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2021年人教版七年级数学下册第五章相交线与平行线5.1.1相交线同步测试含答案.pdf

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1、人教版七年级数学下册第五章相交线与平行线5.1.1 相交线同步测试含答案小试牛刀1.下列图形中,1 与 2 互为邻补角的是()答案D根据邻补角定义可得D中 1 与 2 是邻补角.2.如图,直线 AB和 CD相交于点O,AOD+BOC=200,则AOC 的度数为()A.120B.100C.90D.80答案D AOD+BOC=200,BOC=AOD,BOC=100,AOC=180 -100=80.故选D.3.如图,直线 a,b 相交于点O,若1 等于 50,则2 等于()A.50B.40C.140D.130答案A 由题图可知1和2 是对顶角,根据对顶角相等可得 2=1=50,故选A.4.如图,直线

2、 AB、CD相交于点 O,若1=150,则2=.答案30 解析直线 AB、CD相交于点O,1=150,2=180-1=180-150=30.精品w o r d 可编辑资料-第 1 页，共 7 页-5.如图,直线AB 与CD 相交于点O,OE 平分 AOC,若 EOC=25,则 BOD 的度数为.答案50解析OE平分 AOC,EOC=25,AOC=2 EOC=25 2=50.由对顶角相等可知:BOD=AOC=50.能力提升1.下列图形中1和2 是对顶角的是()答案D 互为对顶角的两个角有公共顶点,且一个角的两边分别是另一个角两边的反向延长线.满足条件的只有D.2.如图所示,AB,

3、CD,EF 交于点 O,1=20,2=60,则BOC 的度数为.答案803.如图所示,下列判断正确的是()A.图(1)中1 与2 是一组对顶角B.图(2)中1 与2 是一组对顶角精品w o r d 可编辑资料-第 2 页，共 7 页-文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL

4、6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE1

5、0S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码

6、:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1

7、HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 Z

8、E10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档

9、编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2C.图(3)中1 与2 是一组

10、邻补角D.图(4)中1 与2 是一组邻补角答案D 根据对顶角和邻补角的定义可知只有D正确.4.下面各图中,1 和2 是对顶角的是()答案B 有公共顶点且一角的两边分别是另一角两边的反向延长线的两个角称为对顶角,A,C中1 与2 都没有公共顶点,A,C 错误,D.1与2 的一边重合,不互为反向延长线,D 错误,B.1 与2 满足对顶角的定义,B 正确,故选 B.5.如图,直线 AB,CD相交于点O,EOB=90,COB=145,则DOE=.答案55解析COB=145,DOB=180-145=35,EOB=90,DOE=90-35=55.6.如图,直线 AB,CD相交于 O,射线 OM 平分 AO

11、C,若BOD=80,求BOM 的度数.解析BOD=80,AOC=80,BOC=100,OM平分 AOC,MOC=40,BOM=BOC+MOC=140.8.如图,直线AB,CD 相交于O,已知 AOC=75,OE把 BOD 分成两部分,且BOE EOD=2 3,求AOE.精品w o r d 可编辑资料-第 3 页，共 7 页-文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:

12、CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 H

13、L6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE

14、10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编

15、码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1

16、 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1

17、ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文

18、档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2解析BOD=AOC=75(对顶角相等),BOE EOD=2 3,BOE=75 22+3=30,AOE=180 -BOE=180 -30=150.7.如图,某市民广场地面上新安装了一个棱锥造型的艺术装饰品.小明运用所学的知识很快测量出 ABC 的度数,他是怎样测量的?解析如图:他测量出1的度数,由邻补角的性质,得ABC=180 -1.3.如图,直线 a,b,c两两相交,1=23,2=64,求4的度数.解析因为2=64,又由对顶角相等知 1=2,所以 1=64,又因为 1=23,所以 3=32.因为 3=4,所以 4=

19、32.4.如图,直线 AB与直线 CD相交于点O,BOC比AOC的 2 倍大 30.求BOD的度数.精品w o r d 可编辑资料-第 4 页，共 7 页-文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1

20、HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 Z

21、E10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档

22、编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X

23、1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1

24、 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2

25、文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2解析设AOC=x,则BOC=2x+30.依题意得x+2x+30=180,x=50.所以 BOD=AOC=50

26、.8.如图,已知直线AB、CD、EF相交于点O.(1)AOD的对顶角是;EOC的对顶角是;(2)AOC的邻补角是;EOB的邻补角是.答案(1)BOC;DOF(2)AOD和BOC;EOA和BOF解析根据对顶角和邻补角的定义解答.9.如图,OC平分 AOB,反向延长OC至 D,反向延长OA至 E,3=25,求BOE 的度数.解析由对顶角相等,得2=3=25.因为 OC平分 AOB,所以 AOB=2 2=50.又因为 BOE与AOB互为邻补角,所以 BOE=180 -AOB=180 -50=130.10.如图,直线 AB,CD相交于点 O,OE平分 BOD,OF平分 COE,AOD BOE=4 1,

27、求AOF的度数.解析由已知可设 BOE=x,则AOD=4x.精品w o r d 可编辑资料-第 5 页，共 7 页-文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9

28、J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E

29、8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4

30、Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2

31、G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW

32、5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6

33、H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2因为 OE平分 BOD,所以 BOD=2 BOE=2x.因为 AOD+BOD=180,所以4x+2x=180,解得 x=30,所以 BOE=30,所

34、以 DOE=30,所以 COE=150.因为 OF平分 COE,所以 EOF=12COE=75,所以 BOF=EOF-BOE=45,所以 AOF=180 -BOF=135.11.(如图所示,直线 AB与 CD相交于点 O,OE平分 AOD,BOC=80,求BOD 和AOE的度数.解析因为BOD与BOC是邻补角,BOC=80,所以 BOD=180 -BOC=100.因为 AOD与BOC是对顶角,所以 AOD=BOC=80.又因为 OE平分 AOD,所以 AOE=12BOC=40.12.如图,已知直线AB 与 CD 交于点O,ON平分 DOB.若BOC=110,则AON的度数为度.答案145 解析

35、直线AB 与CD 交于点O,BOC=110,BOD=70,ON为BOD 的平分线,BON=DON=35,易知 AOD=BOC=110,AON=AOD+DON=145.精品w o r d 可编辑资料-第 6 页，共 7 页-文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H

36、4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S

37、2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:C

38、W5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL

39、6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE1

40、0S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码

41、:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y213.已知 AOB与BOC互补,且

42、两个角有公共顶点和一条公共边,AOB=3 BOC,求这两个角的平分线夹角的度数.解析分两种情况:若AOB和BOC互为邻补角,则其情形如图所示:射线 OD,OE分别平分 AOB 和BOC,由一对邻补角的平分线互相垂直可知DOE=90.若AOB和BOC只是互为补角但不是邻补角,则其情形如图:射线 OD,OE分别是 AOB和BOC的平分线,可设 BOC=x,则AOB=3x,可得x+3x=180,解得 x=45.则AOB=135.则DOE=12AOB-12BOC=12135-1245=45.综上可知,所求夹角的度数为90或 45.精品w o r d 可编辑资料-第 7 页，共 7 页-文档编码:CW5

43、E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H

44、4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S

45、2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:C

46、W5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL

47、6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE1

48、0S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2文档编码:CW5E8X3R1X1 HL6H4Q2I2I1 ZE10S2G9J3Y2

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关键词：: 2021 年人教版七年级数下册第五相交平行线 5.1 同步测试答案

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