期中综合测试(原卷+答案)2025-2026学年数学人教版七年级下册

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名称 期中综合测试(原卷+答案)2025-2026学年数学人教版七年级下册
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更新时间 2026-03-29 00:00:00

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期中综合测试
(时间:120分钟 满分:120分)
一、选择题(本大题共10小题,每小题3分,共30分.在每小题给出的四个选项中,只有一项是符合题目要求的)
1.下列运算正确的是( ).
A.=±4 B.=-3
C.=-2 D.=2+3
2.在-2,,,3.14,,这6个数中,无理数共有( ).
A.4个 B.3个
C.2个 D.1个
3.下列说法正确的有( ).
①0的算术平方根是0;②8的算术平方根是4;③±是11的平方根;④-5是25的一个平方根;⑤±2是8的立方根;⑥81的平方根是9.
A.1个 B.2个
C.3个 D.4个
4.下列说法正确的是( ).
A.过一点有且只有一条直线与已知直线平行
B.垂直于同一条直线的两条直线互相平行
C.从直线外一点到这条直线的垂线段,叫作这点到这条直线的距离
D.在平面内过一点有且只有一条直线与已知直线垂直
5.在平面直角坐标系的第三象限内有一点P,它到x轴的距离为3,到y轴的距离为2,则点P的坐标为( ).
A.(-2,-3) B.(-3,-2)
C.(2,-3) D.(3,-2)
6.如图,M,N,P,Q是数轴上的四个点,其中最适合表示的点是( ).
A.点M B.点N
C.点P D.点Q
7.如图,一艘船在A处遇险后向相距25 n mile,位于B处的救生船报警求助.船员应用方向和距离描述遇险船相对于救生船的位置为( ).
A.南偏西25°方向
B.南偏西25°方向,距离为25 n mile
C.北偏东25°方向
D.北偏东25°方向,距离为25 n mile
8.如图,已知a∥b,小华把三角板的直角顶点放在直线b上.若∠1=40°,则∠2的度数为( ).
A.100° B.110°
C.120° D.130°
9.如图,要修建一条公路,从A村沿北偏东75°方向到B村,从B村沿北偏西25°方向到C村,从C村到D村的公路平行于从A村到B村的公路,则∠DCB的度数为( ).
A.100° B.80°
C.75° D.50°
10.如图,在平面直角坐标系上有一个质点A0(-1,0),质点A0第一次跳动至点A1(1,1),第二次跳动至点A2(-2,1),第三次跳动至点A3(2,2),第四次跳动至点A4(-3,2),……依此规律跳动下去,则点A2 025与点A2 026之间的距离是( ).
A.2 023 B.2 025
C.2 027 D.2 029
二、填空题(本大题共5小题,每小题3分,共15分)
11.已知≈1.766,≈5.586,则≈________.
12.如图,将周长为8的三角形ABC沿BC方向向右平移2个单位长度,得到三角形DEF,连接AD,则四边形ABFD的周长为____________.
13.如图,直线m∥n,若∠1=110°,∠2=100°,则∠3=________.
14.的整数部分是a,小数部分是b,则a-2b=________.
15.如图,已知AP平分∠BAC,CP平分∠ACD,∠1+∠2=90°,则下列结论:①AB∥CD;②∠ABE+∠CDF=180°;③AC∥BD;④若∠ACD=2∠E,则∠CAB=2∠F.其中正确的有____________(填序号).
三、解答题(一)(本大题共3小题,每小题7分,共21分)
16.计算:-+|1-|+.
17.已知点P(m+2,3),Q(-5,n-1),根据以下条件确定m,n的值.
(1)点P在y轴上,点Q在x轴上;
(2)PQ∥x轴,且点P与点Q的距离为3.
18.已知2a-1的平方根是±3,的算术平方根是b,求的值.
四、解答题(二)(本大题共3小题,每小题9分,共27分)
19.如图,B,C,E三点在同一条直线上,A,F,E三点在同一条直线上,∠1=∠2=∠E,∠3=∠4.求证:AB∥CD.
【证明】因为∠2=∠E,
所以AD∥BC,
所以∠3=∠CAD.
又因为∠3=∠4,所以∠4=∠CAD.
因为∠1=∠2,
所以∠1+∠CAF=∠2+∠CAF,
即∠BAF=∠CAD,
所以∠4=∠BAF,所以AB∥CD.
20.如图,直线AB,CD相交于点O,OE平分∠BOD.
(1)若∠EOF=55°,OD⊥OF,求∠AOC的度数;
(2)若OF平分∠COE,∠BOF=15°,求∠DOE的度数.
21.如图,已知火车站的坐标为(2,1),文化宫的坐标为(-1,2).
(1)请你根据题目条件,画出平面直角坐标系;
(2)写出体育场、市场、超市、宾馆的坐标;
(3)请将原点O、宾馆B和文化宫C看作三点用线段连接起来,得到三角形OBC,然后将此三角形向右平移2个单位长度,再向下平移3个单位长度,画出平移后的三角形O1B1C1,并求出其面积.
五、解答题(三)(本大题共2小题,第22题13分,第23题14分,共27分)
22.【探究性问题】
对于平面内的∠M和∠N,若存在一个常数k>0,使得∠M+k∠N=360°,则称∠N为∠M的k系补周角.例如∠M=90°,∠N=45°,∠M+6∠N=360°,则∠N为∠M的6系补周角.
【初步探究】
(1)若∠H=120°,则∠H的4系补周角的度数为________°.
【解决问题】
(2)在平面内AB∥CD, 点E是平面内一点,连接BE,DE.
①如图1,若∠D=60°,∠B是∠E的3系补周角,求∠B的度数.
②如图2,∠ABE和∠CDE均为钝角,点F在点E的右侧,且满足∠ABF=n∠ABE,∠CDF=n∠CDE(其中n为常数,且n>1),点P是∠ABE的平分线BG与∠CDE的平分线DP的交点,且∠BPD是∠F的k系补周角,请写出此时k的值(用含n的式子表示),并说明理由.
eq \o(\s\up7(),\s\do5(图1))     eq \o(\s\up7(),\s\do5(图2))
23.如图1,在平面直角坐标系中,A(a,0),C(b,2),且满足(a+2)2+|b-2|=0,过点C作CB⊥x轴于点B.
(1)求三角形ABC的面积;
(2)如图2,若过点B作BD∥AC交y轴于点D,且AE,DE分别平分∠CAB,∠ODB,求∠AED的度数;
(3)在y轴上存在点P,使得三角形ABC和三角形ACP的面积相等,求出点P的坐标.
eq \o(\s\up7(),\s\do5(图1))   eq \o(\s\up7(),\s\do5(图2))   eq \o(\s\up7(),\s\do5(备用图))期中综合测试
(时间:120分钟 满分:120分)
一、选择题(本大题共10小题,每小题3分,共30分.在每小题给出的四个选项中,只有一项是符合题目要求的)
1.下列运算正确的是( B ).
A.=±4 B.=-3
C.=-2 D.=2+3
2.在-2,,,3.14,,这6个数中,无理数共有( C ).
A.4个 B.3个
C.2个 D.1个
3.下列说法正确的有( C ).
①0的算术平方根是0;②8的算术平方根是4;③±是11的平方根;④-5是25的一个平方根;⑤±2是8的立方根;⑥81的平方根是9.
A.1个 B.2个
C.3个 D.4个
4.下列说法正确的是( D ).
A.过一点有且只有一条直线与已知直线平行
B.垂直于同一条直线的两条直线互相平行
C.从直线外一点到这条直线的垂线段,叫作这点到这条直线的距离
D.在平面内过一点有且只有一条直线与已知直线垂直
5.在平面直角坐标系的第三象限内有一点P,它到x轴的距离为3,到y轴的距离为2,则点P的坐标为( A ).
A.(-2,-3) B.(-3,-2)
C.(2,-3) D.(3,-2)
6.如图,M,N,P,Q是数轴上的四个点,其中最适合表示的点是( D ).
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A.点M B.点N
C.点P D.点Q
7.如图,一艘船在A处遇险后向相距25 n mile,位于B处的救生船报警求助.船员应用方向和距离描述遇险船相对于救生船的位置为( B ).
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A.南偏西25°方向
B.南偏西25°方向,距离为25 n mile
C.北偏东25°方向
D.北偏东25°方向,距离为25 n mile
8.如图,已知a∥b,小华把三角板的直角顶点放在直线b上.若∠1=40°,则∠2的度数为( D ).
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A.100° B.110°
C.120° D.130°
9.如图,要修建一条公路,从A村沿北偏东75°方向到B村,从B村沿北偏西25°方向到C村,从C村到D村的公路平行于从A村到B村的公路,则∠DCB的度数为( B ).
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A.100° B.80°
C.75° D.50°
10.如图,在平面直角坐标系上有一个质点A0(-1,0),质点A0第一次跳动至点A1(1,1),第二次跳动至点A2(-2,1),第三次跳动至点A3(2,2),第四次跳动至点A4(-3,2),……依此规律跳动下去,则点A2 025与点A2 026之间的距离是( C ).
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A.2 023 B.2 025
C.2 027 D.2 029
二、填空题(本大题共5小题,每小题3分,共15分)
11.已知≈1.766,≈5.586,则≈________.
【答案】55.86
12.如图,将周长为8的三角形ABC沿BC方向向右平移2个单位长度,得到三角形DEF,连接AD,则四边形ABFD的周长为____________.
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【答案】12
13.如图,直线m∥n,若∠1=110°,∠2=100°,则∠3=________.
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【答案】150°
14.的整数部分是a,小数部分是b,则a-2b=________.
【答案】12-2
15.如图,已知AP平分∠BAC,CP平分∠ACD,∠1+∠2=90°,则下列结论:①AB∥CD;②∠ABE+∠CDF=180°;③AC∥BD;④若∠ACD=2∠E,则∠CAB=2∠F.其中正确的有____________(填序号).
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【答案】①②④
三、解答题(一)(本大题共3小题,每小题7分,共21分)
16.计算:-+|1-|+.
【答案】2+
17.已知点P(m+2,3),Q(-5,n-1),根据以下条件确定m,n的值.
(1)点P在y轴上,点Q在x轴上;
(2)PQ∥x轴,且点P与点Q的距离为3.
【答案】(1)m=-2,n=1
(2)n=4,m=-4或m=-10
18.已知2a-1的平方根是±3,的算术平方根是b,求的值.
【答案】3
四、解答题(二)(本大题共3小题,每小题9分,共27分)
19.如图,B,C,E三点在同一条直线上,A,F,E三点在同一条直线上,∠1=∠2=∠E,∠3=∠4.求证:AB∥CD.
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【证明】因为∠2=∠E,
所以AD∥BC,
所以∠3=∠CAD.
又因为∠3=∠4,所以∠4=∠CAD.
因为∠1=∠2,
所以∠1+∠CAF=∠2+∠CAF,
即∠BAF=∠CAD,
所以∠4=∠BAF,所以AB∥CD.
20.如图,直线AB,CD相交于点O,OE平分∠BOD.
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(1)若∠EOF=55°,OD⊥OF,求∠AOC的度数;
(2)若OF平分∠COE,∠BOF=15°,求∠DOE的度数.
【解】(1)因为OE平分∠BOD,
所以∠BOE=∠DOE.
因为∠EOF=55°,OD⊥OF,
所以∠DOE=35°,
所以∠BOE=35°,
所以∠BOD=70°,
所以∠AOC=70°.
(2)因为OF平分∠COE,
所以∠COF=∠EOF.
因为∠BOF=15°,
设∠DOE=∠BOE=x,
则∠COF=x+15°,
所以x+15°+x+15°+x=180°,
解得x=50°,
故∠DOE的度数为50°.
21.如图,已知火车站的坐标为(2,1),文化宫的坐标为(-1,2).
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(1)请你根据题目条件,画出平面直角坐标系;
(2)写出体育场、市场、超市、宾馆的坐标;
(3)请将原点O、宾馆B和文化宫C看作三点用线段连接起来,得到三角形OBC,然后将此三角形向右平移2个单位长度,再向下平移3个单位长度,画出平移后的三角形O1B1C1,并求出其面积.
【答案】(1)图略
(2)体育场(-2,4),市场(6,4),超市(4,-2),宾馆(4,3).
(3)图略 
五、解答题(三)(本大题共2小题,第22题13分,第23题14分,共27分)
22.【探究性问题】
对于平面内的∠M和∠N,若存在一个常数k>0,使得∠M+k∠N=360°,则称∠N为∠M的k系补周角.例如∠M=90°,∠N=45°,∠M+6∠N=360°,则∠N为∠M的6系补周角.
【初步探究】
(1)若∠H=120°,则∠H的4系补周角的度数为________°.
【解决问题】
(2)在平面内AB∥CD, 点E是平面内一点,连接BE,DE.
①如图1,若∠D=60°,∠B是∠E的3系补周角,求∠B的度数.
②如图2,∠ABE和∠CDE均为钝角,点F在点E的右侧,且满足∠ABF=n∠ABE,∠CDF=n∠CDE(其中n为常数,且n>1),点P是∠ABE的平分线BG与∠CDE的平分线DP的交点,且∠BPD是∠F的k系补周角,请写出此时k的值(用含n的式子表示),并说明理由.
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【解】(1)60
(2)①过点E作EF∥AB,如图.
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所以∠B=∠BEF.
因为AB∥CD,∠D=60°,
所以EF∥CD,
所以∠DEF=∠D=60°,
所以∠B+∠D=∠BEF+∠DEF,
即∠B+60°=∠BED.
因为∠B是∠BED的3系补周角,
所以∠BED=360°-3∠B,
所以∠B+60°=360°-3∠B,
所以∠B=75°.
②k=2n.理由如下:
连接BD,PF,如图.
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则∠DBF+∠BFD+∠BDF=180°.
因为AB∥CD,
所以∠ABD+∠CDB=180°,
所以∠ABF+∠BFD+∠CDF=∠ABD+∠DBF+∠BFD+∠FDB+∠CDB=360°.
因为∠ABE的平分线与∠CDE的平分线相交于点P,
所以∠ABP=∠ABE,∠CDP=∠CDE.
由①可得∠BPD=∠ABP+∠CDP,
所以∠BPD=∠BPF+∠DPF=(∠ABE+∠CDE).
因为∠ABF=n∠ABE,∠CDF=n∠CDE,
所以∠BPD=(∠ABF+∠CDF),
所以∠BPD=(∠ABF+∠CDF),
所以∠BPD=(360°-∠BFD),
所以∠BFD+2n∠BPD=360°,
所以∠BPD是∠BFD的k系补周角,
此时k=2n.
23.如图1,在平面直角坐标系中,A(a,0),C(b,2),且满足(a+2)2+|b-2|=0,过点C作CB⊥x轴于点B.
(1)求三角形ABC的面积;
(2)如图2,若过点B作BD∥AC交y轴于点D,且AE,DE分别平分∠CAB,∠ODB,求∠AED的度数;
(3)在y轴上存在点P,使得三角形ABC和三角形ACP的面积相等,求出点P的坐标.
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【解】(1)因为(a+2)2+|b-2|=0,
所以a+2=0,b-2=0,
所以a=-2,b=2,
所以A(-2,0),C(2,2).
因为CB⊥AB,所以B(2,0),
所以AB=4,CB=2,
则S三角形ABC=×4×2=4.
(2)如图,过点E作EF∥AC.
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因为CB⊥x轴,
所以CB∥y轴,∠CBA=90°,
所以∠ODB=∠6.
又因为BD∥AC,所以∠CAB=∠5,
所以∠CAB+∠ODB=∠5+∠6=180°-∠CBA=90°.
因为BD∥AC,EF∥AC,
所以BD∥AC∥EF,
所以∠1=∠3,∠2=∠4.
因为AE,DE分别平分∠CAB,∠ODB,
所以∠3=∠CAB,∠4=∠ODB,
所以∠AED=∠1+∠2=∠3+∠4=(∠CAB+∠ODB)=45°.
(3)①当点P在y轴正半轴上时,如图.
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设点P(0,t),分别过点P,A,B作MN∥x轴,AN∥y轴,BM∥y轴,MN与AN,BM分别交于点N,M,则AN=t,CM=t-2,MN=4,PM=PN=2.
因为S三角形ABC=4,所以S三角形ACP=S梯形MNAC-S三角形ANP-S三角形CMP=4,
即×4(t-2+t)-×2t-×2(t-2)=4,解得t=3,
所以点P的坐标为(0,3).
②当点P在y轴负半轴上时,如图,同①作辅助线.
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设点P(0,q),则AN=-q,CM=-q+2=2-q,MN=4,PM=PN=2.因为S三角形ACP=S梯形MNAC-S三角形ANP-S三角形CMP=4,
即×4(-q+2-q)-×2(-q)-×2(2-q)=4,解得q=-1,
所以点P的坐标为(0,-1).
综上所述,点P的坐标为(0,-1)或(0,3).
同课章节目录