泸州市八年级下期末抽考试题

1泸州市八年级下期末抽考试题一、选择题：（本大题共12个小题，每小题3分，共36分）1．函数1yx中自变量x的取值范围是A．1xB.1x≥C.1xD.1x≤2.下列四个二次根式中是最简二次根式的是A．8B．12C．4D．1a3.如图，为使人字型梯子使用时更牢固，需要在两边侧面中点处固定一根木条，若人字型梯子的两着地端点A、B间距离为1米，则需固定的木条的长（忽略接头）是A．0.4米B．0.6米C．0.5米D．不能确定4.若点(2,)Am在正比例函数12yx的图象上，则m的值是A．1B．14C．14D．15．在△ABC中，90Co，1AC，3BC，则AB的长是A．2B．4C．2D．16．如图，□ABCD的对角线相交于点O，20ACBD，8AB，则AOB△的周长为A．28B．18C．18D．247．某射击队对要选派一名队员参赛，对甲、乙、丙、丁四人进行射击测试，每人10次射击的平均成绩恰好都是9.5环，方差分别是20.90S甲、21.22S乙、20.43S丙、21.68S丁，根据本次射击测试成绩，教练应该选派的队员是A．甲B．乙C．丙D．丁8．矩形的一条对角线长为4，两对角线的一个交角为120o，则该矩形的周长为A．443B．223C．43D．89．菱形具有而平行四边形不一定具有的性质是A．两组对边分别平行B．对角线互相平分C．两组对角分别相等D．对角线互相垂直10．下列曲线不能表示y是x的函数的图象的是2A．B.C.D.11．如图，直线111ykxb与222ykxb相交于点(1,2)，则下列结论中不正确的是A．方程组111222ykxbykxb的解是12xyB．当1x时，12yyC．当1x时，21yyD．12120kkbb12．如图，正方形ABCO和正方形DEFO的顶点A、E、O在同一直线l上，且22EF，AB=6，给出下列结论：①10AE；②45CODo；③6COFS△；④217CFBD．其中正确的是A．①②③B．②③④C．①②④D．①③④二、填空题（每小题3分，共12分）13．因式分解：24m▲．14．已知等腰三角形的周长为16，则腰长y关于底边长x的函数关系式为▲；15．一个平行四边形的一条边长是9，两条对角线的长分别是12和65，则这个四边形的面积为▲．16．在△ABC中，90Co，AB边上的高为2，则2211ACBC的值是▲．三、解答题（52分）17．（本题满分6分）计算：2021()(2)22．18．（本题满分6分）计算：127(243)65．第11题图第12题图第16题图319．（本题满分6分）化简：21(2)1xxxxx.20．（本题满分7分）酒城家园装饰公司共20名员工，员工基本工资的平均数为4100元．现就其各岗位每人的基本工资情况和各岗位人数，绘制了下列尚不完整的统计图表，根据统计图表，回答下列问题：各岗位每人的基本工资情况统计表（1）将各岗位人数统计图补充完整；（2）求该公司监理每人的基本工资；（3）该公司所有员工基本工资的中位数是元，众数是元；你认为用基本工资的平均数和中位数来代表该公司员工基本工资的一般水平，哪一个更恰当？请说明理由．21．（本题满分7分）如图，每个小正方形的边长都为1.（1）求四边形ABCD的面积；（2）判断ABC△的形状并证明．22．（本题满分9分）如图，在平行四边形ABCD中，8AB，12BC，AC的垂直平分线交AD于点E，交AC于点O，交BC于点F.（1）求CDE△的周长；（2）猜想四边形AECF的形状，并证明你的猜想．岗位经理设计师监理施工员杂工基本工资20000600030002000第22题图第21题图BCDA423．（本题满分11分）如图，在平面直角坐标系xOy中，矩形ABCO顶点A、C分别在y轴和x轴上，已知3cmOA，5cmOB．（1）求直线OB的解析式；（2）若射线OB上有一点(,)Pxy，OCP△面积为S，求S与x的函数关系式，并求23cmS时，点P的坐标；（3）在（2）的条件下，在x轴上找一点Q，使PQBQ最小，求点Q的坐标．第23题图QPyxOACB5八年级数学抽考参考答案及评分意见一．选择题：题号123456789101112答案BDCAABCCDBDA二．填空题：13．(2)(2)mm；14.182yx；15.365；16.12．三．解答题：17．解：原式11144·····················································································································5分1．·····························································································································6分18．解：原式133(266)25····································································································3分533625·················································································································4分922．·······················································································································6分19．解：原式2221()1xxxxxx····································································································1分22211xxxxx········································································································2分211xxxx················································································································3分(1)(1)1xxxxx·······································································································5分1x．························································································································6分20．解：（1）如图；………………………………1分（2）该公司监理每人的基本工资为：[4100×20－(20000＋6000×2＋3000×8＋2000×5)]÷4=4000；…………………………………3分（3）3000；3000；………………………5分用基本工资的中位数来代表该公司员工基本工资的一般水平更恰当.因为该公司有一半以上的人员(13人)的工资未超过3000元.…………………………………7分21．解：（1）四边形ABCD的面积为:168(24266323)262；………2分（2）ABC△是直角三角形，证明如下:第21题图BCDA6因为224225AB，……………………3分2262210BC，………………………4分228165AC，…………………………5分222ABBCAB,………………………………6分所以ABC△是90Bo的直角三角形．……7分22．解：（1）因为AC的垂直平分线交AD于点E，所以AEEC，…………………………………………1分所以CDE△的周长为：20CD+CE+DE=AD+CD=AB+BC=，……………3分（2）四边形AECF是菱形，证明如下：·········································································4分因为ABCD是平行四边形，所以AD//BC，·············································································································5分所以EACACF，AEFCFE，又AOCO，···········································6分所以AOECOF△△，································································································7分所以AEFC，又//AEFC，····················································································8分所以ABCD是平行四边形，又AEEC，所以四边形AECF是菱形．·······················································································9分23．解：（1）因为ABO△是直角三角形，3cmOA，5cmOB，所以224cmABOBOA，······························································1分所以点B的坐标为(4,3)，····································································2分所以直线OB的解析式为：34yx；······················································4分（2）S与x的函数关系式为：1334242Sxx，·········································5分当23cmS时，由332x得：2x，·····················································6分所以点P的坐标为3(2,)2；···································································7分（3）设点P关于x轴的对称点为3(2,)2P，因为PQBQPQBQPB≥，当且仅当B、Q、P在同一直线上时取等号，····························································8分设经过点P、B的一次函数解析式为：第22题图7ykxb，则：32243kbkb，······················································