北师大版数学八年级上册期末模拟试卷02（含答案）

这是一份北师大版数学八年级上册期末模拟试卷02（含答案），共11页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
北师大版数学八年级上册期末模拟试卷一、选择题1. 25的平方根是A．5    B．－5    C．±    D．±52．下列图形中，是中心对称图形的是3．某射击小组有20人，教练根据他们某次射击的数据绘制成如图所示的统计图，则这组数  据的众数和中位数分别是A. 7,  7      B. 8,  7.5     C. 7,  7.5      D. 8,  6.54．如图，两个较大正方形的面积分别为225、289，则字母A所代表的正方形的面积为   A．4    B．8      C．16    D．645．化简÷的结果是   A．    B.     C.         D. 2(x＋1)6.不等式组的解集在数轴上表示为   7．如果关于x的不等式（a＋1）x＞a＋1的解集为x＜1，则a的取值范围是    A.a＜0    B.a＜－1    C.a＞1    D.a＞－18．实数a在数轴上的位置如图所示，则＋化简后为A．  7    B．  －7      C．2a－15       D．无法确定9．若方程＋＝那么A、B的值    A.2,1    B.1,2    C.1,1    D.－1, －110．已知长方形ABCD中，AB＝3，AD＝9，将此长方形折叠，使点B与点D重合，折痕为EF,则△ABE的面积为A．6    B．8    C．10    D．1211．如图，△ABC绕点A顺时针旋转45°得到△AB′C′，若∠BAC＝90°,AB＝AC＝，则图中阴影部分的面积等于A．2-      B.1       C．       D. -l12．如图，△ABC中，∠ACB＝90°，AC＞BC，分别以△ABC的边AB、BC、CA为一边内△ABC外作正方形ABDE、BCMN、CAFG，连接EF、GM、ND，设△AEF、△BND、△CGM的面积分别为S1、S2、S3，则下列结论正确的是    A．Sl＝S2＝S3    B．S1＝S2＜S3      C．Sl＝S3＜S2    D．S2＝S3＜Sl二、填空题13．计算：一＝______________.14．分解因式：a2－6a＋9＝______________.15．当x＝______时，分式的值为0.16.已知a＋b＝3，a2b＋ab2＝1，则ab＝____________·17．如图，一只蚂蚁沿着边长为2的正方体表面从点4出发，经过3个面爬到点B，如果它运动的路径是最短的，则最短路径的是长为__________________.18．如图，在四边形ABCD中，AD＝4，CD＝3, ∠ABC＝∠ACB＝∠ADC＝45°，则BD的长为______________. 三、解答题19．计算：(1)－3                       (2)÷   20．(1)因式分解：m3n―9mn.   (2)求不等式≤的正整数解   21．(1)解方程：＝2＋    (2)解不等式组，并把解集在数轴上表示出来     22.(1)如图1，△ABC是边长为2的等边三角形，将△ABC沿直线BC向右平移，使点B与点C重合，得到△DCE，连接BD，交AC于点F．求线段BD的长．  (2)一次环保知识竞赛共有25道题，规定答对一道题得4分，答错或不答一道题扣1分.在这次竞赛中，小明被评为优秀（85分或85分以上），小明至少答对了几道题？      23．济南与北京两地相距480千米，乘坐高铁列车比乘坐普通快车能提前4小时到达．已知高铁列车的平均行驶速度是普通快车的3倍，求高铁列车的平均行驶速度.       24．先化简再求值：(x＋1一)×，其中x＝－       25．某公司需招聘一名员工，对应聘者甲、乙、丙从笔试、面试、体能三个方面进行量化考核，甲、乙、丙各项得分如下表： 笔试面试体能甲837990乙858075丙809073    (1)根据三项得分的平均分，从高到低确定三名应聘者的排名顺序．(2)该公司规定：笔试，面试、体能得分分别不得低于80分，80分，70分，并按60%，30%，10%的比例计入总分，根据规定，请你说明谁将被录用．     26．如图，在四边形ABCD中，对角线AC，BD交于点E，∠BAC＝90°，∠CED＝45°，∠DCE＝30°，DE＝，BE＝2．  (1)求CD的长：(2)求四边形ABCD的面积                 27．已知，点D是等边△ABC内的任一点，连接OA，OB，OC.    (1)如图1，己知∠AOB＝150°，∠BOC＝120°，将△BOC绕点C按顺时针方向旋转60°得△ADC．①∠DAO的度数是_______________    ②用等式表示线段OA，OB，OC之间的数量关系，并证明；    (2)设∠AOB＝α，∠BOC＝β.    ①当α，β满足什么关系时，OA＋OB＋OC有最小值？请在图2中画出符合条件的图形，并说明理由；    ②若等边△ABC的边长为1，直接写出OA＋OB＋OC的最小值．               参考答案一、选择题题号123456789101112答案DBCDCBBACADA二、填空题13.14. ( a－3) 215. －316. 17. 18. 三．解答题：19.解：(1) =···························································1分= ·······················································2分   =1··························································3分(2) =···························································5分=···························································6分20.解：(1) m3n－9mn．=···························································1分=···························································2分=···························································3分 (2)解：3(x－2)≤2(7－x) ·····································4分3x－6≤14－2x5x≤20x≤4·············································5分∴这个不等式的正整数解为1、2、3、4.·····························6分21.(1) ······················································1分······················································2分 ···················································3分经检验是增根，原方程无解···································4分（2），解：解不等式①得：x＞1，·······································5分解不等式②得：x＞5，··········································6分∴不等式组的解集为x＞5，·······································7分在数轴上表示不等式组的解集为：．··························································8分22. (1)解：∵正△ABC沿直线BC向右平移得到正△DCE∴ BE=2BC=4, BC=CD,DE=AC=2,∠E=∠ACB=∠DCE=∠ABC=60°·············2分∴∠DBE=∠DCE =30°···········································3分∴∠BDE=90°·················································4分在Rt△BDE中，由勾股定理得····························································5分(2)解：设小明答对了x道题，·····································6分4x－(25－x) ≥85··············································8分x≥22···············································9分所以，小明至少答对了22道题.·····································10分23. 解：设普通快车的速度为xkm/h，由题意得：······················1分····························································3分=4··························································4分x=80························································5分经检验x=80是原分式方程的解·····································6分3x=3×80=240 ················································7分答：高铁列车的平均行驶速度是240km/h．···························8分24.解：=······················································1分=······················································2分=······················································3分=······················································4分当=时·······················································5分原式==······················································6分25. 解：(1) =（83+79+90）÷3=84，=（85+80+75）÷3=80，=（80+90+73）÷3=81．········································ 3分从高到低确定三名应聘者的排名顺序为：甲，丙，乙；··················4分(2)∵该公司规定：笔试，面试、体能得分分别不得低于80分，80分，70分，∴甲淘汰，···················································· 5分乙成绩=85×60%+80×30%+75×10%=82.5，···························7分丙成绩=80×60%+90×30%+73×10%=82.3，··························· 9分∴乙将被录取．  ·············································10分26解： (1)过点D作DH⊥AC，······································1分∵∠CED=45°，∴∠EDH=45°，∴∠HED=∠EDH，∴EH=DH，····················································3分∵EH2+DH2=DE2，DE=，∴EH2=1，  ∴EH=DH=1，··················································5分又∵∠DCE=30°，∠DHC=90°，∴DC=2 ······················································6分(2)∵在Rt△DHC中，············································7分∴12+HC2=22，∴HC=，······················································8分∵∠AEB=∠CED=45°，∠BAC=90°，BE=2，∴AB=AE=2，··················································9分∴AC=2+1+=3+，···············································10分∴S四边形ABCD=S△BAC+S△DAC···············································11分=×2×（3+）+×1×（3+）=··························································12分   27. 解：（1）①90°. ········································2分 ②线段OA，OB，OC之间的数量关系是.  ·····························3分如图1，连接OD.···············································4分∵△BOC绕点C按顺时针方向旋转60°得△ADC，∴△ADC≌△BOC，∠OCD=60°.∴CD = OC，∠ADC =∠BOC=120°，AD= OB.∴△OCD是等边三角形，·········································5分∴OC=OD=CD，∠COD=∠CDO=60°，∵∠AOB=150°，∠BOC=120°，∴∠AOC=90°，∴∠AOD=30°，∠ADO=60°.∴∠DAO=90°.················································6分在Rt△ADO中，∠DAO=90°，∴.∴.·························································7分        (2)①如图2，当α=β=120°时，OA+OB+OC有最小值. ···················8分作图如图2，··················································9分如图2，将△AOC绕点C按顺时针方向旋转60°得△A’O’C，连接OO’. ∴△A′O′C≌△AOC，∠OCO′=∠ACA′=60°.∴O′C= OC, O′A′ = OA，A′C = BC, ∠A′O′C =∠AOC.∴△OC O′是等边三角形.·······································10分∴OC= O′C = OO′，∠COO′=∠CO′O=60°.∵∠AOB=∠BOC=120°，∴∠AOC =∠A′O′C=120°.∴∠BOO′=∠OO′A′=180°.∴四点B，O，O′，A′共线.∴OA+OB+OC= O′A′ +OB+OO′ =BA′ 时值最小.···················· 11分②当等边△ABC的边长为1时，OA+OB+OC的最小值A′B=. ················12分