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顺义区2013届初三第一次练习

一、选择题（本题共32分，每小题4分）下面各题均有四个选项，其中只有一个．．是符合题意的． 1．3-的倒数是（）

A ． 13

B ． 13

C ． 3-

D ．3

2．据2013年4月1日《CCTV —10讲述》栏目报道，2012年7月11日，一位26岁的北京小伙樊蒙，推

着坐在轮椅上的母亲，开始从北京到西双版纳的徒步旅行，圆了母亲的旅游梦，历时93天，行程3 359公里．请把3 359用科学记数法表示应为（）

A ．233.5910?

B ．43.35910?

C ．33.35910?

D ．433.5910? 3．下面四个几何体中，俯视图为四边形的是（）

4．我区某一周的最高气温统计如下表：

则这组数据的中位数与众数分别是（） A ．17，17 B ． 17，18 C ．18，17 D ．18，18

5．下列计算正确的是（）

A ．235a a a +=

B ．236a a a ?= C. 235()a a = D. 6．如图，A B ∥

D ，点

E 在B C 上，68B E D ∠=?，38D ∠=?，则B ∠的度数为（） A ． 30? B ． 34? C ． 38? D ．68? 7．若x y ，

为实数，且30x ++

=，则2013

y x ??

的值为（）

A ．1

B ． 1-

C ． 2

D ． 2-

8．如图，AB 为半圆的直径，点P 为AB 上一动点，动点P 从点A 出发，沿AB 匀速运动到点B ，运动时间为t ，分别以AP 和PB 为直径作半圆，则图中阴影部分的面积S 与时间t 之间的函数图象大致为

A ．

B ．

C ．

D ．

二、填空题（本题共16分，每小题4分）

9．分解因式：231212ab ab a -+= ．

10．袋子中装有3个红球和4个黄球，这些球除颜色外均相同．在看不到球的条件下，随机从袋中摸出一

个球，则摸出红球的概率是_____________．

A B C

532

a a a

÷=

12．如图，边长为1的菱形A B C D 中，60D A B ∠=°，则菱形A B C D 的面积是，连结对角线

A C ，以A C 为边作第二个菱形11AC C D ，使160D AC ∠=°；连结1AC ，再以1AC 为边作第三个菱形

122AC C D ，使2160D AC ∠=°；……，按此规律所作的第n 个菱形的面积为___________．

三、解答题（本题共30分，每小题5分）

．计算：10

()4sin 60( 3.14)3

π-+?---．

14．解不等式组312(1)312

x x x -<+??

?+?

?，≥，并把解集在数轴上表示出来．

15．已知：如图，C A 平分B C D ∠, 点E 在A C 上，B C E C =，A C D C =．

求证：A D ∠=∠ ．

16．已知2

320a a +-=，求代数式2

31()9

a a a +

-+-的值．

E D C

C 1

D 1

D 2

C 2

D C

A B

图

17．如图，已知(2,2)A --，(,4)B n 是一次函数y kx b =+的图象和反比例函数m y x

=的图象的两个交点．

（1）求反比例函数和一次函数的解析式；（2）求A O B ?的面积.

18．某商店销售一种旅游纪念品，3月份的营业额为2000元，4月份该商店对这种纪念品打

8折销售，结果销售量增加30件，营业额增加800元，求该种纪念品3月份每件的销售价格是多少？

四、解答题（本题共20分，每小题5分）

19．已知：如图，四边形ABCD 中，对角线AC 、BD 相交于点E ，

BD D C ⊥，45A B D ∠=?，30A C D ∠=?

，AD C D ==AC 和BD 的长.

20．如图，已知A B C △，以A C 为直径的O 交A B 于点D ，点E 为

A D 的中点，连结C E 交A

B 于点F ，且B F B

C =.

（1）判断直线B C 与⊙O 的位置关系，并证明你的结论；（2）若O 的半为2,3cos 5

B =

,求C E 的长.

21．某课外实践小组的同学们为了解2012年某小区家庭月均用水情况，随机调查了该小区部分家庭，并将调查数据进行如下整理，

请解答以下问题：

（1）表中m = ，n = ；

（2）把频数分布直方图补充完整；

（3）求该小区用水量不超过15t 的家庭占被调查家庭总数的百分比；

（4）若该小区有1500户家庭，根据调查数据估计，该小区月均用水量超过20t 的家庭大约有多少户？

22．如图1，在四边形A B C D 中，A B C D =，E F 、分别是B C A D 、的中点，连结E F 并延长，分别与B A C D 、的延长线交于点M N 、，则B M E C N E ∠=∠（不需证明）．

小明的思路是：在图1中，连结B D ，取B D 的中点H ，连结H E H F 、，根据三角形中位线定理和平行

C 月用水量

频数（户）

124252015105

830

问题：如图2，在A B C

、的中点，

、分别是B C A D

=，E F

△中，A C A B

>，D点在A C上，A B C D

连结E F并延长，与B A的延长线交于点G，若60

∠=°，连结G D，判断A G D

△的形状并证明．

E F C

五、解答题（本题共22分，第23题7分，第24题7分，第25题8分）

23．已知关于x的方程2(32)220

-+++=

m x m x m

（1）求证：无论m取任何实数时，方程恒有实数根.

（2）若关于x的二次函数2(32)22

y m x m x m

=-+++的图象与x轴两个交点的横坐标均为正整数，且m为整数，求抛物线的解析式.

24．如图1，将三角板放在正方形A B C D上，使三角板的直角顶点E与正方形A B C D的顶点A重合．三角板的一边交C D于点F，另一边交C B的延长线于点.

（1）求证：E F E G

=；

（2）如图2，移动三角板，使顶点E始终在正方形A B C D的对角线A C上，其他条件不变，（1）中的结论是否仍然成立？若成立，请给予证明；若不成立，请说明理由；

（3）如图3，将（2）中的“正方形A B C D”改为“矩形A B C D”，且使三角板的一边经过点B，其他条

件不变，若A B a

=，B C b

=，求E F

E G

的值．

25．如图，已知抛物线23

=++与y轴交于点A，且经过(1,0)(5,8)

y ax bx

、两点，点D是抛物线顶

B C

点，E是对称轴与直线A C的交点，F与E关于点D对称．

（1）求抛物线的解析式；

（2）求证：A F E C F E

∠=∠；

（3）在抛物线的对称轴上是否存在点P，使A F P

?相似．若有，请求出所有符合条件的点P的

?与F D C

坐标；若没有，请说明理由．

顺义区2013届初三第一次统一练习数学试题参考答案及评分参考

一、选择题

13．解：原式=3412

-- …………………………………………4分

=2 ……………………………………………… 5分 14．解：解不等式312(1)x x -<+，得3x <． ………………………………… 1分

解不等式

312

x +≥，得1x -≥． ………………………………… 2分

∴不等式组的解集为13x -<≤． ………………………………… 4分

在数轴上表示其解集为如图所示

…………………………………5分

15．证明：∵C A 平分B C D ∠

∴ A C B D C E ∠=∠ ……………………………………………1分在A B C ?和D E C ?中

∵BC EC AC B D C E AC D C =??

∠=∠??=?

……………………………………………3分 ∴A B C ?≌D E C ? …………………………………………… 4分 ∴A D ∠=∠ ……………………………………………5分

16．解：原式=2

333(

)(3)(3)

(3)(3)

a a a a a a a

--+

+-+- ………………………2分

3(3)(3)

a a a a a

………………………………………… 3分

=1(3)

a a +

13a a

+ ……………………………………………… 4分

∵ 2320a a +-=

∴ 232a a += ∴原式=

………………………………………………5分

17．解：（1）将(2,2)A --代入m y x

中，得4m =．

∴4y x

． …………………………………………………………………1分

将(,4)B n 代入4y x

中，得．1n = ………………………………2分

将(2,2)A --，(1,4)B 代入y kx b

=+中

22,

4.k b k b -+=-??+=?

………3分解得2,2.

k b =??=? ∴22y x =+． ……………………………………………4分（2）设直线AB 与y 轴交于点C 当0x =时，2y =．

∴2O C =．

∴11222132

A O

B A O

C B O C S S S ???=+=

??+

??= ………………………5分

18．解：设该种纪念品3月份每件的销售价格为x 元， ……………………………1分根据题意，列方程得

20002000800

300.8x

- ………………………………………………3分

解之得50x =． …………………………………………………………4分

经检验50x =是所得方程的解．

答：该种纪念品3月份每件的销售价格是50元． …………………………5分解法二：设3月份销售这种纪念品x 件，则4月份销售（x +30）件 …………1分

根据题意，列方程得

420002000800530

x +?=

+ ……………………………………………3分

解之得40x =． ………………………………………………4分经检验40x =是所得方程的解

答：该种纪念品3月份每件的销售价格是

20005040

=（元）…………5分

19解：∵ BD D C ⊥

∴ 90B D C ∠=?

∵ 30A C D ∠=?，AD C D == ∴ 60,30,DEC DAC ACD ∠=?∠=∠=?

t a n302

D E C D

=??==

∴24

E C D E

==，30

A D E

∠=?…………………………………………1分∴2

AE D E

==………………………………………………………2分- ∴246

A C A E E C

=+=+=………………………………………………3分过点A作AM BD

⊥,垂足为M

∵60

A E

B D E C

∠=∠=?

∴

sin602

AM AE

=??=?=

c o s6021

M E A E

=?=?=………………………………………………4分∵45

A B D

∠=?

∴BM AM

∴

123

BD BM M E DE

=++=+=+…………………………5分20．⑴B C与⊙O相切

证明：连接A E，

∵A C是O

的直径

∴90

∠=

∴90

E A D A

F E

∠+∠=?

∵B F B C

∴BC E BFC

∠=∠

又∵E为 A D的中点

∴EAD AC E

∠=∠…………………………1分

∴90

BC E AC E

∠+∠=?

即A C B C

⊥

又∵A C是直径

∴B C是O

的切线…………………………2分

（2）∵O

的半为2

∴4

A C=,

∵

cos

由（1）知，90

ACB

∠= ，

∴5

A B= ,3

B C=

∴3

B F= ,2

AF=………………………… 3分

∵EAD AC E

∠=∠, E E

∠=∠

∴AEF

?∽C E A

?，

∴

E A A F

E C C A

设 ,2EA x EC x ==

由勾股定理 22416x x +=

，5

x =± （舍负）

∴

C E =

…………………………5分

21．解：（1）表中填12m =；0.08n =. …………………………2分

(2）补全的图形如下图.

- …………………………3分

（3）0.120.240.320.68++=.

即月均用水量不超过15t 的家庭占被调查的家庭总数的68%.

…………………………4分（4）(0.080.04)1500180+?=.

所以，该小区月均用水量超过20t 的家庭大约有180户. ………………5分 22．判断A G D ?是直角三角形

证明：如图连结B D ，取B D 的中点H ，连结H F H E 、，……………………1分

F 是A D 的中点，

∴H F AB ∥，12

H F A B =，………………… 2分

∴13∠=∠．

同理，12H E C D H E C D =∥，， ∴2E F C ∠=∠．

A B C D = ，

∴H F H E =，

∴12∠=∠． …………………………………………3分

60EFC ∠= °，

∴360E F C A F G ∠=∠=∠=°，

∴A G F ?是等边三角形．………………………………4分 AF FD = ，

∴G F F D =，

∴30F G D F D G ∠=∠=° 30

805101520254

12频数（户）

月用水量

A B

D F G

H E

2 3

即A G D △是直角三角形．…………………………… 5分

23．（1）证明：①当0m =时，方程为220x -+=，所以 1x =，方程有实数根.…… 1分 ②当0m ≠时, []2

(32)4(22)m m m ?=-+-+ =22912488m m m m ++-- =244m m ++

=2(2)0m +≥ ………………………………2分所以,方程有实数根

综①②所述，无论m 取任何实数时，方程恒有实数根 …………3分（2）令0y =，则2(32)220mx m x m -+++= 解关于x 的一元二次方程，得11x = ，222x m

……………………5分

二次函数的图象与x 轴两个交点的横坐标均为正整数，且m 为整数，

所以m 只能取1，2

所以抛物线的解析式为254y x x =-+或2286y x x =-+………………7分 24．

（1）证明：∵9090G EB BEF D EF BEF ∠+∠=∠+∠=°，°，

∴.D EF G EB ∠=∠

又∵E D B E =，

∴R t R t F E D G E B △≌△.

∴.E F E G = ………………………………………………………2分

（2）成立.

证明：如图，过点E 分别作B C C D 、的垂线，垂足分别为H I 、，

则90E H E I H E I =∠=，°.

∵9090G EH H EF IEF H EF ∠+∠=∠+∠=°，°， ∴.IE F G E H ∠=∠

∴R t R t FEI G EH △≌△.

∴.E F E G = …………………………………4分

（3）解：如图，过点E 分别作B C C D 、的垂线，垂足分别为M N 、，则90M E N ∠=°，

.E M A B E N A D ∥，∥

∴

.E M C E E N

A B

C A

A D =

∴.E M A D a

E N A B b ==

…………………………………5分 ∴9090G M E M EF FEN M EF ∠+∠=∠+∠=°，°，

∴.M E N G E M ∠=∠

∴R t R t F E N G E M △∽△. ∴

.E F E N b

…………………………………7分

25．解：（1）将点(1,0)(5,8)B C 、代入2y ax bx =+30

25538a b a b ++=??

++=? ……………………1解之得14a b =??=-?

，

所以抛物线的解析式为2

43y x x =-+

……………………2分

（2）由（1）可得抛物线顶点(2,1)D -

……………………3分

直线A C 的解析式为3y x =+

由E 是对称轴与直线A C 的交点，则(2,E 由F 与E 关于点D 对称，则(2,7)F - ……………………4分

证法一：

从点,A C 分别向对称轴作垂线,AM CN ，交对称轴于,M N 在R t F A M ?和R t F C N ?中

90AMF CNF ∠=∠=，

21310

A M C N M F

N F

所以R t F A M ?∽R t F C N ?

所以A F E C F E ∠=∠

…………………………………5分

证法二：直线A F 的解析式为53y x =-+ 点 (5,8)C 关于对称轴的对称点是(1,8)Q - 将点(1,8)Q -代入53y x =-+可知点Q 在直线A F 所以A F E C F E ∠=∠

（3）在F D C ?中，三内角不等，且C D F ∠为钝角

若点P 在点F 下方时，

在A F P ?中，AFP ∠为钝角

因为A F E C F E ∠=∠，00

180,180AFE AFP CFE CDF ∠+∠=∠+∠<

所以AFP ∠和C D F ∠不相等

所以，点P 在点F 下方时，两三角形不能相似 …………………… 6分 20 若点P 在点F 上方时，

只需A F P F

C F

D F

=（点P在D F之间）或

A F P F

D F C F

=（点P在F D的延长线上）

解得点P的坐标为(2,3)

-或(2,19)………………………………………8分

2017年中考英语一模试题

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