晋文源初三摸底检测试题数学第Ⅰ卷选择题（共30分）一、选择题（在每小题给出的四个选项中，只有一项符合题目要求，请选出并在答题卡上将该项涂黑.本大题共10个小题，每小题3分，共30分）1．数轴上点A ，B 表示的数分别是5，－2，它们之间的距离可以表示为A ．|－2－5|B ．－2－5C ．－2+5D ．|－2+5|2．下面四个手机应用图标中是轴对称图形的是A. B. c. D.3．在一个不透明的袋子中装有形状、大小、质地完全相同的5个球，其中3个黑球、2个白球，从袋子中一次摸出3个球，下列事件是必然事件的是A ．摸出的是3个白球B ．摸出的是3个黑球C ．摸出的球中至少有1个是黑球D ．摸出的是2个白球、1个黑球4．下列运算正确的是A ．515422=÷-）（B ．14322-=-x x C ．523)1575(=÷-D ．632)(x x =--5．不等式4262+>-x x 的解集是A ．x ＜－5B ．x ＞－5C ．x ＞5D ．x ＜56．由若干个相同的小正方体搭成的一个几何体的主视图和俯视图如图所示，则组成这个几何体的小正方体的个数最多有A ．8B ．7C ．6D ．5（第6题图）7．生物活动小组的同学们观察某植物生长，得到该植物高度y （单位：cm ）与观察时间x （单位：天）的关系，并画出如图所示的图象（CD ∥x 轴），该植物最高的高度是A ．50cmB ．20cmC ．16cmD ．12cm（第7题图）8．如图，有一个边长为2cm 的正六边形纸片，若在该纸片上沿虚线剪一个最大圆形纸片，则这个圆形纸片的半径是A ．3cmB ．2cmC ．32cmD ．4cm（第8题图）9．如图，已知在平面直角坐标系中，点O 是坐标原点，△AOB 是直角三角形，∠AOB =90°，OB =2OA ，点B 在反比例函数x y 2=上，若点A 在反比例函数x k y =上，则k 的值为A ．21B ．21-C ．41D ．41-（第9题图）10．如图，点A 在x 轴上，∠OAB =90°，∠B =30°，OB =6，将△OAB 绕点O 按顺时针方向旋转120°得到△OA ′B ′，则点B ′的坐标是A ．（33，－3）B ．（3，33）C ．（33，3）D ．（3，33 ）（第10题图）第Ⅱ卷(非选择题90分)二、填空题（本大题共5个小题，每小题3分，共15分）11-1．据2020年3月公布的《山西省2019年国民经济和社会发展统计公报》显示，经初步核算，2019年我省实现地区生产总值17026.68亿元，比上年增长6.2%.数据17026.68亿元用科学记数法表示为元．11-2.我们规定把同一副扑克牌中的红桃A，黑桃A，梅花A三张牌背面朝上放在桌子上，将扑克牌洗匀后从中随机抽取一张，记下扑克牌的花色后放回，洗匀后再随机抽取一张，则两次抽取的扑克牌为同一张的概率为．（第11-2题）11-3.杨辉，字谦光，钱塘（今浙江杭州）人，南宋杰出的数学家和数学教育家，杨辉一生留下了大量的著述.下面是杨辉在1275年提出的一个问题（选自杨辉所著《田亩比类乘除算法》）：直田积（矩形面积）八百六十四步（平方步），只云阔（宽）不及长一十二步（宽比长少一十二步），问阔及长各几步．解答这个问题可知长为步.11-4.如图，在□ABCD 中，AH ⊥BC 于点H ，点E 在AD 上，∠EBC =45°，BE 交AH 于点F ，连接CF ，CF ⊥CD ．若BH =1，AB =10，则EF 的长为.（第11-4题）11-5.如图，在□ABCD 中，AB =BC =2，∠ABC =60°，过点D 作DE ∥AC ，DE =21AC ，连接AE ，则△ADE 的周长为.（第11-5题）三、解答题（本大题共8个小题，共75分.解答题应写出文字说明、证明过程或演算步骤）12.（每小题5分，共10分）（1）解方程组：⎪⎩⎪⎨⎧=+-=+--.232),1(32)1(4y x y y x （2）已知实数a 满足a 2+2a －9=0，求12)2)(1(121122+-++÷-+-+a a a a a a a 的值．13.（本题7分）如图，在Rt △ABC 中，∠C =90°，点D 是CB 的中点，将△ACD 沿AD 折叠后得到△AED ，过点B 作BF ∥AC 交AE 的延长线于点F .求证：BF =EF .14.（本题6分）阅读理解，并解决问题：“整体思想”是中学数学中的一种重要思想，贯穿于中学数学的全过程，比如整体代入，整体换元，整体约减，整体求和，整体构造，…，有些问题若从局部求解，采取各个击破的方式，很难解决，而从全局着眼，整体思考，会使问题化繁为简，化难为易，复杂问题也能迎刃而解．例：当代数式x2+3x+5的值为7时，求代数式3x2+9x－2的值．解：因为x2+3x+5=7，所以x2+3x=2．所以3x2+9x－2=3（x2+3x）－2=3×2－2=4.以上方法是典型的整体代入法.请根据阅读材料，解决下列问题：（1）已知a2+3a－2=0，求5a3+15a2－10a+2020的值．（2）我们知道方程x2+2x－3=0的解是x1=1，x2=－3，现给出另一个方程（2x+3）2+2（2x+3）－3=0，则它的解是.15.（本题9分）某社区组织了以“奔向幸福，‘毽’步如飞”为主题的踢毽子比赛活动，初赛结束后有甲、乙两个代表队进入决赛，已知每队有5名队员，按团体总数排列名次，在规定时间内每人踢100个以上（含100）为优秀．下表是两队各队员的比赛成绩．1号2号3号4号5号总数甲队1031029810097500乙队979910096108500经统计发现两队5名队员踢毽子的总个数相等，按照比赛规则，两队获得并列第一.学习统计知识后，我们可以通过考查数据中的其它信息作为参考，进行综合评定：（1）甲、乙两队的优秀率分别为，；（2）甲队比赛数据的中位数为个；乙队比赛数据的中位数为个；（3）分别计算甲、乙两队比赛数据的方差；（4）根据以上信息，你认为综合评定哪一个队的成绩好？简述理由．16.（本题8分）如图1，一辆汽车从A地出发去往C地，A，C两地相距273km.由于A，C之间某路段正在修路.驾驶员临时改变路线，先由A地开往B地，再由B地开往C地，如图2是从该场景中抽象出来的示意图，已知∠A=30°，∠C=45°，则这样的行驶路程比原来路程273km远了多少？（结果精确到1km，参考数据：2≈1.41，3≈1.73）17.（本题9分）“十三五”以来，山西省共解决372个村、35.8万农村人口的饮水型氟超标问题，让农村群众真正喝上干净水、放心水、安全水.某公司抓住商机，根据市场需求代理A，B两种型号的净水器，已知每台A型净水器比每台B型净水器进价多200元，用5万元购进A型净水器与用4.5万元购进B型净水器的数量相等.（1）求每台A型，B型净水器的进价各是多少元？（2）该公司计划购进A，B两种型号的净水器共55台进行试销，其中A型净水器为m 台，购买两种净水器的总资金不超过10.8万元．则最多可购进A型号净水器多少台？（A 型）（B 型）18.（本题12分）综合与实践正方形内“奇妙点”及性质探究定义：如图1，在正方形ABCD 中，以BC 为直径作半圆O ，以D 为圆心，DA 为半径作⌒AC ，与半圆O 交于点P .我们称点P 为正方形ABCD 的一个“奇妙点”.过奇妙点的多条线段与正方形ABCD 无论是位置关系还是数量关系，都具有不少优美的性质值得探究.性质探究：如图2，连接DP 并延长交AB 于点E ，则DE 为半圆O 的切线.证明：连接OP ，OD .由作图可知，DP =DC ，OP =OC ，又∵OD =OD .∴△OPD ≌△OCD .（SSS ）∴∠OPD =∠OCD =90°．∴DE 是半圆O 的切线.问题解决：（1）如图3，在图2的基础上，连接OE .请判断∠BOE 和∠CDO 的数量关系，并说明理由；（图1）（图2）（图3）（图4）（图5）（2）在（1）的条件下，请直接写出线段DE ，BE ，CD 之间的数量关系；（3）如图4，已知点P 为正方形ABCD 的一个“奇妙点”，点O 为BC 的中点，连接DP 并延长交AB 于点E ，连接CP 并延长交AB 于点F ，请写出BE 和AB 的数量关系，并说明理由；（4）如图5，已知点E ，F ，G ，H 为正方形ABCD 的四个“奇妙点”.连接AG ，BH ，CE ，DF ，恰好得到一个特殊的“赵爽弦图”.请根据图形，探究并直接写出一个不全等的几何图形面积之间的数量关系.19.（本题14分）综合与探究：在平面直角坐标系xOy 中，已知抛物线32332632++-=x x y 与x 轴交于A ，B 两点（点B 在点A 的右侧），与y 轴交于点C ，它的对称轴与x 轴交于点D ，直线l 经过C ，D 两点，连接AC .（1）求A ，B 两点的坐标及直线l 的函数表达式；（2）探索直线l 上是否存在点E ，使△ACE 为直角三角形，若存在，求出点E 的坐标；若不存在，说明理由；（3）若点P 是直线l 上的一个动点，试探究在抛物线上是否存在点Q ：①使以点A，C，P，Q为顶点的四边形为菱形，若存在，请直接写出点Q的坐标；若不存在，说明理由；②使以点A，C，P，Q为顶点的四边形为矩形，若存在，请直接写出点Q的坐标；若不存在，说明理由.找准方向　事半功倍临门一脚　决战中考体现最新中考改革细微变化，新方向新素材、新主题查漏补缺、强化训练全仿真模拟多角度命题全方位猜押多维度预测晋文源教育隆重出品时政热点　中考冲刺时政热点事件化高频考点集训化中考疑点问题化考政重点习题化扫码关注轻松提分上市时间：5月10日上市时间：6月1日上市时间：6月25日晋文源初三摸底检测试题数学参考答案及评分标准一、选择题：题号12345678910答案A D C D A B C A B D二、填空题：11-1． 1.702668×101211-2．3111-3．3611-4．2211-5．73+三、解答题：12．解：（1）原方程可化为⎩⎨⎧=+=-②①，.122354y x y x ····································2分①×2+②得11x =22.解得x =2.····················································································4分把x =2代入①得y =3.所以，这个方程组的解为⎩⎨⎧==.3,2y x ·························································5分（2）解：原式=)2)(1()1()1)(1(2112++-⋅-++-+a a a a a a a ···························7分=2)1(111+--+a a a =2)1(2+a .·······································································8分∵a 2+2a －9=0，∴（a +1）2=10.·············································································9分∴原式=51102)1(22==+a ．································································10分13．证明：如答图，连接DF ，答图∵D 是CB 的中点，∴CD =BD .·····················································································1分∵将△ACD 沿AD 折叠后得到△AED ，∴CD =ED ，∠AED =∠C =90°.·······························································2分∴BD =ED ，∠DEF =90°.······································································3分∵BF ∥AC ，∠C =90°，∴∠CBF =90°.∴∠DBF =∠DEF =90°.·········································································4分在Rt △DBF 和Rt △DEF 中，⎩⎨⎧==,ED BD DF DF ，∴Rt △DBF ≌Rt △DEF （HL ）.···································································6分∴BF =EF .······················································································7分14.（1）（方法不唯一）例如，解：5a 3+15a 2－10a +2020=5a （a 2+3a －2）+2020.···································································2分∵a 2+3a －2=0.∴原式=0+2020=2020.············································································3分∴5a 3+15a 2－10a +2020的值为2020．·····················································4分（2）x 1=－1，x 2=－3．········································································6分15.解：（1）60%，40%；····································································2分（2）100，99；··················································································4分（3）甲、乙两队比赛数据的平均数均为500÷5=100（个）.·························5分5265)10097()100100()10098()100102()100103(222222=-+-+-+-+-=甲s .·······································································································6分185)100108()10096()100100()10099()10097(222222=-+-+-+-+-=乙s .·······································································································7分（4）综合评定甲队的成绩好.································································8分理由如下：因为甲队的优秀率比乙队高；甲队的中位数比乙队大；甲班的方差比乙班低，比较稳定，综合评定甲队比较好．·········································································9分16．解：如答图，过点B 作BD 垂直于AC 于点D ，答图在Rt △ABD 中，设BD=x ，AD=BD 330tan =︒，x BD AB 230sin =︒=，·······································································1分在Rt △BDC 中，CD=BD=x ，︒=45sin BD BC =2.································2分∵AD+CD=AC ，∴2733=+x x ，·············································································3分∴10013273≈+=x .············································································4分∴1412,2002≈=≈=x BC x AB ，····················································5分∴341141200≈+≈+BC AB （km ），··············································6分∴68273341≈-≈-+AC BC AB （km ）.···································7分答：这样的行驶路程比原来路程273km 远了68km.（67也给分）·················8分17．解：（1）设每台B 型净水器的进价是x 元.······································1分根据题意，得x x 4500020050000=+.·························································2分解得x =1800.·······················································································3分经检验，x =1800是原分式方程的解，且符合题意.·······································4分∴x +200=2000．答：每台A 型净水器的进价是2000元，每台B 型净水器的进价是1800元；···5分（2）购进A 型净水器m 台，则购进B 型净水器（55－m ）台.···················6分依题意，得2000m +1800（55－m ）≤108000.············································7分解得m ≤45．·····················································································8分答：最多可购进A 型净水器45台．························································9分18．解：（1）∠BOE=∠CDO .···································································1分理由如下：∵△OPD ≌△OCD .∴∠OPD =∠OCD =90°，∠POD=∠COD ，∠CDO =∠PDO=21∠PDC .∴∠POC +∠PDC =360°－∠OPD －∠OCD =180°.············································2分∵∠POC +∠BOP =180°，∴∠BOP =∠PDC .··················································································3分在Rt △POE 和Rt △BOE 中∵OE =OE ，OP =OB （由作图得出）.∴△POE ≌△BOE .∴∠POE=∠BOE =21∠BOP .·····································································4分∵∠CDO =∠PDO=21∠PDC .∴∠BOE=∠CDO .··················································································5分（2）线段DE ，BE ，CD 之间的数量关系是DE =BE+CD .··································7分（3）如答图，连接OE ，OD ，答图由（1）可知，∠BOE=∠CDO .又∵∠B=∠OCD =90°，点O 为BC 的中点，∴tan ∠BOE=tan ∠CDO .∴21==DC OC BO BE .············································8分∴BE =21BO =2121⨯BC =41BC .·······························9分∵四边形ABCD 是正方形，∴AB =BC .∴BE =41AB .·························································································10分（4）答案不唯一，例如，△ABH 的面积等于正方形EFGH 的面积；正方形EFGH 的面积等于正方形ABCD 面积的51等等.·····························································12分19．解：（1）当y =0时，032332632=++-x x .解得1x =－2，2x =6.∴点A 的坐标为（－2，0），点B 的坐标为（6，0）.······························2分（∵32332632++-=x x y =338)2(632+--x .（可以不写））∴抛物线的对称轴为直线x =2.∴点D 的坐标为（2，0）.·································································3分当x =0时，y =32.∴点C 的坐标为（0，32）.································································4分设直线l 的表达式为y =kx ＋b ，则⎩⎨⎧=+=.0232b k b 解得⎪⎩⎪⎨⎧=-=.323b k ∴直线l 的表达式为323+-=x y .···················································5分（2）直线l 上存在点E ，使△ACE 为直角三角形.·······································6分∵点A 的坐标为（－2，0），点D 的坐标为（2，0），∴AD =4.又∵点C 的坐标为（0，32），CO ⊥AD ，∴AC =CD =4)32(222=+.∴AC =CD =AD .∴△ACE 为等边三角形.∴∠ADC =∠CAD =60°.···········································································7分分两种情况：①当∠AE 1C =90°时，。