七年级上《第四章基本平面图形》单元测试卷有答案

七年级上《第四章基本平面图形》单元测试卷有答案
七年级上《第四章基本平面图形》单元测试卷有答案

北师版七年级数学上册单元测试卷

班级姓名

第四章基本平面图形

A卷(共100分)

一、选择题(每小题3分,共30分)

1.下列关于作图的语句中叙述正确的是()

A.画直线AB=10 cm

B.画射线O B=10 cm

C.已知A,B,C三点,过这三点画一条直线

D.延长线段AB到点C

2.如图所示,线段共有()

A.4条B.5条C.6条D.7条

第2题图第3题图

3.如图所示,从点A到点F的最短路线是()

A.A→D→E→F B.A→C→E→F

C.A→B→E→F D.无法确定

4.如图所示,∠1+∠2=()

A.60°B.D.180°

第4题图第5

5.如图,O A是北偏东30°方向的一条射线,若射线O B与射线O A垂直,则O B的方向角是()

A.北偏西30°B.北偏西60°

C.东偏北30°D.东偏北60°

6.如图,下列关系错误的是()

A.∠A O C=∠A O B+∠B O C

B.∠A O C=∠A O D-∠C O D

C.∠A O C=∠A O B+∠B O D-∠B O C

D.∠A O C=∠A O D-∠B O D+∠B O C

7.已知线段AB=2 cm,BC=8 cm,则A、C两点间的距离为()

A.6 cm B.10 cm

C.6 cm或10 cm D.不超过10 cm

8.如图,O B是∠A O C的平分线,O D是∠C O E的平分线.如果∠A O B=50°,∠C O E=60°,则下列结论错误的是()

A.∠A O E=110°B.∠B O D=80°

C.∠B O C=50°D.∠D O E=30°

9.如图,长度为12 cm的线段AB的中点为M,点C将线段MB分成MC∶CB =1∶2的两部分,则线段AC的长度为()

A.2 cm B.8 cm

C.6 cm D.4 cm

10.不相等的有理数a,b,c在数轴上的对应点分别是A,B,C,如果|a-b|+|b-c|=|a-c|,那么点B()

A.在A,C点的左边

B.在A,C点的右边

C.在A,C点之间

D.上述三种均可能

二、填空题(每小题4分,共16分)

11.如图,点O是直线AB上的点,O C平分∠A O D,∠B O D=40°,则∠A O C =______°.

12.如图,点C是线段AB的中点,AB=6 cm,如果点D是线段AB上一点,且BD=1 cm,那么CD=____cm.

13.时钟表面3时30分时,时针与分针的夹角的度数是____;8时20分时,时针和分针的夹角的度数是____.

三、解答题(本大题共6小题,共54分)

15.(6分)一串有趣的图案按一定的规律排列:

按此规律求出第2 017个图案.

16.(8分)[2016春·翔安区期末]如图是按规律摆放在墙角的一些小正方体,从上往下分别记为第一层,第二层,第三层,…,第n层.

(1)第三层有________个小正方体;

(2)从第四层至第六层(含第四层和第六层)共有________个小正方体;

(3)第n层有________个小正方体;

(4)若每个小正方体边长为a分米,共摆放了n层,则要将摆放的小正方体能看到的表面部分涂上防锈漆,则防锈漆的总面积为________平方分米.

17.(10分)如图,已知点C是线段AB上一点,AC<CB,D,E分别是AB,CB的中点,AC=8,EB=5,求线段DE的长.

18.(10分)如图,已知点C=CD=DB,点P是线段CD的中点.

(1)图中共有几条线段?

(2)已知线段PD=2 cm,求线段AB的长.

19.(10分)如图,直线AB,CD相交于点O,且∠D O E=3∠C O E,∠E O B =90°,求∠A O D的度数.

20.(10分)如图,∠A O B=76°,O C为∠A O B内部一条射线,O M,O N分别平分∠B O C,∠A O C,求∠M O N的大小.

B卷(共50分)

四、填空题(本大题共5个小题,每小题4分,共20分)

21.根据图中箭头的指向规律,从2 013到2 014再到2 015,箭头的方向是________.

22.如图,AB=BC=________.

23.3时40分时,时针与分针所夹的角是________度.

24.如图,已知O E是∠A O C的平分线,O D是∠B O C的平分线,若∠A O C =110°,∠B O C=30°,则∠D O E=____.

第24题图第25题图

25.如图,平面内有共端点的六条射线O A ,O B ,O C ,O D ,O E ,O F ,从射线O A 开始按逆时针方向依次在射线上写出数字1,2,3,4,5,6,7,…,则“17”在射线____上,“2 017”在射线____上.

五、解答题(本大题共3个小题,共30分)

26.(8分)某数学活动小组的20位同学站成一列做报数游戏,规则是:从前

面第一位同学开始,每位同学依次报自己顺序的倒数加1,第1位同学报? ??

??11+1,第2位同学报? ??

??

12+1,…,求这样得到的20个数的积.

27.(9分)如图,一根5 m 长的绳子,一端拴在90°的围墙墙角的柱子上,另一端拴着一只小羊(羊只能在草地上活动),求小羊在草地上的最大活动区域的面积.

28.(13分)如图,B 是线段AD 上一动点,沿A →D →A 以2 cm/s 的速度往返运动1次,C 是线段BD 的中点,AD =10 cm ,设点B 的运动时间为t 秒(0≤t≤10).

(1)当t =2时, ①AB =____cm ;

②求线段CD 的长度;

(2)用含t 的代数式表示运动过程中AB 的长;

(3)在运动过程中,若AB 的中点为E ,则EC 的长是否变化?若不变,求出EC 的长;若发生变化,请说明理由.

参考答案

1. D

2. C

3. C

4. B

5. B

6. C

7. D

8. A

9. B 10. C

11. 70 12.2 13.75° 130°

【解析】 3点30分时针转过的角度为3.5×30°=105°,分针转过的角度为30×6°

=180°,时针与分针夹角为180°-105°=75°;8点20分时针转过的角度为? ?

?

?

?8+13×30°=240°+10°=250°,分针转过的角度为20×6°=120°,时针和分针的夹角为250°-120°=130°. 14.12

15. 解4为周期的,2 017=4×504+1,则第2 017

与第116.(1)6(2)46(3)n (n +1)2(4)32a 2

n (n +1) 17. 解:∵E 是CB 的中点,

∴CB =2EB =10. 又∵AC =8,

∴AB =AC +CB =18. ∵D 是AB 的中点,

∴DB =1

2AB =9. ∴DE =DB -EB =4.

18.解:(1)图中有线段AC ,AP ,AD ,AB ,CP ,CD ,CB ,PD ,PB ,DB ,共10条.

(2)因为AC =CD =DB =2PD =2×2=4(cm), 所以AB =3×4=12(cm).

19. 解:由∠D O E =3∠C O E ,且∠D O E +∠C O E =180°,

可得∠C O E =45°,∠E O D =135°. 又因为∠B O E =90°,所以∠B O D =∠E O D -∠B O E =135°-90°=45°. 所以∠A O D =180°-∠B O D =180°-45°=135°. 20. 解:因为O M ,O N 分别平分∠B O C ,∠A O C ,

所以∠M O C =12∠B O C ,∠N O C =1

2∠A O C ,

所以∠M O N =∠M O C +∠N O C =12∠B O C +12∠A O C =1

2(∠B O C +∠A O C )=1

2∠A OB .

因为∠A O B =76°,所以∠M O N =1

2×76°=38°.

21. 由图可知,每4个数为一个循环依次循环,2 012÷4=503,即0到2 011共2 012个数,构成前面503个循环,∴2 012是第504个循环组的第1个数,2 013是第504个循环组的第2个数,∴从2 013到2 014再到2 015,箭

22.10 23.130 24.40°25. O E O A

26. 解:由题意得,20个数的积=21×32×43×…×21

20=21. 27. 解:如答图所示,大扇形的圆心角是90°,半径是5 m .

答图

所以大扇形的面积为90360×π×52=25π

4(m 2), 小扇形的圆心角是180°-120°=60°, 半径是5-4=1(m),

则小扇形的面积为60360×π×12=π

6(m 2).

所以小羊A 在草地上的最大活动区域的面积为25π4+π6=77π12(m 2

). 28.

解:(1)①当t =2时,AB =2t =2×2=4(cm); ②∵AD =10 cm ,AB =4 cm , ∴BD =10-4=6(cm). ∵C 是线段BD 的中点,

∴CD =12BD =1

2×6=3(cm).

(2)∵B 是线段AD 上一动点,沿A →D →A 以2 m/s 的速度往返运动, ∴0≤t≤5时,AB =2t cm ;5

∵AB 的中点为E ,C 是线段BD 的中点,

∴EC =12(AB +BD )=12AD =1

2×10=5(cm).

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