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2012-2013学年北京市北方交大附中八年级下学期期中考试题

2012-2013学年北京市北方交大附中八年级下学期期中考试题
2012-2013学年北京市北方交大附中八年级下学期期中考试题

北方交大附中2012—2013学年度第一学期期中练习

初二数学

考试时长:90分钟满分:100分班级姓名

一、选择题:(每小题3分,共24分,下面各题均有四个选项,其中只有一个

..符合题意)1.下列各式中,分式的个数有()

3

1

-

x

1

2

+

a

b

π

y

x+

2

2

1

-

-

m

、a

+

2

1

2

2

)

(

)

(

y

x

y

x

+

-

x

1

2-、

11

5

-

(A) 2个 (B) 3个 (C) 4个 (D) 5个

2.下列图形中,是中心对称图形,但不是轴对称图形的是().

(A)正方形(B)矩形(C)菱形(D)平行四边形

3.如图,有一块长方形花圃,有少数人为了避开拐角走“捷径”,在花圃内走出了一条“路”,他们仅仅少走了()m的路,却踩伤了花草.

(A) 5 (B) 4 (C) 3 (D) 2

4.如果矩形的面积为62

cm,那么它的长y cm与宽x cm之间的函数关系用图

象表示大致为()

(A) (B) (C) (D) 5.如下图,E、F、G、H分别是四边形ABCD四条边的中点,要使四边形EFGH为矩形,四边形ABCD应具备的条件是().

(A)一组对边平行而另一组对边不平行(B)对角线相等

(C)对角线互相平分(D)对角线互相垂直

6.如上图,已知点A是函数x

y=与

x

y

4

=的图象在第一象限内的交点,点B在x轴负半轴上,且OB

OA=,则AOB

?的面积为()

(A) 2

2 (B) 2 (C)2 (D)4

7.已知四边形ABCD,给出下列4个条件①AB∥CD②AD∥BC ③AB=CD

④∠BAD=∠DCB,从以上4个条件中任选2个条件为一组,能推出四边形ABCD为

平行四边形的有()

(A) 3组 (B) 4组 (C) 5组(D)6组

D

C

B

A

H

G

F

E

x

A

B

C

D

E

G

8.任何一个正整数n 都可以进行这样的分解:t s n ?=(s 、t 是正整数,且s ≤t ),如果q p ?在n 的所有这种分解中两因数之差的绝对值最小,我们就称q p ?是最佳分解,并规定q

p

F n =)(.例如:18可以分解成1×18,2×9,3×6,这时就有2163)==n F (。结合以上信息,给出下列)n F (的说法:①2

12=)(F ;②8

3

24=

)(F ;③327=)(F ;④若n 是一个完全平方数,则1)=n F (,其中正确的序号是(). (A) ①④ (B) ①②③④ (C) ①②④ (D) ②③

二、填空题(每空2分,共18分) 9.已知分式

5

-x x

, 当x 时,分式有意义;当x 时,分式的值为零.

10.已知平行四边形ABCD 的对角线AC,BD 相交于点O ,AB=5,AO=4,BO=3,

则平行四边形的周长是_____,面积是_______.

11. 设双曲线x

y 3

=

上有两点),(11y x A 和),(22y x B ,且当210x x <<时,则21___y y (填<, = , >).

12.点A 是反比例函数图象上一点,它到原点的距离为10,到x 轴的距离为8,则此函数

表达式可能为_________________.

13.如图,把矩形ABCD 沿EF 折叠,使点C 落在点A 处,点D 落在点G 处,若

60=∠CFE ,且DE=1,则边BC 的长为 .

14.如图所示,矩形ABCD 的面积为102

cm ,它的两条对角线交于点1O ,以AB 、1AO 为邻边作平行四边形11O ABC ,平行四边形

11O ABC 的对角线交于点2O ,同样以AB 、2AO 为邻边作平行

四边形22O ABC ,……,依次类推,则平行四边形55O ABC 的面积为_________2

cm ,平行四边形n n O ABC 的面积为_________2

cm .

2

三、解答题(共58分)

15.(6分)已知2

10a a --=,求代数式11

1

a a --的值.

解:

16.(8分)先化简32222222

32a b a b a ab

ab a ab b a b +--÷++-, 然后请你自取一组,a b 的值代入求值.

解:

17. (6分) 如图,在△ABC 中,AD 是BAC ∠的平分线,DE ∥AC 交AB 于E ,DF ∥AB

交AC 于F ,求证:四边形AEDF 是菱形.

证明:

18.(5分)反比例函数x

k

y =

的图象与直线y =x -2交于点A ,且A 纵坐标为1,求该反比例函数的解析式. 解:

19.(6分) 某校师生到距学校20千米的公路旁植树,甲班师生骑自行车先走,45分钟后, 乙班师生乘汽车出发,结果两班师生同时到达,已知汽车的速度是自行车速度的2.5倍, 求两种车的速度每小时各是多少千米? 解:

F

C B

D E

A

20.(3分)现有一张正方形纸片,将它折两次(第一次折后也可打开铺平再折第二次),使得折痕 将纸片分为面积相等且不重叠的四部分(称为一个操作),如图甲(虚线表示折痕). 除图甲外,请你再给出三个不同的操作,分别将折痕画在图①至图③中.

(规定:一个操作得到的四个图形和另一个操作得到的四个图形,如果能够“配对”得到四组全等的图形,那么就认为是相同的操作,如图乙和图甲是相同的操作)

21 .(5分)在△ABC 中,AB=AC ,∠BAC=120°,过点C 作CD ∥AB ,且CD=2AB ,连结BD ,BD=2.求△ABC 的面积.

解:

22.(6分)如图1,P 为Rt △ABC 所在平面内任一点(不在直线AC 上),∠ACB=90°,M 为AB 的中点.操作:以P A 、PC 为邻边作平行四边形P ADC ,连结PM 并延长到点E ,使ME=PM ,连结DE .(1)请你猜想与线段DE 有关的三个结论,并证明你的猜想; (2)若将“Rt △ABC ”改为“任意△ABC ”,其他条件不变,利用图2操作,并写出与线段DE 有关的结论(直接写答案).

解:

图1

E M P

D

C

B

A

图2

B

A D C

B A 图甲 图② 图③ 图乙 图①

23.(8分) 已知反比例函数x

k

y =

图象过第二象限内的点),2(m A -,x AB ⊥轴于B ,Rt △AOB 面积为3, 若直线b ax y +=经过点A ,并且经过反比例函数x

k y =的图象上另一点)23

,(-n C .

(1)求反比例函数的解析式及直线b ax y +=的解析式;

(2)在y 轴上是否存在一点P ,使PAO ?为等腰三角形,若存在,请直接写出P 点坐标,若不存在,

说明理由. 解:

24.(5分)已知:如图①,正方形ABCD 中,E 为对角线BD 上一点,过E 点作EF ⊥BD 交BC 于F ,连接DF ,G 为DF 中点,连接EG ,CG . (1)求证:EG =CG ;

(2)将图①中△BEF 绕B 点逆时针旋转45o,如图②所示,取DF 中点G ,连接EG ,CG . 问(1)中的结论是否仍然成立?若成立,请给出证明;若不成立,请说明理由. (3)将图①中△BEF 绕B 点旋转任意角度,如图③所示,再连接相应的线段,

问(1)中的结论是否仍然成立?通过观察你还能得出什么结论?(均不要求证明)

解:

北方交大附中2012-2013学年度第二学期期中练习

数学答案及评分标准2013. 4

一、选择题(本题共24分,每小题3分)

二、填空题(本题共18分,每空2分)

9. 0;5=≠x x 10. 20; 24 11. <

12.

x y 48±

=(写对一个给1分) 13. 3 14. )

25

(210;1651-n n 或写成

三、解答题(共58分)

15(6分)

解:原式=)1()1(1--

--a a a

a a a ……………………………………………………………2分 =)1(1---a a a

a …………………………………………………………………………3分 =)1(1--

a a =a a --2

1

……………………………………………………………4分

1

0122=-∴=--a a a a

……………………………………………………………………5分

1-=∴原式……………………………………………………………6分

16(8分)

解:原式=

)())(()()(32

2b a a b a b a b a b a b a ab --+?++-………………………………………………4分 (两个公式用对各1分,两个提取公因式共1分,颠倒相乘1分)

=ab ab -3……………………………………………………………………………5分 =ab 2…………………………………………………………………………………6分 取2,1==b a 代入上式,原式=4……………………………………………………8分 注:取值时要求

≠≠a b a 且,取错一个都不给分

17(6分):

证明: DE ∥DF AC ,∥AB

为平行四边形四边形AEDF ∴.………………………..1分 EDA FAD ∠=∠∴………………………………….2分 AD 是BAC ∠的平分线

FAD EAD ∠=∠∴……………………………………………………….3分 FAD EAD ∠=∠∴……………………………………………………….4分 ED AE =∴………………………………………………………5分

为菱形四边形AEDF ∴……………………………………………………6分

18(5分)

解:21-==x y y 代入把………………………………1分

321=-=x x

……………………………………………………………2分

)1,3(A ∴……………………….………3分(注:没写A 点坐标不扣分)

3131

3=?==

???==k x k y y x 代入把

………………………………………………………………4分

x y 3

=

∴…………………………………………………………5分

19(6分)

解:设自行车速度为x 千米/小时,则汽车速度为x 5.2千米/小时,……………………………1分 由题意可列方程为x x 5.220604520=

-……………………………3分

解得16=x ,………………………………………………………………………………4分

经检验,16=x 适合题意,……………………………………………………………………………5分

F C

B

D

E A

405.2=∴x

答:自行车速度为16千米/小时,汽车速度为40千米/小时.……………………………6分 20(3分)

解:画出1解给1分,画出2解给2分,画出3解给3分下面各图供参考:

21

(5分)

解法1: 过点B 作F CB AF A E CD AC BE 于作,过点于交⊥

//

AC AB AB CD =,// 是菱形四边形ABEC ∴

2

,260,60,301200

000

===∴?∴=∴==∠=∠=∠∴=∠==∴BD BE AB BED BE

DE AB CD BED ABE ABC BAC AB CE BE 是等边三角形,

,,中在003090,=∠=∠?ABC AFB ABF 1,3==∴AF BF 332的面积为ABC BC ?∴=∴

(或是连接AE ,由ABEC 为菱形,得AE ⊥BC ,其余同上)

解法2: 如图,在Rt △DFB 中求得高BF ,其余同解法1

解法3:如图,过C 作CE ∥DB ,交BA 延长线于E ,过C 作CF ⊥AE 于F , 可证CEBD 为平行四边形,CE=DB=2

再证:△CEA 为等边三角形,可得AB=AC=CE=2 ,

在Rt △CFA 中求得高CF=3,

∴33221

21=??=?=

?CF AB S ABC

解法4:过A 作AE ⊥CD 于E ,过B 作BF ⊥CD 于F ,

在Rt △AEC 中求得CE=21AC=21

AB

又由ABFE 为矩形,可得AB=EF ,AE=BF , ∵CD=2AB

∴DF= 21

AB=CE

可证△AEF ≌△BFD

∴AC=DB=2,可求得: AB=2,AE=3

33221

21=??=?=

?CF AB S ABC

辅助线1分,证出菱形1分,证出等边三角形1分,运用勾股定理1分,求出面积1分,共5分。

22(6分)

解:(1)DE ∥BC,DE=BC,DE ⊥AC. …………………1分 连结BE.延长ED 与AC 交于点F. PM=ME,AM=BM,∠PMA=∠EMB, ∴△PMA ≌△EMB. ………………2分

∴PA=EB,∠MPA=∠MEB.

∴PA ∥BE.

四边形APCD 是平行四边形,

∴PA ∥CD,PA=CD. ∴BE ∥CD,BE=CD. ∴四边形DCBE 是平行四边形.

∴DE ∥BC,DE=BC. ………………………………3分

∴∠EFA=∠ACB.

∠ACB=90°, ∴∠EFA=90°.

∴DE ⊥AC. ……………………………………4分

(2)

DE ∥BC,DE=BC. ………………………………………………6分

23(8分) 解:(1)

3=?AO B S

3221

321=?=?∴m BO AB ,即

3=∴m ………………………………………………1分

F

图1E M P D

C

B

A A

6

)3,2(==-k x k y A 得:代入把

x y 6

-

=∴反比例函数解析式为………………………………………………2分

46

)23,(=-

=-n x

y n C 得代入把

………………………………………………3分 23

,43)2

3

,4(3,2-=

-=+=-b a b

ax y C A 得:代入)、(把

23

43+

-=∴x y 直线解析式为………………………………………………4分

(2)

)

613

0(1,时,当P PO PA =;)60(2,时,当P AO PA = )130()130(43-=,或,时,当P P PO OA ……………………………………………8分

(注:每个P 点各1分)

24(5分) 解:(1)证明:如图①,在Rt △FCD 中, ∵G 为DF 的中点,

∴ CG=1

2FD .…………………………………………..1分 同理,在Rt △DEF 中,EG=1

2FD .

∴ CG=EG .…………………………………………….2分

(2)(1)中结论仍然成立,即EG=CG .(只写结论,没证出来不给分)

证法一:如图②(一),连接AG ,过G 点作MN ⊥AD 于M ,与EF 的延长线交于N 点. 在△DAG 与△DCG 中,

∵ AD=CD ,∠ADG=∠CDG ,DG=DG ,

∴△DAG ≌△DCG .

∴ AG=CG . 在△DMG 与△FNG 中,

∵∠DGM=∠FGN ,FG=DG ,∠MDG=∠NFG ,

∴△DMG ≌△FNG .

∴ MG=NG

F A D C E

G M N N 图②(一)

在矩形AENM 中,AM=EN . 在Rt △AMG 与Rt △ENG 中, ∵ AM=EN , MG=NG , ∴△AMG ≌△ENG . ∴ AG=EG .

∴ EG=CG .……………………………………………………4分

证法二:如图②(二),延长CG 至M ,使MG=CG , 连接MF ,ME ,EC ,

在△DCG 与△FMG 中,

∵FG=DG ,∠MGF=∠CGD ,MG=CG , ∴△DCG ≌△FMG . ∴MF=CD ,∠FMG =∠DCG . ∴MF ∥CD ∥AB .

∴EF MF ⊥.

在Rt △MFE 与Rt △CBE 中,

∵ MF=CB ,EF=BE , ∴△MFE ≌△CBE ..

∴MEF CEB ∠=∠.

∴∠MEC =∠MEF +∠FEC =∠CEB +∠CEF =90°. ∴△MEC 为直角三角形.

∵ MG = CG ,∴ EG=21

MC .

∴EG CG =.……………………………………………4分

(3)如图③,(1)中的结论仍然成立,即EG=CG . 其他的结论还有:EG ⊥CG .………………………..5分 (两个结论都正确给1分)

F

B

A

D

C

E

图③

G

F A D C E

G M

图②(二)

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