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1-2007年山东省临沂市初中毕业与高中招生考试

发布时间:2014-01-30 09:53:56  

2007年临沂市初中毕业与高中招生考试试题

一、选择题(本大题共14小题,每小题3分,共42分)在每小题所给的四个选项中,只有一项是符合题目要求的.

1.?5的绝对值是( )

A.?5 B.5 C.1 5D.?1 5

2.据了解,我市每年用于校舍维护维修的资金约需7300万元,用科学记数法表示这一数据为( )

A.7.3?10 元 6B.73?10元 6 C.7.3?10元 7 D.73?10元 7

3.下列运算正确的是( )

A.x?x?x 358B.(x)?x

?329 C.x?x?x 437 D.(x?3)?x?9 224.如图,点D,E分别在AB,AC上,则∠( ) ∠A?50,△ABC中,1?∠2的大小为

A.130 ? B.230 ? C.180 ? D.310 ?A5

的结果是( ) A.6 B

.C

.6 D.12 (

第4题图

)

6.如图表示一个用于防震的L形的包装用泡沫塑料,当俯视这一物体时看到的图形形状是( )

A. B. C. D.

(第6题图) 7.若a?b?0,则下列式子:

①a?1?b?2; a?1; b

③a?b?ab; 11④?中,正确的有( ) ab②

A.1个 B.2个 C.3个 D.4个

8.已知反比例函数y?

k的图象在第二、四象限内,函数图象上有两点Ay1),x

B(5,y2),则y1与y2的大小关系为( )

A.y1?y2 B.y1?y2 C.y1?y2 D.无法确定

9.直线l1:y?k1x?b与直线l2:y?k2x在同一平面直角坐标系中的图象如图所示,则

第 1 页 共 11 页

关于x的不等式k1x?b?k2x的解为( )

A.x??1 B.x??1 C.x??2 D.无法确定

Cx x A B ?b (第10题图) (第9题图)

10.如图,在△ABC中,AB?2,AC?1,以AB为直径的圆与AC相切,与边BC交于点D,则AD的长为( )

C

D

11.如图,矩形ABCD中,AB?1,AD?2,M是CD的中点,点P在矩形的边上沿A

B

则△APM的面积y与点P经过的路程x之间的函数关系用图象A?B?C?M运动,

表示大致是下图中的( )

C D A.

B.

AB

(第11题图)

C. D.

12.小明随机地在如图所示的正三角形及内部区域投针,则针扎到其内切圆(阴影)区域的概率为( )

A.1 2

? 9 B

.? 6 ? (第12题图) C

. D

13.如图,客轮在海上以30km/h的速度由B向C航行,在B处测得灯塔A的方位角为北偏东80,测得C处的方位角为南偏东25,航行1小时后到达C处,在C处测得A的方位角为北偏东20,则C到A的距离是( )

???

第 2 页 共 11 页

A

. B

. C

.km D

.km

(第14题图

)

(第13题图)

14.如图,某厂有许多形状为直角梯形的铁皮边角料,为节约资源,现要按图中所示的方法从这些边角料上截取矩形(阴影部分)铁皮备用,当截取的矩形面积最大时,矩形两边长x,y应分别为( )

A.x?10,y?14

C.x?12,y?15 B.x?14,y?10 D.x?15,y?12

二、填空题(本大题共5小题,每小题3分,共15分)把答案填在题中横线上.

a?a2?9?3a?? 15.计算:???a?3a?3?a

16.从数字1,2,3中任取两个不同数字组成一个两位数,则这个两位数大于21的概率是 .

17.如图,在四边形ABCD中,E,F,G,H分别是AB,BD,CD,AC的中点,要使四边形EFGH是菱形,四边形ABCD还应满足的一个条件是 .

B E (第17题图) 图2

图1 (第18题图)

18.有如图所示的两种广告牌,其中图1是由两个等腰直角三角形构成的,图2是一个矩形,从图形上确定这两个广告牌面积的大小关系,并将这种大小关系用含字母a,b的不等式表示为 .

19.如果一个数等于它的不包括自身的所有因数之和,那么这个数就叫完全数.例如,6的不包括自身的所有因数为1,2,3.而且6?1?2?3,所以6是完全数.大约2200多年前,欧几里德提出:如果2?1是质数,那么2nn?1?(2n?1)是一个完全数,请你根据这个结论写出6之后的下一个完全数是 .

三、开动脑筋,你一定能做对!(本大题共3小题,共20分)

20.(本小题满分6分)

第 3 页 共 11 页

某校为了解全校2000名学生的课外阅读情况,在全校范围内随机调查了50名学生,得到他们在某一天各自课外阅读所用时间的数据,将结果绘制成频数分布直方图(如图所示).

(1)这50名学生在这一天课外阅读所用时间的众数是多少?

(2)这50名学生在这一天平均每人的课外阅读所用时间是多少?

(3)请你根据以上调查,估计全校学生中在这一天课外阅读所用时间在1.0小时以上(含1.0小时)的有多少人? 人数(人)

(小时) 时间

(第20题图

)

21.(本小题满分6分)

“种粮补贴”惠农政策的出台,大大激发了农民的种粮积极性,某粮食生产专业户去年计划生产小麦和玉米共18吨,实际生产了20吨,其中小麦超产12%,玉米超产10%,该专业户去年实际生产小麦、玉米各多少吨?

22.(本小题满分8分)

如图,已知矩形ABCD.

(1)在图中作出△CDB沿对角线BD所在的直线对折后的△C?DB,C点的对应点为C?(用尺规作图,保留清晰的作图痕迹,简要写明作法);

(2)设C?B与AD的交点为E,若△EBD的面积是整个矩形面积的

数. A

C

(第22题图)

四、认真思考,你一定能成功!(本大题共2小题,共19分)

23.(本小题满分9分) 1,求∠DBC的度3D

∠ADC?120,如图,已知点A,B,C,D均在已知圆上,AD∥BC,AC平分∠BCD,

四边形ABCD的周长为10.

(1)求此圆的半径;

(2)求图中阴影部分的面积.

?BC (第23题图)

第 4 页 共 11 页

24.(本小题满分10分)

某工程机械厂根据市场需求,计划生产A,B两种型号的大型挖掘机共100台,该厂所筹生产资金不少于22400万元,但不超过22500万元,且所筹资金全部用于生产此两型挖掘机,

(1)该厂对这两型挖掘机有哪几种生产方案?

(2)该厂如何生产能获得最大利润?

(3)根据市场调查,每台B型挖掘机的售价不会改变,每台A型挖掘机的售价将会提高m万元(m?0),该厂应该如何生产可以获得最大利润?(注:利润=售价-成本)

五、相信自己,加油呀!(本大题共2小题,共24分)

25.(本小题满分11分)

如图1,已知△ABC中,AB?BC?1,把一块含30角的直角三角板DEF∠ABC?90,

的直角顶点D放在AC的中点上(直角三角板的短直角边为DE,长直角边为DF),将直角三角板DEF绕D点按逆时针方向旋转.

(1)在图1中,DE交AB于M,DF交BC于N.

①证明DM?DN;

②在这一旋转过程中,直角三角板DEF与△ABC的重叠部分为四边形DMBN,请说明四边形DMBN的面积是否发生变化?若发生变化,请说明是如何变化的?若不发生变化,求出其面积;

(2)继续旋转至如图2的位置,延长AB交DE于M,延长BC交DF于N,DM?DN是否仍然成立?若成立,请给出证明;若不成立,请说明理由;

(3)继续旋转至如图3的位置,延长FD交BC于N,延长ED交AB于M,DM?DN是否仍然成立?请写出结论,不用证明.

E

图2 图3

26.(本小题满分13分) (第25题图) ??

如图1

,已知抛物线的顶点为A(2,1),且经过原点O,与x轴的另一个交点为B.

(1)求抛物线的解析式;

(2)若点C在抛物线的对称轴上,点D在抛物线上,且以O,C

,D,B四点为顶点的四边形为平行四边形,求D点的坐标;

第 5 页 共 11 页

(3)连接OA如图2,在x轴下方的抛物线上是否存在点P,使得△OBP与△OAB,AB,

相似?若存在,求出P点的坐标;若不存在,说明理由.

x

图1

(第26题图)

图2

数学试题参考答案及评分标准

注:第三、四、五题给出了一种解法或两种解法,考生若用其它解法,应参照本评分标准给分.

15.2a?12;

16.

1; 2

17.AD?BC或四边形ABCD等腰梯形(本小

18.

题答案不唯一,填出一个即得满分);

1212

a?b?ab; 22

19.28.

三、开动脑筋,你一定能做对!(共20分)

20.解:(1)众数是1.0小时; ·······

·························

····················································· (2分) (2)x?

1

(0.5?15?1.0?20?1.5?10?2.0?5) 50

, ?1.05(小时)

这50名学生这一天平均每人的课外阅读时间是1.05小时; ······································· (4分)

第 6 页 共 11 页

(3)2000?35?1400. 50

全校学生中这一天课外阅读时间在1.0小时以上(含1.0小时)的约为1400人.

···················································································································· (6分)

21.解:设原计划生产小麦x吨,生产玉米y吨,

根据题意,得

?x?y?18, ····························································································· (3分) ?12%x?10%y?20?18.?

解得??x?10, ·················································································································· (5分) y?8.?

,8?(1?10%)?8.8(吨). 10?(1?12%)?11.2(吨)

答:该专业户去年实际生产小麦11.2吨,玉米8.8吨. ·············································· (6分)

22.(1)作图正确,痕迹清晰. ···················································································· (2分) 作法:

①作∠MBD?∠CBD;

②在BM上截取BC??BC,连结C?D,

则△C?BD就是所求作的三角形; ················································································ (4分)

(2)解:由S△BED?12S梯形ABCD,得S△BED?S△ABD. 33

?S△BED?2S△ABE ········································································································· (5分) ?AD∥BC,?∠CBD?∠EDB.

又?∠EBD?∠DBC,?∠EBD?∠EDB. ························································· (6分)

··································································································· (7分) ?BE?ED?2AE. ·

又?∠A?90,?∠ABE?30. ??

1··············································································· (8分) ?∠CBD?(90??30?)?30?. ·2

四、认真思考,你一定能成功!(共19分)

23.解:(1)?AD∥BC,∠ADC?120,

········································································································· (1分) ?∠BCD?60?. ·

又?AC平分∠BCD, ?

?∠DAC?∠ACB?∠DCA?30?. ········································································· (2分)

?,∠B?60?. ·??AB??AD?CD··············································································· (3分)

?∠BAC?90?,

··············································································· (4分) ?BC是圆的直径,BC?2AB. ·

第 7 页 共 11 页

?四边形ABCD的周长为10,

?AB?AD?DC?2,BC?4.

······································································································ (5分) ?此圆的半径为2. ·

(2)设BC的中点为O,由(1)可知O即为圆心. ················································· (6分) 连接OA,OD,过O作OE?AD于E.

在Rt△AOE中,∠AOE?30,

?

?OE?OA?cos30??

1······················································································ (8分)

?S△AOD??2? ·2

?S阴影?S扇形AOD?S△OAD60???222?······································ (9分) ?? ·???3

24.解:(1)设生产A型挖掘机x台,则B型挖掘机可生产(100?x)台,

由题意知22400≤200x?240(100?x)≤22500, ················································· (1分) 解得37.5≤x≤40. ··································································································· (2分) ?x取非负整数,?x为38,39,40.

?有三种生产方案:

······· (3分) A型38台,B型62台;A型39台,B型61台;A型40台,B型60台. ·

(2)设获得利润W(万元),

由题意知W?50x?60(100?x)?6000?10x. ························································ (4分)

, ?当x?38时,W最大?5620(万元)

即生产A型38台,B型62台时,获得利润最大. ···················································· (5分)

(3)由题意知W?(50?m)x?60(100?x)?6000?(m?10)x, ·························· (6分) ?当0?m?10,则x?38时,W最大,

即A型挖掘机生产38台,B型挖掘机生产62台; ···················································· (7分) 当m?10时,m?10?0,三种生产方案获得利润相等; ········································· (8分) 当m?10,则x?40时,W最大, ············································································ (9分) 即A型挖掘机生产40台,B型挖掘机生产60台. ·················································· (10分)

五、相信自己,加油呀!(共24分)

25.(1)①证明:连结DB.

在Rt△ABC中,?AB?BC,AD?DC.

········································································ (1分) ?DB?DC?AD,∠BDC?90?. ·

方法一:

?∠ABD?∠C?45?.

?∠MDB?∠BDN?∠CDN?∠BDN?90?,

第 8 页 共 11 页

?∠MDB?∠NDC.

?△BMD≌△CND.

·············································································································· (3分) ?DM?DN.

方法二:

?∠A?∠DBN?45?.

?∠ADM?∠MDB?∠BDN?∠MDB?90?.

?∠ADM?∠BDN.

?△ADM≌△BDN.

·············································································································· (3分) ?DM?DN.

②四边形DMBN的面积不发生变化; ········································································· (4分) 由①知:△BMD≌△CND,

?S△BMD?S△CND.

?S四边形DMBN?S△DBN?S△DMB?S△DBN?S△DNC?S△DBC?11·········· (6分) S△ABC?. ·24

(2)DM?DN仍然成立,························································································· (7分) 证明:连结DB.

在Rt△ABC中,?AB?BC,AD?DC,

?DB?DC,∠BDC?90?.

?∠DCB?∠DBC?45?.

?∠DBM?∠DCN?135?.

?∠NDC?∠CDM?∠BDM?∠CDM?90?,

?∠CDN?∠BDM.

?△CDN≌△BDM.

·············································································································· (9分) ?DM?DN.

(3)DM?DN. ······································································································ (11分)

26.解:(1)由题意可设抛物线的解析式为y?a(x?2)?1. ································ (1分) 2

?抛物线过原点,

?0?a(0?2)2?1.

1?a??. 4

1?抛物线的解析式为y??(x?2)2?1, 4

12即y??x?x. ········································································································· (3分) 4

(2)如图1,当四边形OCDB是平行四边形时,

第 9 页 共 11 页

∥. CD 1由?(x?2)2?1?0, 4

得x1?0,x2?4, x 图1 ?B(4,0),OB?4.

································································································· (5分) ?D点的横坐标为6. ·

1将x?6代入y??(x?2)2?1, 4

1得y??(6?2)2?1??3, 4

················································································································ (6分) ?D(6,?3); ·

根据抛物线的对称性可知,在对称轴的左侧抛物线上存在点D,使得四边形ODCB是平行四边形,此时D点的坐标为(?2,······································································ (7分) ?3), ·

当四边形OCBD是平行四边形时,D点即为A点,此时D点的坐标为(2,····· (8分) 1). ·

(3)如图2,由抛物线的对称性可知:

AO?AB,∠AOB?∠ABO.

若△BOP与△AOB相似,

x 必须有∠POB?∠BOA?∠BPO. ························

9分) 设OP交抛物线的对称轴于A?点,

显然A?(2,?1),

1············································································ (10分) ?直线OP的解析式为y??x. ·2

112由?x??x?x,得x1?0,x2?6. 24

················································································································ (11分) ?P(6,?3).

过P作PE?x轴,

在Rt△BEP中,BE?2,PE?3,

?PB???4.

?PB?OB.?∠BOP?∠BPO.

···················································································· (12分) ?△PBO与△BAO不相似, ·

同理可说明在对称轴左边的抛物线上也不存在符合条件的P点.

所以在该抛物线上不存在点P,使得△OBP与△OAB相似. ······························· (13分)

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