九年级数学综合练习一_3

九年级数学综合练习一

一、选择题(共8道小题，每小题4分，共32分) 1．－2的相反数等于( )． A．2

B．－2

C．

1 2D．?1 22．2009年，全国普通高校本、专科共计划招生6290000人，将6290000用科学记数法表示应为( )． A．6.29×105 B．62.9×105 C．6.29×106 D．0.629×107 3．下图是由五个相同的小正方体搭成的几何体，它的主视图是( )

4．若一个多边形的内角和为1080°，则这个多边形的边数为( )． A．5 B．6 C．7 D．8

5．2004～2008年社会消费品零售总额及增长速度情况如图所示，那么社会消费品零售总额比上年增长最快的年份是( )．

A．2005年 B．2006年 C．2007年 D．2008年

6．如图，AB∥DF，AC⊥BC于C，BC与DF交于点E，若∠A＝20°，则∠CEF等于( )．

A．110° B．100° C．80° D．70°

7．如图，在边长为1的等边三角形ABC中，若将两条含120°圆心角的则S与△ABC面积的比等于( )．

、

及边AC所围成的阴影部分的面积记为S，

A．

1 2B．

1 3C．

1 4D．

1 68．若m、n(m＜n)是关于x的方程1－(x－a)(x－b)＝0的两根，且a＜b，则a、b、m、n的大小关系是( )． A．m＜a＜b＜n B．a＜m＜n＜b C．a＜m＜b＜n D．m＜a＜n＜b 二、填空题(共4道小题，每小题4分，共16分) 9．在函数y?1中，自变量x的取值范围是______．

2x?410．若x?1?(y?4)2?0，则xy的值等于______．

11．如图，△ABC中，∠ABC的平分线交AC于E，BE⊥AC，DE∥BC交AB于D，若BC＝4，则DE＝______．

12．在Rt△ABC中，∠ACB＝90°，BC＜AC，若BC?AC?三、解答题(本题共30分，每小题5分)

1AB2则∠A＝______°． 4113．计算：12?(3?π)0?()?1?2sin60?.

2

?2(x?1)?3?3x,?14．解不等式组?x?2在数轴上表示它的解集，并求它的整数解．

?4?x,??3

15．已知：如图，△ABC中，AB＝AC，BC为最大边，点D、E分别在BC、AC上，BD＝CE，F为BA延长线上一点，BF＝

CD．求证：∠DEF＝∠DFE．

16．解方程：

x2?1?2? x?2x?4

17．已知抛物线y＝－x2＋(m＋2)x＋3m－20经过点(1，－3)，求抛物线与x轴交点的坐标及顶点的坐标．

18．已知：如图，在梯形ABCD中，AD∥BC，AB＝AD＝2，∠A＝60°，BC＝4．求CD的长．

四、解答题(本题共20分，第19题5分，第20题5分，第21题6分，第22题4分)

19．已知：如图，AB为⊙O的弦，过点O作AB的平行线，交⊙O于点C，直线OC上一点D满足∠D＝∠ACB．

(1)判断直线BD与⊙O的位置关系，并证明你的结论； (2)若⊙O的半径等于4，tan?ACB?4，求CD的长． 3

20．有三个完全相同的小球，上面分别标有数字1、－2、－3，将其放入一个不透明的盒子中摇匀，再从中随机摸球两次(第一

次摸出球后放回摇匀)，设第一次摸到的球上所标的数字为m，第二次摸到的球上所标的数字为n，依次以m、n作为点M的横、纵坐标．

(1)用树状图(或列表法)表示出点M(m，n)的坐标所有可能的结果； (2)求点M(m，n)在第三象限的概率． 21．某运输公司用10辆相同的汽车将一批苹果运到外地，每辆汽车能装8吨甲种苹果，或10吨乙种苹果，或11吨丙种苹果．公

司规定每辆车只能装同一种苹果，而且必须满载．已知公司运送了甲、乙、丙三种苹果共100吨，且每种苹果不少于一车． (1)设用x辆车装甲种苹果，y辆车装乙种苹果，求y与x之间的函数关系式，并写出自变量x的取值范围； (2)若运送三种苹果所获利润的情况如下表所示： 苹果品种 每吨苹果所获利润/万元 甲 0.22 乙 0.21 丙 0.2 设此次运输的利润为W(万元)，问：如何安排车辆分配方案才能使运输利润W最大，并求出最大利润．

22．已知：如图，△ABC中，AC＜AB＜BC．

(1)在BC边上确定点P的位置，使∠APC＝∠C．请画出图形，不写画法； (2)在图中画出一条直线l，使得直线l分别与AB、BC边交于点M、N，并且沿直线l将△ABC剪开后可拼成一个等腰梯形．请画出直线l及拼接后的等腰梯形，并简要说明你的剪拼方法． 说明：本题只需保留画图痕迹，无需尺规作图．

五、解答题(本题共22分，第23题7分，第24题8分，第25题7分) 23．已知：反比例函数y?与y?288和y?在平面直角坐标系xOy第一象限中的图象如图所示，点A在y?的图象上，AB∥y轴，xxx228的图象交于点B，AC、BD与x轴平行，分别与y?、y?的图象交于点C、D． xxx

(1)若点A的横坐标为2，求梯形ACBD对角线的交点F的坐标；

(2)若点A的横坐标为m，比较△OBC与△ABC面积的大小，并说明理由； (3)若△ABC与以A、B、D为顶点的三角形相似，请直接写出点A的坐标．

324．已知：如图，在平面直角坐标系xOy中，直线y??x?6与x轴、y轴的交点分别为A、B，将∠OBA对折，使点O的对

4应点H落在直线AB上，折痕交x轴于点C．

(1)直接写出点C的坐标，并求过A、B、C三点的抛物线的解析式； (2)若抛物线的顶点为D，在直线BC上是否存在点P，使得四边形ODAP为平行四边形?若存在，求点P的坐标；若不存在，说明理由；

(3)设抛物线的对称轴与直线BC的交点为T，Q为线段BT上一点，直接写出|QA－QO|的取值范围．

25．已知：PA?2，PB＝4，以AB为一边作正方形ABCD，使P、D两点落在直线AB的两侧．

(1)如图，当∠APB＝45°时，求AB及PD的长；

(2)当∠APB变化，且其他条件不变时，求PD的最大值，及相应∠APB的大小．

参考答案

九年级数学综合练习一

1．A． 2．C． 3．B． 4．D 5．D． 6．A． 7．B． 8．A． 9．x?－2． 10．1． 11．2． 12．15． 13．解：12?(3?π)?()?2sin60

012?1??23?1?2?2?3 ······························································ 4分 2··············································································· 5分 ?3?1 ·

?2(x?1)?3?3x,①?14．?x?2

?4?x,②??3由①得x?1． ················································································ 1分 由②得x?5． ··············································································· 2分

不等式组的解集在数轴上表示如下：

································· 3分

所以原不等式组的解集为1?x?5． ·················································· 4分 所以原不等式组的整数解为1，2，3，4． ············································ 5分 15．证明：如答图1，

答图1

∵AB＝AC， ∴∠B＝∠C． ················································································ 1分 在△BDF和△CED中， ?BD?CE,???B??C, ?BF?CD,?∴△BDF≌△CED． ········································································ 3分 ∴DF＝ED． ·················································································· 4分 ∴∠DEF＝∠DFE． ········································································ 5分 16．解：去分母，得x(x＋2)－(x2－4)＝2． ················································ 1分

去括号，得x2＋2x－x2＋4＝2． ························································· 2分 整理，得2x＝－2． ········································································· 3分 解得x＝－1． ················································································ 4分 经检验，x＝－1是原方程的解． ························································ 5分 17．解：∵抛物线y＝－x2＋(m＋2)x＋3m－20经过(1，－3)点，

∴－12＋(m＋2)＋3m－20＝－3． 整理，得4m－19＝－3． 解得m＝4． ·················································································· 1分 ∴二次函数的解析式为y＝－x2＋6x－8． ············································· 2分 令y＝0，可得－x2＋6x－8＝0． 解得x1＝2，x2＝4． ········································································ 3分 ∴抛物线与x轴的交点坐标为(2，0)，(4，0)． ······································ 4分 ∵y＝－x2＋6x－8＝－(x－3)2＋1， ∴抛物线的顶点坐标为(3，1)． ·························································· 5分 18．解：连结BD，作DE⊥BC于点E．(如答图2) ······································· 1分

答图2

∵AB＝AD＝2，∠A＝60°，

∴△ABD为等边三角形，BD＝2，∠ADB＝60°． ································· 2分 ∵AD∥BC，

∴∠DBC＝60°． ··········································································· 3分 在Rt△BDE中，∠BED＝90°，∠DBE＝60°，

∴DE＝BD·sin60°＝3，BE＝BD·cos60°＝1． ······························ 4分 在Rt△CDE中，∠CED＝90°，CE＝BC－BE＝3，

······························································ 5分 ?CD?DE2?CE2?23. ·

解法二：作DE∥AB交BC于E，作EF⊥CD于F； 解法三：连结BD，并延长BA、CD交于E． 19．解：(1)直线BD与⊙O相切．

证明：如答图3，连结OB． ··················································· 1分

答图3

∵∠OCB＝∠CBD＋∠D，∠1＝∠D， ∴∠2＝∠CBD．

∵AB∥OC，∴∠2＝∠A．∴∠A＝∠CBD． ∵OB＝OC，∴∠BOC＋2∠3＝180°． ∵∠BOC＝2∠A，∴∠A＋∠3＝90°． ∴∠CBD＋∠3＝90°． ∴∠OBD＝90°． ································································· 2分 ∴直线BD与⊙O相切． ························································· 3分 (2)解：∵∠D＝∠ACB，tan?ACB?44···························· 4分 ,?tanD?． ·

334， 3在Rt△OBD中，∠OBD＝90°，OB＝4，tanD?4OB?sinD?,OD??5.

5sinD∴CD＝OD－OC＝1． ··························································· 5分

20．解：(1)组成的点M(m，n)的坐标的所有可能性为：

或列表如下：

第一次 第二次 1 －2 －3 1 (1，1) (1，－2) (1，－3) －2 (－2，1) (－2，－2) (－2，－3) －3 (－3，1) (－3，－2) (－3，－3) ··································································································· 3分

(2)落在第三象限的点有(－2，－2)，(－2，－3)，(－3，－2)，(－3，－3)，因此点M落在第三象限的概率为

分

21．解：(1)∵8x＋10y＋11(10－x－y)＝100， ·············································· 1分

∴y与x之间的函数关系式为y＝－3x＋10． ······························· 2分 ∵y≥1，解得x≤3．

∵x≥1，10－x－y≥1，且x是正整数，

∴自变量x的取值范围是x＝1或x＝2或x＝3． ·························· 3分 (2)W＝8x×0.22＋10y×0.21＋11(10－x－y)×0.2＝－0.14x＋21． ········· 4分 因为W随x的增大而减小，所以x取1时，可获得最大利润， 此时W＝20.86(万元)． ··························································· 5分

获得最大运输利润的方案为：用1辆车装甲种苹果，用7辆车装乙种苹果，2辆车装丙种苹果． 6分

22．解：(1)答案见答图4(任选一种即可)． ················································· 2分

4． 59

答图4

(2)答案见答图5． ····································································· 3分

答图5

剪拼方法：取AB的中点M，过点M作AP的平行线l，与BC交于点N，过点A作BC的平行线，与l交于点H，将△BMN绕点M顺时针旋转180°到△AMH，则四边形ACNH为拼接后的等腰梯形． 4分

23．解：(1)如答图6，当点A的横坐标为2时，点A、B、C、D的坐标分别为

A(2，4)，B(2，1)，C(

1，4)，D(8，1)． ·································· 1分 2221解法一：直线CD的解析式为y??x?? ······························· 2分

55

答图6

∵AB∥y轴，F为梯形ACBD的对角线的交点，

22117∴x＝2时，y???2???

555∴点F的坐标为(2，

17)．······················································ 3分 53解法二：AC?,BD?6,AB?3.

2∵梯形ACBD，AC∥BD，F为梯形ACBD的对角线的交点， ∴△ACF∽△BDF．

?AFAC1??? BFBD4

17AF13． ································ 2分 ?,AF?，点F的纵坐标为

AB55517∴点F的坐标为(2，)． ······················································ 3分

5(2)如答图7，作BM⊥x轴于点M作CN⊥x轴于点N．当点A的横坐标为m时，点A、B、C、D的坐标分别为A(m，

?82m82)，B(m，)，C(，)，D(4m，)． mm4mm113m69S?ABC??AC?AB????? ··································· 4分

224m4S△OBC＝S梯形CNMB＋S△OCN－S△OBM

1823m15＝S梯形CNMB＝(?)?··········································· 5分 ?? ·

2mm44

答图7

∴S△OBC＞S△ABC． ··································································· 6分 (3)点A的坐标为(2，4)． ···························································· 7分

24．解：(1)点C的坐标为(3，0)． ···························································· 1分

∵点A、B的坐标分别为A(8，0)，B(0，6)，

∴可设过A、B、C三点的抛物线的解析式为y＝a(x－3)(x－8)．

将x＝0，y＝6代入抛物线的解析式，得a?∴过A、B、C三点的抛物线的解析式为y?(2)可得抛物线的对称轴为x?1··························· 2分 ? ·41211············ 3分 x?x?6. ·

44112511，顶点D的坐标为(,?)，设抛物线的对称轴与x轴的交点为G．

2162直线BC的解析式为y＝－2x＋6． ············································· 4分 设点P的坐标为(x，－2x＋6)．

解法一：如答图8，作OP∥AD交直线BC于点P，连结AP，作PM⊥x轴于点M．

答图8

∵OP∥AD，

∴∠POM＝∠GAD，tan∠POM＝tan∠GAD．

25?2x?6PMDG，即?16? ??11xOMGA8?2解得x?1616是原方程的解． ?经检验x?771610此时点P的坐标为(,). ······················································ 5分

77

但此时OM?165,GA?,OM?GA. 72OMGA,AD?,?POM?∠GAD，

cos?POMcos?GAD∴OP＜AD，即四边形的对边OP与AD平行但不相等， ∴直线BC上不存在符合条件的点P． ········································ 6分

解法二：如答图9，取OA的中点E，作点D关于点E的对称点P，作PN⊥x轴于点N．则∠PEO＝∠DEA，PE＝D E．可得△PEN≌△DEG． ?OP?

答图9

由OE?OA?4，可得E点的坐标为(4，0)． 23525NE?EG?,ON?OE?NE?,NP?DG??

2216525∴点P的坐标为(,). ·························································· 5分

216∵x＝

5525时，－2x＋6＝－2×＋6＝1≠， 2216∴点P不在直线BC上．

∴直线BC上不存在符合条件的点P． ········································ 6分 (3)|QA－QO|的取值范围是0≤|QA－QO|≤4. ···································· 8分

说明：如答图10，由对称性可知QO＝QH，|QA－QO|＝|QA－QH|．当点Q与点B重合时，Q、H、A三点共线，|QA－QO|取得最大值4(即为AH的长)；设线段OA的垂直平分线与直线BC的交点为K，当点Q与点K重合时，|QA－QO|取得最小值0．

答图10

25．解：(1)①如答图11，作AE⊥PB于点E． ············································ 1分

答图11

∵△APE中，∠APE＝45°，PA?2,

?AE?PA?sin?APE?2?PE?PA?cos?APE?2?2?1, 22?1. 2∵PB＝4，∴BE＝PB－PE＝3． ················································ 2分 在Rt△ABE中，∠AEB＝90°，

?AB?·············································· 3分 AE2?BE2?10. ·

②解法一：如答图12，因为四边形ABCD为正方形，可将△PAD绕点A顺时针旋转90°得到△P′AB，可得△PAD

≌△P′AB，PD＝P′B，PA＝P′A．

答图12

∴∠PAP′＝90°，∠APP′＝45°，∠P′PB＝90°．

∴PP′＝2PA＝2． ·························································· 4分 ····················· 5分 ?PD?P?B?PP?2?PB2?22?42?25. ·

解法二：如答图13，过点P作AB的平行线，与DA的延长线交于F，设DA的延长线交PB于G．

答图13

AE10AE?, ?3cos?EAGcos?ABE12EG?,PG?PB?BE?EG??

33在Rt△PFG中，可得PF＝PG·cos∠FPG

在Rt△AEG中，可得AG??PG?cos?ABE?1010·························· 4分 ,FG?? ·

515在Rt△PDF中，可得PD?PF2?(AD?AG?FG)2 10210102)?(10??)?20?25. 5315 ········································································· 5分 ?((2)如答图14所示，将△PAD绕点A顺时针旋转90°得到△P′AB，PD的最大值即为P′B的最大值．

答图14

∵△P′PB中，P′B＜PP′＋PB，PP′＝2PA＝2，PB＝4，

且P、D两点落在直线AB的两侧，

∴当P′、P、B三点共线时，P′B取得最大值(见答图15)．

答图15

此时P′B＝PP′＋PB＝6，即P′B的最大值为6． ············· 6分 此时∠APB＝180°－∠APP′＝135°． ····························· 7分

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