南师附中集团2017-2018学年第一学期八年级数学期中试题(含答案)

1南京清江花苑严老师2017－2018学年度上学期期中考试数学学科试题（考试时间100分钟，试卷总分100分）一、选择题（本大题共6小题，每小题2分，共12分）1．剪纸是我国最普及的民间艺术．下列剪纸作品中，不是轴对称图形的是（）A．B．C．D．2．下列实数中，为有理数的是（）A．πB．5C．39D．03．下列长度的三条线段能组成直角三角形的是（）A．3，4，5B．5，13，15C．7，14，25D．8，12，204．△ABC是锐角三角形，若AB＝2，∠A＝45°，则AC的长可能是（）A．1B．2C．3D．45．从一个等腰三角形纸片的顶角顶点出发，能将其剪成两个等腰三角形纸片，则原等腰三角形纸片的顶角等于（）A．90°B．72°C．108°D．90°或108°6．如图1，在△ABC中，∠C＝90°，M为AB中点．将△ACM沿CM翻折，得到△DCM(如图2)，P为CD上一点，再将△DMP沿MP翻折，使得D与B重合(如图3)，给出下列四个命题：①BP∥AC；②△PBC≌△PMC；③PC⊥BM；④∠BPC＝∠BMC．其中真命题的个数是（）A．1B．2C．3D．4FABCMDABCMDPABCM（图1）（图2）（图3）（第6题）2南京清江花苑严老师二、填空题（本大题共10小题，每空2分，共20分）7．81的平方根是．8．小明体重为48.96kg，用四舍五入法将48.96kg精确到0.1kg可得近似值kg．9．若等边三角形的边长是2cm，则它的高为cm．10．如图，在△ABC中，AD是它的角平分线，若S△ABD：S△ACD＝3：2，则AB：AC＝．11．如图，在△ABC中，∠ACB＝90°，AC＝4，BC＝3，点M在AB上，且∠ACM＝∠BAC，则CM的长为．12．如图，OP＝1，过P作PQ1⊥OP且PQ1＝1，以O为圆心，OQ1为半径画弧，交OP的延长线于P1；再过P1作P1Q2⊥OP1且P1Q2＝1，以O为圆心，OQ2为半径画弧，交OP的延长线于P2，则OP2的长为．13．比较大小：392（填“＞”，“＜”或“＝”号）．14．如图，△ABC≌△ADE，且E在BC上．若∠DEA＝80°，则∠BED的度数为．15．如图，在△ABC中，∠BAC＝90°，分别以AC，BC为边长，在三角形外作正方形ACFG和正方形BCED，AH⊥DE，分别交DE，BC于点H，P．若BP＝2，CP＝4，则正方形ACFG的面积为．16．如图，在△ABC中，∠BAC＝90°，AC＝6，AB＝8，点P是BC上一动点，PQ⊥BC，△A'B'C'与△ABC关于PQ成轴对称，若重合部分是等腰三角形，则BP的长应该满足的条件是．ABCDE（第14题）ABCDEFGH（第15题）PABCA'B'C'PQ（第16题）ABCD（第10题）ABCM（第11题）OPP1P2Q1Q2（第12题）3南京清江花苑严老师三、解答题（本大题共10题，共68分）17．（6分）求下列各式中的x（1）(2x)2＝4；（2）x3－4＝－12．18．（8分）计算（1）38＋0－14；（2）||2－3＋(π－3)0＋||1－3．19．（6分）如图，AB⊥BD，CD⊥BD，AD＝BC．求证：AD∥BC．20．（6分）如图，P为∠AOB的角平分线上的一点，PH⊥OA，垂直为H．M为PH上一点，MN⊥OB，与OC，OB的交点分别为Q，N．求证：MP＝MQ．ABCOHMNQP（第20题）ABCD（第19题）4南京清江花苑严老师21．（6分）同学们在学习“探索三角形全等的条件”时，发现“有两边和其中一边的对角对应相等的两个三角形全等”是假命题．说明一个命题是假命题，只需要画出反例即可．如图，已知△ABC和A'B'，A'B'＝AB．请用直尺和圆规在图（2）中作△A'B'C'，使得∠A'＝∠A，B'C'＝BC，且△A'B'C'与△ABC不全等．（保留作图痕迹，不写作法）22．（6分）如图，在△ABC中，∠C＝90º，AB＝8，点D是BC上一点，AD＝BD＝5，求CD的长．23．（6分）阅读理解：求107的近似值．解：设107＝10＋x，其中0＜x＜1，则107＝(10＋x)2，即107＝100＋20x＋x2．因为0＜x＜1，所以0＜x2＜1，所以107≈100＋20x，解之得x≈0.35，即107的近似值为10.35．理解应用：利用上面的方法求97的近似值（结果精确到0.01）．ABCABCA'B'（第21题）（图1）（图2）ABCD（第22题）5南京清江花苑严老师24．（8分）【经典回顾】(教材P25)如图，PC＝PD，QC＝QD，PQ，CD相交于点E．求证：PQ⊥CD．【数学思考】已知三个点A，B和C，只允许用圆规作点D，使得C，D两点关于AB所在的直线对称．25．（8分）如图，等腰直角三角形ABC中，点D在斜边BC上，以AD为直角边作等腰直角三角形ADE．（1）求证：△ABD≌△ACE；（2）求证：BD2＋CD2＝2AD2．ABCDE（第25题）ABCPDCQE（第24题）6南京清江花苑严老师26．（8分）已知Rt△ABC≌Rt△DEF，∠BAC＝90°，AB＝3，BC＝5，两个三角形按图1所示的位置放置，点B，F重合，且点E，B，F，C在同一条直线上．如图2，现将△DEF沿直线BC以每秒1个单位向右平移，当F点与C点重合时，运动停止．设运动时间为t秒．（1）若t＝2时，则CF的长是；（2）当t为何值时，△ADB是等腰三角形．ABCDE（F）（图1）ABCDEF（图2）ABCDEF（备用图）7南京清江花苑严老师八年级数学参考答案一、选择题（本大题共6小题，每小题2分，共12分）题号123456答案BDABDB二、填空题（本大题共10小题，每小题2分，共20分）7．9；8．49.0；9．3；10．3:2；11．52；12．3；13．；14．20；15．24；16．BP05或.BP6410．三、解答题（本大题共10小题，共68分）17．(本题6分)解：（1）(2x)2＝42x＝±2···················································································2分x＝±1x＝1，x2＝－1．······································································3分（2）x3－4＝－12x3＝－8··················································································4分x＝－2···················································································6分18．(本题8分)解：（1）38＋0－14＝2＋0－12···················································································3分＝32····························································································4分（2）||2－3＋(π－3)0＋||1－3＝2－3＋1＋3－1······································································7分＝2．·························································································8分19．(本题6分)证明：∵AB⊥BD，CD⊥BD，∴∠ABD＝∠CDB＝90°．在Rt△ABD和Rt△CDB中，AD＝CB，BD＝DB．∴Rt△ABD≌Rt△CDB(HL)．·······························································4分∴∠ADB＝∠CBD．··········································································5分ABCD8南京清江花苑严老师∴AD∥BC．····················································································6分20．(本题6分)证明：∵P为∠AOB的角平分线上的一点，∴∠AOC＝∠BOC．∵PH⊥OA，MN⊥OB，∴∠PHO＝∠MNO＝90°．∴∠AOC＋∠HPO＝∠BOC＋∠NQO．∴∠HPO＝∠NQO．·······································································4分又∵∠MQP＝∠NQO．∴∠HPO＝∠MQP．∴MP＝MQ．················································································6分21．(本题6分)（1）作出∠A'＝∠A．········································································3分（2）作出B'C'＝BC．·········································································6分则△A'B'C'是所求作的三角形．22．(本题6分)解：∵∠C＝90º，∴△ACD和△ACB是直角三角形．在Rt△ACD中，∵AC2＋CD2＝AD2，∴AC2＝AD2－CD2．在Rt△ACB中，∵AC2＋CB2＝AB2，∴AC2＝AB2－CB2．∴AD2－CD2＝AB2－CB2．设CD＝x，则BC＝5＋x，52－x2＝82－(5＋x)2．············································································4分∴x＝1.4．即CD的长为1.4．···············································································6分23．(本题6分)解：设97＝10－x，其中0＜x＜1，则97＝(10－x)2，即97＝100－20x＋x2．·············2分因为0＜x＜1，所以0＜x2＜1，所以97≈100－20x，··············································································4分解之得x≈0.15，即97的近似值为9.85．···················································6分(设97＝9＋x，求出的近似值为9.89也给满分．)24．(本题8分)ABCOHMNQPCA'B'C'ABCD9南京清江花苑严老师（1）证明：∵PC＝PD，∴P在CD的垂直平分线上．····························································2分∵QC＝QD，∴Q在CD的垂直平分线上．···························································4分∴PQ是CD的垂直平分线．∴PQ⊥CD．······························································