝ ‎∴DG＝， GB＝∴CG＝ 在Rt△FDG中，FG2+DG2＝DF2，（－DF）2+（）2＝DF2，解得DF＝，CF＝DF＝……………8分 易证∠HEF=∠DCG，又∵∠EHF＝∠DGC＝90°‎ ‎∴△EHF∽△CGD∴＝∴＝＝，设FH＝x，则EH＝3x，‎ ‎∵EH∥AC，∴△EHB∽△ACB∴＝∴＝解得x＝ ，‎ ‎∴EF＝FH＝…………10分 D E G F A C D B E F A C B A C B 备用 备用 图①‎ D E F H G