2016中考汇编5相交线与平行线

2016中考汇编5相交线与平行线
2016中考汇编5相交线与平行线

2016年全国各地中考数学试题分类解析汇编(第一辑)第5

章相交线与平行线

一.选择题(共20小题)

1.(2016?济宁)如图,将△ABE向右平移2cm得到△DCF,如果△ABE的周长是16cm,那么四边形ABFD的周长是()

A.16cm B.18cm C.20cm D.21cm

2.(2016?娄底)下列命题中,错误的是()

A.两组对边分别平行的四边形是平行四边形

B.有一个角是直角的平行四边形是矩形

C.有一组邻边相等的平行四边形是菱形

D.内错角相等

3.(2016?衡阳)下列命题是假命题的是()

A.经过两点有且只有一条直线

B.三角形的中位线平行且等于第三边的一半

C.平行四边形的对角线相等

D.圆的切线垂直于经过切点的半径

4.(2016?宁波)能说明命题“对于任何实数a,|a|>﹣a”是假命题的一个反例可以是()

A.a=﹣2 B.a=C.a=1 D.a=

5.(2016?大庆)如图,从①∠1=∠2 ②∠C=∠D ③∠A=∠F 三个条件中选出两个作为已知条件,另一个作为结论所组成的命题中,正确命题的个数为()

A.0 B.1 C.2 D.3

6.(2016?深圳)下列命题正确的是()

A.一组对边平行,另一组对边相等的四边形是平行四边形

B.两边及其一角相等的两个三角形全等

C.16的平方根是4

D.一组数据2,0,1,6,6的中位数和众数分别是2和6

7.(2016?贵港)下列命题中错误的是()

A.两组对角分别相等的四边形是平行四边形

B.矩形的对角线相等

C.对角线互相垂直的四边形是菱形

D.对角线互相垂直平分且相等的四边形是正方形

8.(2016?玉林)下列命题是真命题的是()

A.必然事件发生的概率等于0.5

B.5名同学二模的数学成绩是92,95,95,98,110,则他们的平均分是98分,众数是95

C.射击运动员甲、乙分别射击10次且击中环数的方差分别是5和18,则乙较甲稳定

D.要了解金牌获得者的兴奋剂使用情况,可采用抽样调查的方法

9.(2016?齐齐哈尔)下列命题中,真命题的个数是()

①同位角相等

②经过一点有且只有一条直线与这条直线平行

③长度相等的弧是等弧

④顺次连接菱形各边中点得到的四边形是矩形.

A.1个B.2个C.3个D.4个

10.(2016?湖州)定义:若点P(a,b)在函数y=的图象上,将以a为二次项系数,

b为一次项系数构造的二次函数y=ax2+bx称为函数y=的一个“派生函数”.例如:

点(2,)在函数y=的图象上,则函数y=2x2+称为函数y=的一个“派生函数”.现

给出以下两个命题:

(1)存在函数y=的一个“派生函数”,其图象的对称轴在y轴的右侧

(2)函数y=的所有“派生函数”,的图象都进过同一点,下列判断正确的是()

A.命题(1)与命题(2)都是真命题

B.命题(1)与命题(2)都是假命题

C.命题(1)是假命题,命题(2)是真命题

D.命题(1)是真命题,命题(2)是假命题

11.(2016?安顺)已知命题“关于x的一元二次方程x2+bx+1=0,必有实数解”是假命题,则在下列选项中,b的值可以是()

A.b=﹣3 B.b=﹣2 C.b=﹣1 D.b=2

12.(2016?茂名)如图,直线a、b被直线c所截,若a∥b,∠1=60°,那么∠2的度数为()

13.(2016?成都)如图,l1∥l2,∠1=56°,则∠2的度数为()

A.34° B.56° C.124° D.146°

14.(2016?重庆)如图,AB∥CD,直线l交AB于点E,交CD于点F,若∠2=80°,则∠1等于()

A.120° B.110° C.100° D.80°

15.(2016?宿迁)如图,已知直线a、b被直线c所截.若a∥b,∠1=120°,则∠2的度数为()

A.50° B.60° C.120° D.130°

16.(2016?临沂)如图,直线AB∥CD,∠A=40°,∠D=45°,则∠1的度数是()

A.80° B.85° C.90° D.95°

17.(2016?荆州)如图,AB∥CD,射线AE交CD于点F,若∠1=115°,则∠2的度数是()

18.(2016?衡阳)如图,直线AB∥CD,∠B=50°,∠C=40°,则∠E等于()

A.70° B.80° C.90° D.100°

19.(2016?陕西)如图,AB∥CD,AE平分∠CAB交CD于点E,若∠C=50°,则∠AED=()

A.65° B.115° C.125° D.130°

20.(2016?贺州)如图,已知∠1=60°,如果CD∥BE,那么∠B的度数为()

A.70° B.100° C.110° D.120°

2016年全国各地中考数学试题分类解析汇编(第一辑)

第5章相交线与平行线

参考答案与试题解析

一.选择题(共20小题)

1.(2016?济宁)如图,将△ABE向右平移2cm得到△DCF,如果△ABE的周长是16cm,那么四边形ABFD的周长是()

A.16cm B.18cm C.20cm D.21cm

【分析】先根据平移的性质得到CF=AD=2cm,AC=DF,而AB+BC+AC=16cm,则四边形ABFD的周长=AB+BC+CF+DF+AD,然后利用整体代入的方法计算即可

【解答】解:∵△ABE向右平移2cm得到△DCF,

∴EF=AD=2cm,AE=DF,

∵△ABE的周长为16cm,

∴AB+BE+AE=16cm,

∴四边形ABFD的周长=AB+BE+EF+DF+AD

=AB+BE+AE+EF+AD

=16cm+2cm+2cm

=20cm.

故选C.

【点评】本题考查了平移的性质:把一个图形整体沿某一直线方向移动,会得到一

个新的图形,新图形与原图形的形状和大小完全相同.新图形中的每一点,都是由

原图形中的某一点移动后得到的,这两个点是对应点.连接各组对应点的线段平行

且相等.

2.(2016?娄底)下列命题中,错误的是()

A.两组对边分别平行的四边形是平行四边形

B.有一个角是直角的平行四边形是矩形

C.有一组邻边相等的平行四边形是菱形

D.内错角相等

【分析】根据平行四边形、矩形、菱形的判定方法即可判断A、B、C正确.

【解答】解:A、两组对边分别平行的四边形是平行四边形,正确.

B、有一个角是直角的平行四边形是矩形,正确.

C、有一组邻边相等的平行四边形是菱形,正确.

D、内错角相等,错误,缺少条件两直线平行,内错角相等.

故选D.

【点评】本题考查命题与定理、平行四边形的判定、菱形的判定、矩形的判定等知识,解题的关键是熟练掌握特殊四边形的判定方法,属于中考常考题型.

3.(2016?衡阳)下列命题是假命题的是()

A.经过两点有且只有一条直线

B.三角形的中位线平行且等于第三边的一半

C.平行四边形的对角线相等

D.圆的切线垂直于经过切点的半径

【分析】根据直线公理、三角形中位线定理、切线性质定理即可判断A、B、D正确.【解答】解:A、经过两点有且只有一条直线,正确.

B、三角形的中位线平行且等于第三边的一半,正确.

C、平行四边形的对角线相等,错误.矩形的对角线相等,平行四边形的对角线不一定相等.

D、圆的切线垂直于经过切点的半径,正确.

故选C.

【点评】本题考查命题与定理、直线公理、三角形中位线定理、切线性质定理等知识,解题的关键是灵活应用直线知识解决问题,属于中考常考题型.

4.(2016?宁波)能说明命题“对于任何实数a,|a|>﹣a”是假命题的一个反例可以是()

A.a=﹣2 B.a=C.a=1 D.a=

【分析】反例就是符合已知条件但不满足结论的例子.可据此判断出正确的选项.【解答】解:说明命题“对于任何实数a,|a|>﹣a”是假命题的一个反例可以是a=﹣2,故选A.

【点评】本题考查了命题与定理:判断一件事情的语句,叫做命题.许多命题都是由题设和结论两部分组成,题设是已知事项,结论是由已知事项推出的事项,一个命题可以写成“如果…那么…”形式.有些命题的正确性是用推理证实的,这样的真命题叫做定理.任何一个命题非真即假.要说明一个命题的正确性,一般需要推理、论证,而判断一个命题是假命题,只需举出一个反例即可.

5.(2016?大庆)如图,从①∠1=∠2 ②∠C=∠D ③∠A=∠F 三个条件中选出两个作为已知条件,另一个作为结论所组成的命题中,正确命题的个数为()

A.0 B.1 C.2 D.3

【分析】直接利用平行线的判定与性质分别判断得出各结论的正确性.

【解答】解:如图所示:当①∠1=∠2,

则∠3=∠2,

故DB∥EC,

则∠D=∠4,

当②∠C=∠D,

故∠4=∠C,

则DF∥AC,

可得:∠A=∠F,

即?③;

当①∠1=∠2,

则∠3=∠2,

故DB∥EC,

则∠D=∠4,

当③∠A=∠F,

故DF∥AC,

则∠4=∠C,

故可得:∠C=∠D,

即?②;

当③∠A=∠F,

故DF∥AC,

则∠4=∠C,

当②∠C=∠D,

则∠4=∠D,

故DB∥EC,

则∠2=∠3,

可得:∠1=∠2,

即?①,

故正确的有3个.

故选:D.

【点评】此题主要考查了命题与定理,正确掌握平行线的判定与性质是解题关键.

6.(2016?深圳)下列命题正确的是()

A.一组对边平行,另一组对边相等的四边形是平行四边形

B.两边及其一角相等的两个三角形全等

C.16的平方根是4

D.一组数据2,0,1,6,6的中位数和众数分别是2和6

【分析】根据平行四边形的判定定理、三角形全等的判定定理、平方根的概念、中位数和众数的概念进行判断即可.

【解答】解:A.一组对边平行,另一组对边相等的四边形不一定是平行四边形,故错误;

B.两边及其一角相等的两个三角形不一定全等,故错误;

C.16的平方根是±4,故错误,

D.一组数据2,0,1,6,6的中位数和众数分别是2和6,故正确,

故选:D.

【点评】本题考查的是命题的真假判断,正确的命题叫真命题,错误的命题叫做假命题.判断命题的真假关键是要熟悉课本中的性质定理.

7.(2016?贵港)下列命题中错误的是()

A.两组对角分别相等的四边形是平行四边形

B.矩形的对角线相等

C.对角线互相垂直的四边形是菱形

D.对角线互相垂直平分且相等的四边形是正方形

【分析】直接利用平行四边形以及矩形、菱形、正方形的判定方法分别分析得出答案.

【解答】解:A、两组对角分别相等的四边形是平行四边形,正确,不合题意;

B、矩形的对角线相等,正确,不合题意;

C、对角线互相垂直的平行四边形是菱形,故此选项错误,符合题意;

D、对角线互相垂直平分且相等的四边形是正方形,正确,不合题意.

故选:C.

【点评】此题主要考查了命题与定理,正确掌握平行四边形以及矩形、菱形、正方形的判定方法是解题关键.

8.(2016?玉林)下列命题是真命题的是()

A.必然事件发生的概率等于0.5

B.5名同学二模的数学成绩是92,95,95,98,110,则他们的平均分是98分,众数是95

C.射击运动员甲、乙分别射击10次且击中环数的方差分别是5和18,则乙较甲稳定

D.要了解金牌获得者的兴奋剂使用情况,可采用抽样调查的方法

【分析】命题的“真”“假”是就命题的内容而言.任何一个命题非真即假.要说明一个命题的正确性,一般需要推理、论证,而判断一个命题是假命题,只需举出一个反例即可.

【解答】解:A、必然事件发生的概率等于1,错误;

B、5名同学二模的数学成绩是92,95,95,98,110,则他们的平均分是98分,众数是95,正确;

C、射击运动员甲、乙分别射击10次且击中环数的方差分别是5和18,则甲稳定,错误;

D、要了解金牌获得者的兴奋剂使用情况,可采用全面调查的方法,错误;

故选B

【点评】本题考查了命题与定理:判断一件事情的语句,叫做命题.许多命题都是由题设和结论两部分组成,题设是已知事项,结论是由已知事项推出的事项,一个命题可以写成“如果…那么…”形式.有些命题的正确性是用推理证实的,这样的真命题叫做定理.

9.(2016?齐齐哈尔)下列命题中,真命题的个数是()

①同位角相等

②经过一点有且只有一条直线与这条直线平行

③长度相等的弧是等弧

④顺次连接菱形各边中点得到的四边形是矩形.

A.1个B.2个C.3个D.4个

【分析】根据平行线的性质对①进行判断;根据平行公理对②进行判断;根据等弧的定义对③进行判断;根据中点四边的判定方法可判断顺次连接菱形各边中点得到的四边形为平行四边形,加上菱形的对角线垂直可判断中点四边形为矩形.

【解答】解:两直线平行,同位角相等,所以①错误;

经过直线外一点有且只有一条直线与这条直线平行,所以②错误;

在同圆或等圆中,长度相等的弧是等弧,所以③选项错误;

顺次连接菱形各边中点得到的四边形是矩形,所以④正确.

故选A.

【点评】本题考查了命题与定理:判断一件事情的语句,叫做命题.许多命题都是由题设和结论两部分组成,题设是已知事项,结论是由已知事项推出的事项,一个命题可以写成“如果…那么…”形式.有些命题的正确性是用推理证实的,这样的真命题叫做定理.

10.(2016?湖州)定义:若点P(a,b)在函数y=的图象上,将以a为二次项系数,

b为一次项系数构造的二次函数y=ax2+bx称为函数y=的一个“派生函数”.例如:

点(2,)在函数y=的图象上,则函数y=2x2+称为函数y=的一个“派生函数”.现

给出以下两个命题:

(1)存在函数y=的一个“派生函数”,其图象的对称轴在y轴的右侧

(2)函数y=的所有“派生函数”,的图象都进过同一点,下列判断正确的是()

A.命题(1)与命题(2)都是真命题

B.命题(1)与命题(2)都是假命题

C.命题(1)是假命题,命题(2)是真命题

D.命题(1)是真命题,命题(2)是假命题

【分析】(1)根据二次函数y=ax2+bx的性质a、b同号对称轴在y轴左侧,a、b异号对称轴在y轴右侧即可判断.

(2)根据“派生函数”y=ax2+bx,x=0时,y=0,经过原点,不能得出结论.

【解答】解:(1)∵P(a,b)在y=上,

∴a和b同号,所以对称轴在y轴左侧,

∴存在函数y=的一个“派生函数”,其图象的对称轴在y轴的右侧是假命题.

(2)∵函数y=的所有“派生函数”为y=ax2+bx,

∴x=0时,y=0,

∴所有“派生函数”为y=ax2+bx经过原点,

∴函数y=的所有“派生函数”,的图象都进过同一点,是真命题.

故选C.

【点评】本题考查命题与定理、二次函数的性质,理解题意是解题的关键,记住二次函数y=ax2+bx的性质a、b同号对称轴在y轴左侧,a、b异号对称轴在y轴右侧,属于基础题.

11.(2016?安顺)已知命题“关于x的一元二次方程x2+bx+1=0,必有实数解”是假命题,则在下列选项中,b的值可以是()

A.b=﹣3 B.b=﹣2 C.b=﹣1 D.b=2

【分析】根据判别式的意义,当b=﹣1时△<0,从而可判断原命题为是假命题.【解答】解:△=b2﹣4,当b=﹣1时,△<0,方程没有实数解,

所以b取﹣1可作为判断命题“关于x的一元二次方程x2+bx+1=0,必有实数解”是假命题的反例.

故选C.

【点评】本题考查了命题与定理:判断一件事情的语句,叫做命题.许多命题都是由题设和结论两部分组成,题设是已知事项,结论是由已知事项推出的事项,一个命题可以写成“如果…那么…”形式.有些命题的正确性是用推理证实的,这样的真命题叫做定理.任何一个命题非真即假.要说明一个命题的正确性,一般需要推理、论证,而判断一个命题是假命题,只需举出一个反例即可.

12.(2016?茂名)如图,直线a、b被直线c所截,若a∥b,∠1=60°,那么∠2的度数为()

A.120° B.90° C.60° D.30°

【分析】利用两直线平行,同位角相等就可求出.

【解答】解:∵直线被直线a、b被直线c所截,且a∥b,∠1=48°

∴∠2=48°.

故选C.

【点评】本题考查了平行线的性质,应用的知识为两直线平行,同位角相等.

13.(2016?成都)如图,l1∥l2,∠1=56°,则∠2的度数为()

A.34° B.56° C.124° D.146°

【分析】根据平行线性质求出∠3=∠1=50°,代入∠2+∠3=180°即可求出∠2.

【解答】解:∵l1∥l2,

∴∠1=∠3,

∵∠1=56°,

∴∠3=56°,

∵∠2+∠3=180°,

∴∠2=124°,

故选C.

【点评】本题考查了平行线的性质和邻补角的定义,注意:两直线平行,同位角相等.

14.(2016?重庆)如图,AB∥CD,直线l交AB于点E,交CD于点F,若∠2=80°,则∠1等于()

A.120° B.110° C.100° D.80°

【分析】由平行线的性质得出∠1+∠DFE=180°,由对顶角相等求出∠DFE=∠2=80°,即可得出结果.

【解答】解:∵AB∥CD,

∴∠1+∠DFE=180°,

∵∠DFE=∠2=80°,

∴∠1=180°﹣80°=100°;

故选:C.

【点评】本题考查了平行线的性质、对顶角相等的性质;熟记平行线的性质,由对顶角相等求出∠DFE是解决问题的关键.

15.(2016?宿迁)如图,已知直线a、b被直线c所截.若a∥b,∠1=120°,则∠2的度数为()

A.50° B.60° C.120° D.130°

【分析】根据邻补角的定义求出∠3,再根据两直线平行,同位角相等解答.

【解答】解:如图,∠3=180°﹣∠1=180°﹣120°=60°,

∵a∥b,

∴∠2=∠3=60°.

故选:B.

【点评】本题考查了平行线的性质,邻补角的定义,是基础题,熟记性质是解题的关键.

16.(2016?临沂)如图,直线AB∥CD,∠A=40°,∠D=45°,则∠1的度数是()

A.80° B.85° C.90° D.95°

【分析】根据∠1=∠D+∠C,∠D是已知的,只要求出∠C即可解决问题.

【解答】解:∵AB∥CD,

∴∠A=∠C=40°,

∵∠1=∠D+∠C,

∵∠D=45°,

∴∠1=∠D+∠C=45°+40°=85°,

故选B.

【点评】本题考查平行线的性质、三角形的外角的性质等知识,解题的关键是利用三角形的外角等于不相邻的两个内角之和,属于中考常考题型.

17.(2016?荆州)如图,AB∥CD,射线AE交CD于点F,若∠1=115°,则∠2的度数是()

A.55° B.65° C.75° D.85°

【分析】根据两直线平行,同旁内角互补可求出∠AFD的度数,然后根据对顶角相等求出∠2的度数.

【解答】解:∵AB∥CD,

∴∠1+∠F=180°,

∵∠1=115°,

∴∠AFD=65°,

∵∠2和∠AFD是对顶角,

∴∠2=∠AFD=65°,

故选B.

【点评】本题主要考查了平行线的性质,解题的关键是掌握两直线平行,同旁内角互补.

18.(2016?衡阳)如图,直线AB∥CD,∠B=50°,∠C=40°,则∠E等于()

A.70° B.80° C.90° D.100°

【分析】根据平行线的性质得到∠1=∠B=50°,由三角形的内角和即可得到结论.【解答】解:∵AB∥CD,

∴∠1=∠B=50°,

∵∠C=40°,

∴∠E=180°﹣∠B﹣∠1=90°,

故选C.

【点评】本题考查了三角形内角和定理,平行线的性质的应用,注意:两直线平行,同旁内角互补,题目比较好,难度适中.

19.(2016?陕西)如图,AB∥CD,AE平分∠CAB交CD于点E,若∠C=50°,则∠AED=()

A.65° B.115° C.125° D.130°

【分析】根据平行线性质求出∠CAB的度数,根据角平分线求出∠EAB的度数,根据平行线性质求出∠AED的度数即可.

【解答】解:∵AB∥CD,

∴∠C+∠CAB=180°,

∵∠C=50°,

∴∠CAB=180°﹣50°=130°,

∵AE平分∠CAB,

∴∠EAB=65°,

∵AB∥CD,

∴∠EAB+∠AED=180°,

∴∠AED=180°﹣65°=115°,

故选B.

【点评】本题考查了角平分线定义和平行线性质的应用,注意:平行线的性质有:①两条平行线被第三条直线所截,同位角相等,②两条平行线被第三条直线所截,内错角相等,③两条平行线被第三条直线所截,同旁内角互补.

20.(2016?贺州)如图,已知∠1=60°,如果CD∥BE,那么∠B的度数为()

A.70° B.100° C.110° D.120°

【分析】先根据补角的定义求出∠2的度数,再由平行线的性质即可得出结论.

【解答】解:∵∠1=60°,

∴∠2=180°﹣60°=120°.

∵CD∥BE,

∴∠2=∠B=120°.

故选D.

【点评】本题考查的是平行线的性质,用到的知识点为:两直线平行,同位角相等.

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