苏教版九年级下册数学试卷及答案

苏教版年级下册数学卷及UpdatedJack25,2020atam

九年级数摸底试没有比人更高的山，没有比脚更长的路。亲爱的同学们请相信自己，沉着应答，你一定能愉快地完成这次测试之旅，让我们一同走进这次测试吧。祝你成功！考生注：1．试卷含个大题，共25题2．题时，生务必按答要求在题纸规定的置上作，在草稿纸本试卷答题一律无效．3．第一、大题外，其各题如特别说明，必须在题纸的相应置上写证明或计算的主步骤．一、选题：（本大共，每题分，满分）【下列题的四个选中，有只有一个选是正确，选择正确的代号填涂在答题的相应位上．】1计算(

)

的结果是（）Aa

5

B

6

C．a

8

D．

9，2不等式组的解集是（）xA

B

C．

D．x3用换元法解分式方程

x，如果设xx

y，将原方程化为关于的整式方程，那么这个整式方程是（）Ay

2

y

B

2

C

2

y

D

2

4抛物线yx)2（，是常数）的顶点坐标是（）A(mn)

B(

C．(

D(5下列正多边形中，中心角等于内角的是（）A正六边形

B正五边形

C．正四边形

C．正三边形CD

aa6如图，已知ABCDEF那么下列结论正确的是（）A

BCDFCE

B

BCDFCECDBCC．EF

CDD．EFAF二、填题：（本大共，每题4分，满48分【请将果直线入答纸的应位置7分母有理化：．_______8方程的根是．9如果关于的方程

为常数）有两个相等的实数根，那么10已知函数f(x)

11

，那么f(3)．211反比例函数y图像的两支分别在第______限．x12将抛物线y

向上平移一个单位后，得以新的抛物线，那么新的抛物线的表达式是．13如果从小明等名生中任选1作为“博会”愿者，那么小明被选中的概率是．14某商品的原价为，如果经过两次降价，且每次降价的百分率都是m那么该商品现在的价格是_元（结果用含m的代数式表示）．15如图，中，AD是边上的中线，设向量如果用向量a示向量AD那么=_______16在，弦AB长为6它所对应的弦心距为，那么半OA

．

D图

C17在四边形中，对角线AC互相平分，交点在不添加任何辅助线的前提下，要使四边形成为矩形，还需添加一个条件，这个条件可以是_________________18在RtABC中90，M为边上的

点，联结AM（如图3所示）．如果沿直线AM翻

MC图

后，点恰好落在边AC的中点处，那么点到距离是．三、解题：（本大共，满分分）19（本题满分）计算：

2aaaaa

．20（本题满分）解方程组：

22xy.

21（本题满分，每小题满分各5）如图4在梯形ABCD中，BC，ABDC，60联结AC（1求的；（2若MN分别是、DC的中点，联结MN，求线段长．22（本题满分，第（1小题满分，第（2小题满分，第（3小题满分2分，第（）小题满分3分）

为了了解某校初中男生的身体素质状况，在该校六年级至九年级共四个年级的男生中，分别抽取部分学生进行“体向上”试．所有被测试者的“体向上”数情况如表一所示；各年级的被测试人数占所有被测试人数的百分率如图5所示其中六年级相关数据未标出）．次数人数

01

11

22

32

43

54

62

72

82

90

101表一根据上述信息，回答下列问题（直接写出结果）：（1六年级的被测试人数占所有被测试人数的百分率是；（2在所有被测试者中，九年级的人数是；（3在所有被测试者中，“体向上”数不小于人数所占的百分

八年七年

九年六年图率是；（4在所有被测试者的“体向上”数中，众数是．23（本题满分，每小题满分各6）已知线段AC相交于，结、，OB中

D点，F的中点，联结EF（如图示）．

（1添加条件，OFE，

F

C图求证：ABDC．（2分别将“”为①，“”为②，“ABDC”为③，添加条件①、③，以②为结论构成命题1添加条件②、③，以①为结论构成命题．命题是

命题，命题2

命题（选择“”“”入空格）．

24（本题满分，每小题满分各4）在直角坐标平面内O为原点，点的坐标为，坐标为直CMx（如图7示）．点B点A关于原点对称，直线yb为常数）经过点B且与直CM相交于点D，联OD．（1值和点的坐标；

y4

D

M（2设点P在轴的正半轴上，是形，求点P的坐标；（3在（2的条件下，如果以PD为径的

3211

x

等腰三角圆P与圆O外，求O的径．

图

25（本题满分，第（1小题满分，第（2小题满分，第（3小题满分5分）已知，BC，BC，线段上的动点，在线，且PQ满足（如图8所示）．PC（1当AD且Q与B合时（如图9所示），求线段的长；（2在图，联结AP当

32

，且在线段AB上时，设点、之间的距离为x，SAPQSPBC

y

，其中S

APQ

表△的面积，S

△PBC

表PBC面积，求关x的函数解析式，并写出函数定义域；（3当AD，在段延长线上时（如图10所示），求QPC的小．

DA

D

DC

（

C

C

5１51１２5１51１２九年级上学摸底试卷案说明：1．解答列出试题的一种或几种解法．如果考生的解法与所列解法不同，可参照解答中评分标准相应评分；2．第一二大题若无特别说明，每题评分只有满分或零分；3．第三题中各题右端所注分数，表示考生正确做对这一步应得分数；4．评阅卷，要坚持每题评阅到底，不能因考生解答中出现错误而中断对本题的评阅．如果考生的解答在某一步出现错误，影响后继部分而未改变本题的内容和难度，视影响的程度决定后继部分的给分，但原则上不超过后继部分应得分数的一半；5．评分，给分或扣分均以为基本单位．一．选题：（本大共6题满分）1．B；．C；3．A;4．B;5．C;6．A．二．填题：（本大共题，满分48分）7．;8．2．；４12213；6

10；142

；

11一、三；15b；

121612

；17BD

（或ABC9018.2三．解题：（本大共7题满分）2(a1(a19解：原式＝a(

·································（）2＝···································································（）aa1＝··········································································）a＝··········································································（1）20解：由方程①得yx，③··············································（）将③代入②，得2

xx

，·······························

（1）整理，得x20

·················································

（2）解得x·····················································（3）12分别将x入③，得，0，···············（）12所以，原方程组的解为·······························（）y；1221解：（1点A作AE垂足为．······························（分在Rt△ABE，∵60∴BEABBcos········································（分）AEBsin603·········································（）∵12，∴····················································（1）在RtAEC中，tan

AE4．···························1分）2（2在梯形ABCD中，∵AB，60∴60·····························································（1分）过点DBC垂足为F，∵DFC//DF．∵//BC，∴四边形AEFD平行四边形．∴EF．···········（1）在Rt中DCDCF，············（）∴EFEC4．∴．∵MN分别是AB、DC的中点，∴

4．（2分）2222（1·····································································（2）（2；········································································（3分）

（3；·····································································（2分）（45．··········································································（3）23（1明：，OF．······················································1分）∵E为OB的中点，FOC的中点，OBOCOF··································（）．······················································（1分）∵，DOC，∴△AOB△DOC·····················································（2分）DC·································································（1分）（2··········································································（）假．···············································································（3）24解：（1点坐标为点B与关于原点对称，∴点的坐标为

．··················································1分）∵直线yx过点B，·················1分）∵C坐标为，直CM//，∴设点D坐标为1分）∵直线yx与直CM相于点∴x3．D的坐标．…1）（2D的坐标为OD···································1分）当OD时，点P的坐标为；·························（分当POOD时，点的坐标为(5，··························（1分当PD时，设点P的坐标((x，∴(x

，得x，∴点P坐标为(．·（1分）综上所述，所求点的标是、或(（3以为半径的圆与外切时，若点的坐标为(6则圆P的半径PD圆心距6∴的半径r·······················································2分）若点的坐标为，圆的径PD2，圆心距，∴半径r5．·············································2分）综上所述，所求O的半径等1525解：（1∵//BC，．∵ADAB2，ADB．PBC45·······································（）PQ∵，Q与点B重合∴PQPCPBC45···················································）BPC90······························································）在Rt△BPC，BC

．············（）

△△AP（2点作PEBC，PF垂足分别为F．··········（分