初二数学试题

ACBO（7题图）初二数学试题（总分：120分考试时间：120分钟）一、选择题（每小题4分，共32分）1．在平面直角坐标系中，点P（2，–3）在第（）象限．A．一B．二C．三D．四2．下列命题中，真命题是（）A．有两个角相等的梯形是等腰梯形B．对角线互相平分的梯形是等腰梯形C．梯形的两条对角线相等D．对角线相等的梯形是等腰梯形3．若一个图形绕着一个定点旋转一个角（0180）后能够与原来图形重合，那么这个图形叫做旋转对称图形．例如：等边三角形绕着它的中心旋转120（如图），能够与原来的等边三角形重合，因而等边三角形是旋转对称图形．下面①—④四个图形中，旋转对称图形的个数有（）①②③④A．1个B．2个C．3个D．4个4．某产品的生产流水线每小时可生产100件产品，生产前没有产品积压，生产3小时后停止生产并安排工人装箱．若每小时装产品150件，未装箱的产品数值（y）是时间（t）的函数，那么这个函数的大致图象是图中（）ABCD5．如图，直线y=kx的图象如左图所示，则直线ykxk的大致图象是（）6．若关于x的方程233xkxx有增根，则k=（）A．1B．0C．–2D．–37．如图，在锐角△ABC中，50A，AB、AC两边的中垂线相交于点O，则∠BOC的度数为（）xyOy=kxxyOAxyOBxyOCxyOD（5题图）A．50°B．75°C．100°D．115°8．已知坐标平面上的机器人接受指令“[a，A]”（00180aA，）后的行动结果为：在原地顺时针旋转A后，再向面对的方向沿直线行走a个单位．若机器人的位置在原点面向y轴的负半轴，则它完成一次指令[2，60]后，所在位置的坐标为（）A．(13)，B．(31)，C．(31)，D．(13)，二、填空题（每小题3分，共24分）9．函数2yx的自变量x的取值范围是______________．10．如图，已知函数yaxb和ykx的图象交于点P，则根据图象可得关于yaxbykx的二元一次方程组的解为______________．11．已知菱形的边长为5cm，一条对角线长为6cm，则菱形的面积为______________．12．已知：1122272aabbabbaba，则______________．13．已知，如图，在矩形ABCD中，AC、BD交于点O，AE⊥BD于点E，且BE∶DE=1∶3，AB=6cm，则AC=______________．14．已知2222222()2()240ababab，则______________．15．如图，梯形ABCD中，AD∥BC，6030BC，，AD=2，BC=8，则CD=______________．16．在平面直角坐标系中，O是坐标原点，已知点A的坐标为（2，2），请你在y轴上找出点B，使△AOB为等腰三角形，则符合条件的点B共有______________个．三、解答题（共64分）17．分解因式：（每小题5分，共10分）(1)3221632xxx(2)222222()4abcab18．化简求值．（第(1)小题5分，第(2)小题6分，共11分）OABCDE（13题图）ABCD（15题图）xyy=ax+b20–3y=kx（10题图）(1)35(2)22xxxx(2)已知221121()21xxxxxxxx，求的值．19．(8分)作图题如图，方格纸中的每个小方格都是边长为1个单位的正方形，在建立平面直角坐标系后，△ABC的顶点均在格点上，点C的坐标为（4，–1）．(1)把△ABC向左平移6个单位后得到对应的△A1B1C1，画出△A1B1C1．(2)以原点O为对称中心，再画出与△ABC关于原点O对称的△A2B2C2．(3)图中点B1的坐标为__________________；点B2的坐标为___________________；点B1、B2之间的距离为____________．20．(8分)如图，正方形ABCD的边长为1cm，AC为对角线，AE平分∠BAC，交BC于点E，EF⊥AC于点F，求BE的长．ABCDEF21．(9分)甲、乙两人骑自行车同时从学校出发，沿同一路线去博物馆。甲行驶20分钟因事耽误一会儿，事后继续按原速行驶。下图表示甲、乙二人骑自行车行驶的路程y（千米）随时间x（分）变化的图象（全程），根据图象回答下列问题：(1)求线段OD、BC的解析式；(2)乙比甲晚多长时间到达博物馆？(3)甲因事耽误了多长时间？22．(9分)如图，在□ABCD中，E、F分别在AB、CD上，且满足AF=AD，CE=CB，60AFD．(1)证明：四边形AECF是平行四边形．(2)若去掉已知条件中的“60AFD”，而加上“EF平分∠AFC”，试判断四边形AECF的形状，并加以证明．ABCDFE23．(9分)已知：一次函数ykxb的图象与正比例函数ymx的图象相交于点P（a，4），且这两个函数的图象与y轴所围成的三角形面积为6．正比例函数ymx的图象与直线4250xy平行．(1)求a、m的值；(2)求一次函数的解析式．初二数学试题参考答案一、选择题（每小题4分，共32分）题号12345678选项DDCAADCB二、填空题（每小题3分，共24分）9．2x10．32xy11．24cm212．4313．12cm14．615．3316．4三、解答题（共64分）17．(1)解：原式22(816)xxx····································································2分22(4)xx············································································5分(2)解：原式222222(2)(2)abcababcab··········································2分2222[()][()]abcabc························································3分()()()()abcabcabcabc·········································5分18．(1)解：原式3(2)(2)522xxxxx···························································1分3(3)(3)22xxxxx································································3分322(3)(3)xxxxx13x·················································································5分(2)解：原式211[](1)(1)xxxxxx22(1)(1)1(1)xxxxxx·······························································2分21(1)xxx··········································································3分21(1)x·············································································4分当211212(211)x时，原式···········································6分19．解：(1)图略····························································································2分(2)图略····························································································4分(3)点B1的坐标为（14，）································································5分点B2的坐标为（54，）··································································6分点B1、B2之间的距离为22(15)(44)45································8分20．解：∵四边形ABCD为正方形∴9045BACB，，AB=BC=1cm∵EF⊥AC，∴90EFAEFC······················································1分∴△EFC是等腰直角三角形∴EF=FC························································································2分在△ABE和△AFE中，∵90BAEFAEBEFAAEAE∴△ABE≌△AFE（AAS）···································································5分∴AB=AF=1cm，BE=EF在Rt△ABC中，2222112cmACABBC·································7分∴(21)cmFCACAF∴(21)cmBEFC······································································8分21．解：(1)设线段OD的解析式为1ykx当x=60时，y=10∴10=60k1∴116k∴16yx·····················································································1分设线段BC的解析式为2ykxb∵当x=60时，y=10，∴21060kb∵当x=80时，y=15，∴21580kb解得：2154kb，∴154yx············································3分(2)当y=15时，115909080106xx，，（分钟）∴乙比甲晚10分钟达到博物馆·······················································6分(3)15100.257960v甲（千米/分）150.2560（分钟）80–60=20（分钟）∴甲因事耽误了20分钟。·····························································9分22．证明：(1)∵四边形ABCD是平行四边形∴AB//CD，∠B=∠D∴∠AFD=∠EAF，∵60AFD∴60EAF··························1分∵AF=AD∴△ADF是等边三角形∴60D，60B······································