浙教版七年级下册2018经典易错题汇总(Word版,无答案)

浙教版七年级下册2018经典易错题汇总(Word版,无答案)
浙教版七年级下册2018经典易错题汇总(Word版,无答案)

【模块一】翻折

经典易错题汇总

1(. 2018?仙居县一模)如图,把一张长方形纸带沿着直线 GF 折叠,∠CGF=30°

则∠1 的度数是 .

2.(2018 春?莒县期中)如图,生活中将一个宽度相等的纸条按图所示折叠一下如果∠2=100°,那么∠1 的度数为

【模块二】旋转

1.(2017?上海中考)一副三角尺按如图的位置摆放(顶点 C 与 F 重合,边

CA 与边 FE 叠合,顶点 B 、C 、D 在一条直线上).将三角尺 DEF 绕着点 F 按顺时针方向旋转 n°后(0<n <180 ),如果 EF ∥AB ,那么 n 的值是

2.(2017 秋?前郭县期末改编)将一副直角三角尺 ABC 和 CDE 按如图方式放置,其中直角顶点 C 重合,∠D=45°,∠A=30°.将三角形 CDE 绕点 C 旋转若 DE ∥BC ,则直线 AB 与直线 CE 的较大的夹角∠1 的大小为

度.

3.(2018 春?滨海县期中)长江汛期即将来临,防汛指挥部在一危险地带两岸各安置了一探照灯,便于夜间查看江水及两岸河堤的情况.如图,灯 A 射线自 AM 顺时针旋转至 AN 便立即回转,灯 B 射线自 BP 顺时针旋转至 BQ 便立即回转,两灯不停交叉照射巡视.若灯 A 转动的速度是 a°/秒,灯 B 转动的速度是 b°/秒,且 a 、b 满足|a ﹣3b |+(a +b ﹣4)2=0.假定这一带长江两岸河堤是平行的,即 PQ ∥MN ,且∠BAN=45°

(1) 求 a 、b 的值;

(2) 若灯 B 射线先转动 20 秒,灯 A 射线才开始转动,在灯 B 射线到达 BQ 之前,A 灯转动

几秒,两灯的光束互相平行?

(3) 如图,两灯同时转动,在灯 A 射线到达 AN 之前.若射出的光束交于点 C ,过 C 作 CD

⊥AC 交 PQ 于点 D ,则在转动过程中,∠BAC 与∠BCD 的数量关系是否发生变化?若不变, 请求出其数量关系;若改变,请求出其取值范围.

【模块三】压轴题——平行线的性质

1.(2017 春?南沙区期末)已知,直线AB∥DC,点P 为平面上一点,连接AP 与CP.

(1)如图1,点P 在直线AB、CD 之间,当∠BAP=60°,∠DCP=20°时,求∠APC.

(2)如图2,点P 在直线AB、CD 之间,∠BAP 与∠DCP 的角平分线相交于点K,写出∠AKC 与∠APC 之间的数量关系,并说明理由.

(3)如图3,点P 落在CD 外,∠BAP 与∠DCP 的角平分线相交于点K,∠AKC 与∠APC 有何数量关系?并说明理由.

2. (2017 春?武侯区校级期中)如图,已知AB∥CD,CE、BE 的交点为E,现作如下操作:第一次操作,分别作∠ABE 和∠DCE 的平分线,交点为E1,

第二次操作,分别作∠ABE1和∠DCE1的平分线,交点为E2,

第三次操作,分别作∠ABE2和∠DCE2的平分线,交点为E3,…,

第n次操作,分别作∠ABE n

﹣1和∠DCE n

﹣1

的平分线,交点为

E n.若∠E n=1 度,那∠BEC 等于度

3.(2018 春?黄陂区期中)已知直线AB∥CD.

(1)如图1,直接写出∠BME、∠E、∠END 的数量关系为;

(2)如图2,∠BME 与∠CNE 的角平分线所在的直线相交于点P,试探究∠P 与∠E 之间的数量关系,并证明你的结论;

(3)如图3,∠ABM=∠MBE,∠CDN= ∠NDE,直线MB、ND 交于点F,则= .

4.(2017春?丰城市期末)数学思考:(1)如图1,已知AB∥CD,探究下面图形中∠APC和∠PAB、∠PCD 的关系,并证明你的结论

推广延伸:(2)①如图2,已知AA1∥BA1,请你猜想∠A1,∠B1,∠B2,∠A2、∠A3的关系,并证明你的猜想;

、∠A n的关系

②如图3,已知AA1∥BA n,直接写出∠A1,∠B1,∠B2,∠A2、…∠B n

﹣1

拓展应用:(3)①如图4所示,若AB∥EF,用含α,β,γ的式子表示x,应为()

A.180°+α+β﹣γ

B.180°﹣α﹣γ+βC.β+γ﹣αD.α+β+γ

②如图5,AB∥CD,且∠AFE=40°,∠FGH=90°,∠HMN=30°,∠CNP=50°,请你根据上述结论直接写出∠GHM 的度数是.

【模块四】平移

1.如图,已知AM∥BN,∠A=60°.点P是射线AM上一动点(与点A不重合),BC、BD分别平分∠ABP 和∠PBN,分别交射线AM 于点C,D.

(1)求∠CBD 的度数;

(2)当点P 运动时,∠APB 与∠ADB 之间的数量关系是否随之发生变化?若不变化,请写出它们之间的关系,并说明理由;若变化,请写出变化规律.

(3)当点P 运动到使∠ACB=∠ABD 时,∠ABC 的度数是.

【模块五】作图

1.(1)如图1,一个牧童从P 点出发,赶着羊群去河边喝水,则应当怎样选择饮水路线,才能使羊群走的路程最短?请在图中画出最短路线.

(2)如图2,在一条河的两岸有A,B 两个村庄,现在要在河上建一座小桥,桥的方向与河岸方向垂直,桥在图中用一条线段CD 表示.试问:桥CD 建在何处,才能使A 到B 的路程最短呢?请在图中画出桥CD 的位置.

【模块六】几何巩固

1.(2018春?洪山区期中)如图,AB∥DE,∠ABC的角平分线BP和∠CDE的角平分线DK 的反向延长线交于点P 且∠P﹣2∠C=57°,则∠C 等于()

A.24°B.34°C.26°D.22°

第1 题图第2 题图

2.(2018春?高新区校级期中)如图,AB∥CD,∠ABK的角平分线BE的反向延长线和∠DCK 的角平分线CF 的反向延长线交于点H,∠K﹣∠H=27°,则∠K=()

A.76°B.78°C.80°D.82°

3.(2013春?汉阳区期末)如图,AB⊥BC,AE平分∠BAD交BC于点E

AE⊥DE,∠1+∠2=90°,M,N 分别是BA,CD 延长线上的点,∠EAM

和∠EDN 的平分线交于点F.下列结论:

①AB∥CD;②∠AEB+∠ADC=180°;③DE 平分∠ADC;④∠F 为定值

其中结论正确的有()

A.1 个B.2 个C.3 个D.4 个

4.(2018 春?开福区校级期末)学习了平行线后,小敏想出了过已知直线外一点画这条直线的平行线的新方法,她是通过折一张半透明的纸得到的,如图所示,由操作过程可知小敏画平行线的依据可以是.(把所有正确结论的序号都填在横线上)

①如果两条直线和第三条直线平行,那么这两条直线平行;②同位角相等,两直线平行;③ 内错角相等,两直线平行;④同旁内角互补,两直线平行.

5.(2018春?宁波期中)如图(1)所示为长方形纸带,将纸带沿EF折叠成图(2),再沿BF 折叠成图(3),继续沿EF折叠成图(4),按此操作,最后一次折叠后恰好完全盖住∠EFG;整个过程共折叠了9 次,问图(1)中∠DEF 的度数是.

6.(2017春?成都期中)已知AM∥CN,点B为平面内一点,AB⊥BC于B.

(1)如图1,直接写出∠A 和∠C 之间的数量关系;

(2)如图2,过点B 作BD⊥AM 于点D,求证:∠ABD=∠C;

(3)如图3,在(2)问的条件下,点E、F 在DM 上,连接BE、BF、CF,BF 平分∠DBC,BE 平分∠ABD,若∠FCB+∠NCF=180°,∠BFC=3∠DBE,求∠EBC 的度数.

7.(2017春?乐亭县期中)已知,∠AOB=90°,点C在射线OA上,CD∥OE.

(1)如图1,若∠OCD=120°,求∠BOE 的度数;

(2)把“∠AOB=90°”改为“∠AOB=120°”,射线OE 沿射线OB 平移,得O′E,其他条件不变,(如图2所示),探究∠OCD、∠BO′E的数量关系;

(3)在(2)的条件下,作PO′⊥OB 垂足为O′,与∠OCD 的平分线CP 交于点P,若∠BO′E=α,请用含α的式子表示∠CPO′(请直接写出答案).

8.(2017春?碑林区校级期中)探究:如图①,已知直线l1∥l2,直线l3和l1,l2分别交于点

C 和D,直线l3上有一点P.

(1)若点P 在C、D 之间运动时,问∠PAC,∠APB,∠PBD 之间有怎样的关系?并说明理由.

(2) 若点 P 在 C 、D 两点的外侧运动时(点 P 与点 C 、D 不重合),请尝试自己画图,写出

∠PAC ,∠APB ,∠PBD 之间的关系,并说明理由.

(3) 如图②,AB ∥EF ,∠C=90°,我们可以用类似的方法求出∠α、∠β、∠γ之间的关系,

请直接写出∠α、∠β、∠γ之间的关系.

9.(2017 春?锡山区校级月考)将一副三角板中的两块直角三角尺的直角顶点 C 按如图方式叠放在一起(其中,∠A=60°,∠D=30°;∠E=∠B=45°):

(1) 若∠DCE=35°,求∠ACB 的度数;

(2) 猜想∠ACB 与∠DCE 的数量关系,并说明理由;

(3) 请你动手操作,现将三角尺 ACD 固定,三角尺 BCE 的 CE 边与 CA 边重合,绕点 C 顺时

针方向旋转,当 0°<∠ACE <180°且点 E 在直线 AC 的上方时,这两块三角尺是否存在一组边互相平行?若存在,请直接写出∠ACE 角度所有可能的值(不必说明理由);若不存在,请说明理由

10. 如图,把△A B C 纸片沿 D E 折叠,当点 A 落在四边形 B C D E 的外部时,则

与 和

之间有一种数量关系始终保持不变,请试着找一找这个规律,你发现的规律是

A .2∠A = ∠1 - ∠2

C . 3∠A = 2∠1 - ∠2

B .3∠A = 2(∠1 - ∠2)

D . ∠A = ∠1 - ∠

2

+ =3 × + =10 × 11. 小宇同学在一次手工制作活动中,先把一张矩形纸片按图 1 的方式进行折叠,使折痕的

左侧部分比右侧部分短 1cm ;展开后按图 2 的方式再折叠一次,使第二次折痕的左侧部分比右 侧 部 分 长 1cm , 再 展 开 后 , 在 纸 上 形 成 的 两 条 折 痕 之 间 的 距 离 是 ( )

A .0.5cm

B .1cm

C .1.5cm

D .2cm

【模块七】阅读理解

1. 先阅读下列材料,然后解题:

阅读材料:因为(x -2)(x +3)=x 2+x -6,所以(x 2+x -6)÷(x -2)=x +3,即 x 2+x -6 能被 x -2 整除,所以 x -2 是 x 2+x -6 的一个因式,且当 x =2 时,x 2+x -6=0.

(1)类比思考:(x +2)(x +3)= x 2+5x +6 ,所以(x 2+5x +6)÷(x +2)= x +3,即 x 2+5x+6 能被

整除,所以

是 x 2+5x +6 的一个因式,且当 x =

时,x 2+5x +6=0.

(2)拓展探究:根据以上材料,已知多项式 x 2+mx -14 能被 x +2 整除,试求 m 的值. 2.求 1+2+22+23+…+22012 的值,可令 S =1+2+22+23+…+22012,则 2S =2+22+23+24+…+22013,因 此 2S -S =22013-1.仿照以上推理,计算出 1+5+52+53+…+52012 的值为

.

【模块八】求值问题

1.已知 x 2

-5x -1997=0,则代数式(x -2)2-(x -1)2+1的值为( )

x -2

A. 1999

B. 2000

C. 2001

D. -2

2. 若 , 则

n + m

的 值为 .

m n

3.已知(x -1)7

= a + a x + a x 2

+ a x 3

+..... + a x 7

,求 a + a + a + a

的值。

1

2

3

7

1

3

5

7

4. 已 知 2 + 2 = 22 ? 2 ,3 3 2 3,4+ 4 =42× 4 ,…,若 10 a 2 a (a 、b 为正整

3 3

8 8 15 15 b b 数),求分式a 2+2ab +b

2的值.

ab 2+a 2b

5.如果 A 是三次多项式,B 是五次多项式,那么 A +2B 是 次多项式。

【英语】 七年级英语下册任务型阅读易错题(word)

【英语】七年级英语下册任务型阅读易错题(word) 一、七年级英语下册任务型阅读专项目练习(含答案解析) 1.阅读下面短文,在文中填入与文章意思最符合的句子。 Every year, thousands of people get hurt(受伤)or die(死亡)when they are crossing the road.________Old people often get hurt or die because they can't see or hear very well. Children usually meet with accidents(事故)because of their carelessness.________ How can we lessen(减少)traffic accidents? Everybody should follow the traffic rules. For the drivers, they shouldn't drive too fast. If they drive too fast, it will be very difficult to stop the cars in a very short time.________ So, when we walk across the road, we must try to walk on the sidewalk.________ Look left first, next look right, then look left again. Only when we are sure that the road is clear, we can cross it. The right way to cross the road is to walk quickly.________ If people run across the road, they may fall down. We should try to help children, old people or blind people to cross the road, and never play in the street. A. It's not safe to run. B. They forget to look and listen before they cross the road. C. We must stop and look both ways before crossing the road. D. Most of these people are old people and children. E. For the walkers, it's very important to be careful when they are walking on the road. 【答案】 D;B;E;C;A 【解析】【分析】主要讲了怎样减少交通事故。 A. It's not safe to run.跑步不是安全的。 B. They forget to look and listen before they cross the road.在他们过马路前他们忘记看和听。 C. We must stop and look both ways before crossing the road.在通过马路前我们必须停下来看看两边。 D. Most of these people are old people and children.这些人中的大部分是老人和孩子。 E. For the walkers, it's very important to be careful when they are walking on the road.对于步行者来说,当穿过马路时仔细是非常重要的。 (1)根据后句可知主要讲了老人和孩子发生事故的原因,所以前句讲的是发生事故的大部分是老人和孩子,故选D。 (2)根据前句Children usually meet with accidents(事故)because of their carelessness.可知孩子们因为粗心发生交通事故,所以后句讲的是孩子们在通过马路前没有看和听,故选B。 (3)根据前句For the drivers, they shouldn't drive too fast. If they drive too fast, it will be very difficult to stop the cars in a very short time.可知讲的是对于驾驶者来说开车开得太快,所以后句讲的是对于步行者来说过马路时要小心,故选E。 (4)根据前句when we walk across the road, we must try to walk on the sidewalk.可知当我们过马路时要走人行道,所以后句讲的是过马路时怎么做,故选C。 (5)根据后句If people run across the road, they may fall down. 可知如果人们跑着通过马路,他们可能跌到,所以前句讲的是跑是不安全的,故选A。

七年级下册数学易错题整理附答案(超好)

七年级数学下易错题练习答案

第五章相交线与平行线 1.如图,将一张含有30°角的三角形纸片的两个顶点叠放在矩形的两条对边上,若∠2=44°,则∠1的大小为() A.14° B.16° C.90°﹣α D.α﹣44° 【解答】解:如图,∵矩形的对边平行, ∴∠2=∠3=44°, 根据三角形外角性质,可得∠3=∠1+30°, ∴∠1=44°﹣30°=14°, 故选:A. 2.如图,有一块含有30°角的直角三角板的两个顶点放在直尺的对边上.如果∠2=44°,那么∠1的度数是() A.14° B.15° C.16° D.17° 【解答】解:如图,∵∠ABC=60°,∠2=44°, ∴∠EBC=16°,∵BE∥CD, ∴∠1=∠EBC=16°,故选:C. 3.如图,直线a∥b,直线c分别交a,b于点A,C,∠BAC的平分线交直线b于点D,若∠1=50°,则∠2的度数是()

A.50°B.70° C.80° D.110° 【解答】∴∠2=180°﹣50°﹣50°=80°.故选:C. 4.如图把一个直角三角尺的直角顶点放在直尺的一边上,若∠1=50°,则∠2=() A.20°B.30° C.40° D.50° 【解答】解:∵直尺对边互相平行,故选:C. ∴∠3=∠1=50°,∴∠2=180°﹣50°﹣90°=40°. 5.如图,将矩形ABCD沿GH折叠,点C落在点Q处,点D落在AB边上的点E处,若∠AGE=32°,则∠GHC等于() A.112°B.110°C.108°D.106° 【解答】解:∵∠AGE=32°, ∴∠DGE=148°, 由折叠可得,∠DGH=∠DGE=74°, ∵AD∥BC, ∴∠GHC=180°﹣∠DGH=106°, 故选:D. 6.如图,AB∥CD,点E在线段BC上,∠CDE=∠CED.若∠ABC=30°,则∠D为() A.85°B.75° C.60° D.30° 【解答】故选:B.

苏教版七年级上英语易错题精华难题

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