2018-2019学年南京联合体初二上学期数学期中试卷及答案

12018-2019学年南京联合体初二上学期数学期中试卷八年级数学（考试时间100分钟，试卷总分100分）一、选择题（每小题2分，计16分）1．下列图案中，属于轴对称图形的是ABCD2．16的平方根是A．4B.±4C.4D.±43．如图，在数轴上，与表示3的点最接近的点是A．点AB．点BC．点CD．点D4．满足下列条件的△ABC不是直角三角形的是A．BC＝1，AC＝2，AB＝3B．BC＝1，AC＝2，AB＝5C．BC：AC：AB＝3：4：5D．∠A：∠B：∠C＝3：4：55．如图，工人师傅常用“卡钳”这种工具测定工件内槽的宽．卡钳由两根钢条AA′、BB′组成，O为AA′、BB′的中点.只要量出A′B′的长度，由三角形全等就可以知道工件内槽AB的长度．那么判定△OAB≌△OA′B′的理由是A．SASB．ASAC．SSSD．AAS6．如图，有一个池塘，其底面是边长为10尺的正方形，一个芦苇AB生长在它的中央，高出水面部分BC为1尺．如果把该芦苇沿与水池边垂直的方向拉向岸边，那么芦苇的顶部B恰好碰到岸边的B′．则这根芦苇的长度是（第3题）（第5题）（第6题）2A．10尺B．11尺C．12尺D．13尺7．如图，在△ABC中，∠BAC＝106°，EF、MN分别是AB、AC的中垂线，E、N在BC上，则∠EAN的度数是A．74°B．32°C．35°D．30°8．如图，四边形ABCD中，AB＝AD，点B关于AC的对称点B′恰好落在CD上，若∠BAD＝100°，则∠ACB的度数为A．40°B．45°C．60°D．80°二、填空题（每小题2分，共20分）9．比较大小：392．10．下列五个数4，2π，227，38，3.1415926中，是无理数的有_________.11．被誉为“中国天眼”的世界上最大的单口径球面射电望远镜FAST的反射面总面积为249900m2，请将249900精确到万位，并用科学记数法表示为______________.12．如图，在△ABC中，∠B＝∠C，AD平分∠BAC，AB＝5，BC＝6，则AD＝__________.13．如图，已知点A、D、B、F在一条直线上，AC＝EF，AB＝DF，要使△ABC≌△FDE，还需添加一个条件，这个条件可以是．（只需填一个即可）14．如图，在Rt△ABC中，∠C＝90°，以A为圆心，任意长为半径画弧，分别交AC、AB于点M、N，再分别以M、N为圆心，任意长为半径画弧，两弧交于点O，作射线AO交BC于点D，若CD＝2，P为AB上一动点，则PD的最小值为____________.（第13题）（第12题）ABCDFEBDCA（第14题）PDBACOMN（第7题）MAENFBC（第8题）B′CBAD315．如图，在△ABC中，∠ABC与∠ACB的平分线相交于点O，过点O作MN∥BC，分别交AB、AC于点M、N．若△ABC的周长为15，BC＝6，则△AMN的周长为_______.16．如图，∠ABC＝90°，AD∥BC，以B为圆心，BC长为半径画弧，与射线AD相交于点E，连接BE，过点C作CF⊥BE，垂足为F．若AB＝6，BC＝10，则EF的长为__________.17．如图，两块完全一样的含30°角的直角三角板，将它们重叠在一起并绕其较长直角边的中点M转动，使上面一块三角板的斜边刚好过下面一块三角板的直角顶点C．已知AC＝4，则这两块直角三角板顶点A、A′之间的距离等于．18．在△ABC中，∠B＝30°，点D在BC边上，点E在AC边上，AD＝BD，DE＝CE，若△ADE为等腰三角形，则∠C的度数为_______________°．三、解答题(本大题共8小题，共64分)19．(8分)求下列各式中x的值．（1）4x2－9＝0（2）(x＋1)3＝－2720．（6分）已知：如图，∠ABC＝∠DCB，BD、CA分别是∠ABC、∠DCB的平分线．求证：AB＝DC．\21．（7分）如图，在△ABC中，AB＝AC，点D、E、F分别在AB、BC、AC边上，且BE＝CF，BD＝CE．求证：∠ABC＝∠ACB＝∠DEF．AA（第21题）EFDBAC（第20题）CBAD（第17题）（第16题）（第15题）FEBCDABONMACACMB’C’BA’422．（7分）如图，点C在线段AB上，AD∥EB，AC＝BE，AD＝BC，CF平分∠DCE．求证：CF⊥DE．23．（8分）如图，在△ABC中，∠B＝90°，AB＝4，BC＝8．（1）在BC上求作一点P，使PA+PB＝BC；（尺规作图，不写作法，保留作图痕迹）（2）求BP的长．24．（9分）如图，在四边形ABCD中，∠ABC＝∠ADC＝90°，AB＝AD，E是AC的中点．（1）求证：∠EBD＝∠EDB．（2）若∠BED＝120°，试判断△BDC的形状并说明理由．（第23题）EDCABFECBAD（第22题）（第23题）CBA525．（9分）（1）如图①，分别以△ABC的边AB、AC为一边向形外作正方形ABDE和正方形ACGF．求证S△AEF＝S△ABC．（2）如图②，分别以△ABC的边AB、AC、BC为边向形外作正方形ABDE、ACGF、BCHI，可得六边形DEFGHI，若S正方形ABDE＝17，S正方形ACGF＝25，S正方形BCHI＝16，求S六边形DEFGHI．BIDAFECHG（图②）（图①）BCGAFED626．（10分）“面积法”是指利用图形面积间的等量关系寻求线段间等量关系的一种方法．例如：在△ABC中，AB＝AC，点P是BC所在直线上一个动点，过P点作PD⊥AB、PE⊥AC，垂足分别为D、E，BF为腰AC上的高．如图①，当点P在边BC上时，我们可得如下推理：∵S△ABC＝S△ABP＋S△ACP∴12AC▪BF＝12AB▪PD＋12AC▪PE∵AB＝AC∴12AC▪BF＝12AC▪（PD＋PE）∴BF＝PD＋PE（1）【运用】如图②，在上例的条件下，当点P运动到BC的延长线上时，试探究BF、PD、PE之间的关系，并说明理由．（2）【迁移】如图③，点P是等边△ABC内部一点，作PD⊥AB、PE⊥BC、PF⊥AC，垂足分别为D、E、F，若PD＝1，PE＝2，PF＝4．求△ABC的边长.（3）【拓展】若点P是等边△ABC所在平面内一点，且点P到三边所在直线的距离分别为2、3、6．请直接写出等边△ABC的高的所有可能值．BPECFDA（图③）ECFPBDAM（图②）ECFDBAP（图①）72018－2019学年度第一学期期中学情分析样题2八年级数学参考答案及评分标准说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．一、选择题（本大题共8小题，每小题2分，共16分）二、填空题（本大题共10小题，每小题2分，共20分）9．＞10．2π11．2.510512．413．不唯一：∠A＝∠F（AC∥EF、BC＝DE、∠ABC＝∠FDE＝90°等）14．215．916．217．218．40°或20°三、解答题（本大题共8小题，共64分）19．（本题8分）(1)解：4x2＝9，································································································1分x2＝94，·································································································2分x＝±32．································································································4分(2)解：x＋1＝－3，···························································································3分x＝－4．·······························································································4分20．(本题6分)证明：∵BD、CA分别是∠ABC、∠DCB的角平分线，∴∠DBC＝12∠ABC，∠ACB＝12∠DCB··························································1分∵∠ABC＝∠DCB∴∠DBC＝∠ACB·······················································································2分在△ABC和△DCB中∠ABC＝∠DCBBC＝BC∠DBC＝∠ACB∴△ABC≌△DCB·························································································5分∴AB＝DC··································································································6分21．（本题7分）证明：∵AB＝AC∴∠ABC＝∠ACB·························································································1分在△BDE和△CEF中BE＝CF∠ABC＝∠ACBBD＝CE∴△BDE≌△CEF·······························································································4分∴∠BDE＝∠CEF·······························································································5分∵∠ABC＋∠BDE＋∠DEB＝∠DEF＋∠CEF＋∠DEB＝180°·······································6分∴∠ABC＝∠DEF题号12345678答案CBDDADBA8∵∠ABC＝∠ACB∴∠ABC＝∠ACB＝∠DEF··················································································7分22．（本题7分）证明：∵AD∥EB∴∠A＝∠B·······························································································1分在△ADC和△BCE中AC＝BE∠A＝∠BAD＝BC∴△ADC≌△BCE····························································································4分∴DC＝CE······································································································5分∵CF平分∠DCE∴CF⊥DE．···································································································7分