推荐盐城市盐都区2019-2020八年级上期末数学试题及答案

2019-2020学年度第一学期期末质量检测八年级数学试卷（时间：100分钟；满分：120分）号题一二三总分合分人1～8189～2719～分得一、选择题（本大题共8小题，每小题有且只有一个答案正确，请把你认为正确的答案前的字母填入下表相应的空格内，每小题3分，共24分）号题3600195012345678案答1．2的算术平方根是·································································································（▲）A．B．2C．±D．±2222．2013年12月2日，“嫦娥三号”从西昌卫星发射中心发射升空，并于12月14日在月球上成功实施软着陆．月球距离地球平均为384401000米，用四舍五入法取近似值，精确到1000000米，并用科学计数法表示，其结果是··················································································（▲）7788米10．3.810×米DB．3.8×10×米C．3.84A．3.84×10米2221中，无理数的个数有·······················?·················（▲）3．在实数：，，，33.π7B．2个C．3个D．4个A．1个4．在平面直角坐标系中，点P（3，?5）在···························································（▲）AB．第二象限C．第三象限D．第一象限．第四象限5．如图是一个风筝设计图，其主体部分（四边形ABCD）关于BD所在的直线对称，AC与BD相交于点O，且AB≠AD，则下列判断不正确的是······························································（▲）A．△ABD≌△CBDB．△ABC是等边三角形D．△AOD≌△C．△AOB≌△COBCODADOBC第5题第7题b?kx·············0b0＝＜6．一次函数，当，＜时，它的图象大致为·············▲（·）ykyyyyOOOxxxOxCDBA．如图，正方形网格中，已有两个小正方形被涂黑，再将图中其余小正方形涂黑一个，使整个被涂黑71）（▲·······························的图案构成一个轴对称图形，那么涂法共有························种D．6C．5种A．3种B．4种．某物流公司的快递车和货车同时从甲地出发，以各自的速度匀速向乙地行驶，快递车到达乙地后卸83，立即按原路以另一速度返回，直至与货车相遇．已知货车的速度为h完物品再另装货物共用44）之间的函数图象如图所示，现有以下km）与货车行驶时间（h60km/h，两车之间的距离（yxkm/y个结论：A①快递车到达乙地时两车相距120km；120B300km；②甲、乙两地之间的距离为C；③快递车从甲地到乙地的速度为100km/hO31hx/434）．的坐标为（3，75④图中点B8题第4）··（▲···············································其中，正确的结论有············································4个D．3BA．1个．2C．个20分）二、填空题（本大题共10小题，每小题2分，共）到轴的距离是＿＿＿＿＿．9．点P（，3?2?x3）“＝”、＜”＿＿＿＿＿7．（填“＞”、．比较大小：10“4，则它顶角的度数为＿＿＿＿＿．11．已知等腰三角形的一个外角是80°，则斜边上的中线长为＿＿＿＿＿．12．若直角三角形的两条直角边的长分别是6和8ABDCABBC10cmBD6cm13ABCC90AD的距，点到直线＝平分∠．如图，在△，中，∠，那么＝＝°，cm．离是＿＿＿＿＿AEAyDFF1DGH-13421Ox-1BBCAGEBCDC18题题题13题第16第第17第个单位长度到个单位长度，再向下跳2A（?1，0）处向左跳214．在平面直角坐标系中，一青蛙从点的坐标为＿＿＿＿＿．点A′处，则点A′15．写出同时具备下列两个条件的一次函数关系式＿＿＿＿＿．（写出一个即可））．，?2）图像经过点（1（）随的增大而减小；（21yxBCAC、的垂直平分线分别交、16．如图，在△ABC中，AB的垂直平分线分别交ABBC于点D、E，AC．，若∠于点F、GBAC＝100°，则∠EAG＝＿＿＿＿＿°1ax??baxy，则关于的方程的解17．如图，已知直线＝＝＝＿＿＿＿＿．bxx，BCEACD和正三角形同侧分别作正三角形重合）AB上一动点（不与点A、B，在ABC18．如图，为线段；＝BD，连接GH．以下五个结论：①AE交于点，与交于点AE与BDF，AECD交于点GBD与CEH⑤∠GE＝AB②GH∥；③ADDH；④＝HB；AFD＝60°）（填序号即可，一定成立的有＿＿＿＿＿＿＿．76分，解答要求写出文字说明，证明过程或计算步骤）三、解答题（本大题共9小题，共）本题满分8分19．（22032（）计算：．；＝0的值：）求（19x4?2)1)(??8??(x2）近年来，江苏省实施“村村通”工程和农村医疗卫生改革，盐都区计划在张村、6分．（本题满分20，医疗站必须满足下P，张、李两村座落在两相交公路内（如图所示）李村之间建一座定点医疗站列条件：张村①使其到两公路的距离相等；②到张、李两村的距离也相等．李村请你利用尺规作图确定点的位置．P）（不写作法，保留作图痕迹米处，已知木杆原长8．（本题满分6分）如图，一木杆在离地某处断裂，木杆顶部落在离木杆底部21米，求木杆断裂处离地面多少米？16地面8米0，）三点．C（?12，本题满分22．（6分）在平面直角坐标系中，已知A（?15）、B（4，）、的坐标为＿＿＿＿＿，关于B轴的对称点B′关于原点（1）点AO的对称点A′的坐标为＿＿＿＿＿，点xC′的坐标为＿＿＿＿＿；关于点C轴的对称点yB、C′为顶点的△A′′C′的面积．′A2（）求以（1）中的点′、BAD23．（本题满分6分）如图，四边形ABCD是梯形，AD∥BC，∠A＝90°，BD＝CB，CE⊥BD，垂足为E．E（1）求证：△ABD≌△ECB；（2）若∠DBC＝50°，求∠DCE的度数．CB3EC⊥OA，ED⊥OB，垂足分别是是∠AOB的平分线上一点，C、D．本题满分24．（10分）如图，点EB求证：（1）∠EDC＝∠ECD；DOD；2）OC＝（E的垂直平分线．OE是线段CD3（）AOC）阅读下列一段文字，然后回答下列问题．25．（本题满分10分，）、N（，），则这两点间的距离可用下列公式计算：已知平面内两点M（yyxx221122????y?xy??x．MN＝212122????2??3?11＝＝，则这两点间的距离PQ（P．例如：已知（3，1）、Q1，?2）13特别地，如果两点M（，）、N（，）所在的直线与坐标轴重合或平行于坐标轴或垂直于坐yxxy2112x?xy?y．或标轴，那么这两点间的距离公式可简化为MN＝2121（1）已知A（1，2）、B（?2，?3），试求A、B两点间的距离；（2）已知A、B在平行于轴的同一条直线上，点A的横坐标为5，点B的横坐标为?1，试求A、B两x点间的距离；（3）已知△ABC的顶点坐标分别为A（0，4）、B（?1，2）、C（4，2），你能判定△ABC的形状吗？请说明理由．426．（本题满分12分）小华和爸爸上山游玩，爸爸乘电缆车，小华步行，两人相约在山顶的缆车终点会合．已知小华行走到缆车终点的路程是爸爸乘缆车到山顶的线路长的2倍，爸爸在小华出发后50min才乘上电缆车，电缆车的平均速度为180m/min．设小华出发（min）行走的路程为（m），yx图中的折线表示小华在整个行走过程中（m）与（min）之间的函数关系．yx（1）小华行走的总路程是＿＿＿＿＿m，他途中休息了＿＿＿＿＿min；（2）当50≤≤80时，求与的函数关系式；yxx（3）当爸爸到达缆车终点时，小华离缆车终点的路程是多少？y/mO305080x/min5，重合）不与B、C分）已知△ABC为等边三角形，点D为直线BC上的一动点（点D（27．本题满分12按逆时针方向排列），连接CE．为边作等边△ADE（顶点A、D、E以ADCD；CE＝CE，②AC＝＋BC（1）如图1，当点D在边上时，求证：①BD是否成立？若不成立，CDCEBC的延长线上且其他条件不变时，结论AC＝＋2（2）如图，当点D在边之间存在的数量关系，并说明理由；AC、CE、CD请写出CDAC、CE、，当点（3）如图3D在边BC的反向延长线上且其他条件不变时，补全图形，并直接写出之间存在的数量关系．EAAAECCCDBBDDB3图1图图267八年级数学参考答案及评分标准（阅卷前请认真校对，以防答案有误！）一、选择题（每小题3分，共24分）题12345678号答DCBBCDBA案二、填空题（每小题2分，共20分）10．＜．9．3．11．100°．12．5．14．（?3，?2）．1513．4．．答案不唯一，如＝等．y1??x18．20．①②④⑤17．4．．16．三、解答题（共76分）2＝9，····································································································19．（1）·················1分x492，······························································································＝····················2分x43．··············································································································4分＝±x2（2）原式＝1＋2＋2···········································································································3分＝5．···················································································································4分3扣1分；说明：第（1）题答案写成＝x2023的计算分别给1分．、、第（2）题1)?(82)(?20．作出线段垂直平分线，······································································································3分作出角平分线．·················································································································6分21．设木杆断裂处离地面米，由题意得··············································································1分x222．··3分＝·········································································································)x(16?8?x解得＝6····························································································································5分x答：木杆断裂处离地面6米．··························································································6分22．（1）（1，?5）；（4，?2）；（1，0）．·················································································3分115＝＝．·····················································分6（2）S1)5?(4??′C′′△AB2223．（1）∵AD∥BC，∴∠ADB＝∠EBC．∵CE⊥BD，∴∠BEC＝90°．∵∠A＝90°，∴∠A＝∠BEC．·················································································································1分在△ABD和△ECB中，8?A??BEC???ADB??EBC，·2分···········································································································??BD?CB?∴△ABD≌△ECB（AAS）．································································································3分（2）∵BD＝CB，∠DBC＝50°，11∴∠BDC＝＝＝65°．·····················································4分)50)?(180??(180???DBC22??65°＝25°．···························∴在Rt△CDE中，∠DCE＝90°·∠BDC＝90°···············6分24．（1）∵点E是∠AOB的平分线上一点，EC⊥OA，ED⊥OB，∴ED＝EC．·························································································································3分∴∠EDC＝∠ECD．············································································································4分（2）∵EC⊥OA，ED⊥OB，∴∠EDO＝∠ECO＝90°．··································································································5分由（1）知∠EDC＝∠ECD，??∠ECD，即∠ODC＝∠OCD．···········································∴∠EDO∠EDC＝∠ECO··6分∴OC＝OD．························································································································7分（3）∵OC＝OD，∠EOC＝∠EOD，∴OE⊥CD，OE平分CD，即OE是线段CD的垂直平分线．······································10分22????322??1?＝．·····································································25．（1）AB·＝······3分34（2）AB＝＝6．····································································································6分1)??5(（3）△ABC是直角三角形．······························································································7分2222????????2???0?11??442?2＝，＝理由：∵ABBC＝5＝，522????240?4??＝＝，AC20222222＝25．＝＝5＝∴AB25，＋ACBC20)?(5)(222．························································AB·＋AC·＝BC··················································9分∴∴△ABC是直角三角形．································································································10分26．（1）3600，20．·················································································································2分（2）当50≤≤80时，设与的函数关系式为＝，根据题意得···············3分yybkx?xx当＝50时，＝1950；当＝80时，＝3600．·····················································4分yyxx50k?b?1950?∴．?80k?b?3600?k?55?解得．················································································································6分?b??800?yy＝．········································∴与的函数关系式为······································7分80055x?x（3）缆车到山顶的路线长为3600÷2＝1800（m）．·····················································8分缆车到达终点所需时间为1800÷180＝10（min）．······················································9分爸爸到达缆车终点时，小华行走的时间为10＋50＝60（min）．······························10分?800＝2500．···························×＝，得5560＝60＝把代入··········11分yy800x?55x∴当爸爸到达缆车终点时，小华离缆车终点的路程是36002500＝1100（m）····12分?9都是等边三角形，△ADE（1）∵△ABC和27．．＝60°AE，∠BAC＝∠DAE∴AB＝AC＝BC，AD＝??分·1·····························∴∠BAC·∠CAD＝∠DAE·∠CAD，即∠BAD＝∠CAE．··············中，△ACE在△ABD和AC?AB??CAE?BAD??，??AE?AD?3分······························································（∴△ABD≌△ACESAS）．··································4分················································································