2011年中考数学复习高分冲刺经典习题(5);(总分100分时间30分钟);

一、(本部分共10个小题，每小题4分，共40分.;B.;1313C.?D.3;2.经公安部交管局统计，今年5月份全国因道路交通;B.25.591?103C.2.5591?104;D.2.5591?106;3.如图1，平放在台面上的圆锥体的主视图是();4.下列图形中，既是轴对称图形，

2011年中考数学复习高分冲刺经典习题 (5)

(总分100分 时间30分钟)

一、(本部分共10个小题，每小题4分，共40分.每小题给出4个选项，其中只有一个是正确的) ..1.3的倒数是( ) A.?3

B.

1313 C.? D.3

2.经公安部交管局统计，今年5月份全国因道路交通事故造成伤亡共25591人.这个数据用科学记数法可以表示为( ) A.2.5591?105

B.25.591?103 C.2.5591?104

D.2.5591?106

3.如图1，平放在台面上的圆锥体的主视图是( )

4.下列图形中，既是轴对称图形，又是中心对称图形的是( )

图1

A.

B.

C.

D.

A. B. C. D.

5.某烟花爆竹厂从20万件同类产品中随机抽取了100件进行质检，发现其中有5件不合格，那么你估计该厂这20万件产品中合格品约为( ) ...A.1万件 B.19万件 C.15万件 D.20万件 6.化简

x?6x?92x?62的结果是( )

A.

x?32 B.

x?922 C.

x?922 D.

x?32

7.班长去文具店买毕业留言卡50张，每张标价2元，店老板说可以按标价九折优惠，则班长应付( ) A.45元 B.90元 C.10元 D.100元

8.二次函数y?ax?bx?c的图象如图2所示，若点A(1，y1)、B(2，y2)是它图象上的

y 两点，则y1与y2的大小关系是( ) A.y1?y2 B.y1?y2 x=? 3 图2

O x - 1 - 2

C.y1?y2 D.不能确定

?2x?6?6?2x?9.不等式组?3?x的整数解是( )

?2x?1??22 B.1，，23 C.A.1，13A O

B

图3

D E C

1，2 ?x?3 D.0，10.如图3，在矩形ABCD中，DE?AC于E，

?EDC∶?EDA?1∶3，且AC?10，则DE的长度是( )

A.3 B.5 C.52 D.

522

y A C 二、填空题(本题共6个小题，每小题4分，共24分) 11.计算：?y3?2?y? .

?3512.如图4，A为反比例函数y?x第二象限上的任一点，AB?x轴于B，

的图象在 B O x 则矩形ABOC的面积S? . AC?y轴于C.图4

13.为了准备毕业联欢的抽奖活动，小华准备了10个白球，2个红球，8个黄球，每个球除颜色外都相同，把它们放入不透明的口袋中搅匀，规定每位同学每次抽奖，只能从袋中摸出一个球，记下颜色后放回，摸到红球可获钢笔一支.那么小亮抽奖一次得到钢笔的概率是 .

B 14.如图5，小明利用升旗用的绳子测量学校旗杆BC的高度，

他发现绳子刚好比旗杆长11米，若把绳子往外拉直，绳子接触地面A点并与地面形成30°角时，绳子未端D距

A点还有1米，那么旗杆BC的高度为 .

15.下面是按一定规律摆放的图案，按此规律，第2009个图D 案与第1~4个图案中相同的是 .(只填数字)

16.如图6，在Rt△ABC中，?C?90°，点D是BC上一点，

AD?BD，若AB?8，BD?5，则CD? .

A 30° 图5

C …

第1个

第2个

第3个

第4个

第5个

A 第6个

三、解答题(本题共4小题，共36分)

C

D 图6

B

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17.(本题7分)计算：?4?

18.(本题8分)解分式方程：

9?3?2π?? ??2009??.2??0xx?1?31?x ?3.

19.(本题9分)随着网络的普及，越来越多的人喜欢到网上购物.某公司对某个网站2005年到2008年网上商店的数量和购物顾客人次进行了调查.根据调查结果，将四年来该网站网上商店的数量和每个网上商店年平均购物顾客人次分别制成了折线统计图(如图7)和条形统计图(如图8).请你根据统计图提供的信息完成下列填空：

个 90 80 70 60 50 40 30 20 10 0

2005

万人次 50 45 40 35 30 25 20 15 10 5

0 2005

2006 2007 2008 年

2006

图8

2007 2008 年

每年网上商店的数量 每个网上商店年平均购物顾客人次

图7

(1)2005年该网站共有网上商店 个;(2分)

(2)2008年该网站网上购物顾客共有 万人次;(2分)

(3)这4年该网站平均每年网上购物顾客有 万人次.(2分)

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20.(本题10分)如图9，四边形ABCD是正方形，BE?BF，BE?BF，EF与BC交于点G.

(1)求证：△ABE≌△CBF;(4分) (2)若?ABE?50°，求?EGC的大小.(4分)

F 图9 B

E G

C

A

D

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参考答案

第一部分 选择题 题号 1 2 答案 填空题 题号 答案

(注：第13题的答案写成解答题

17.解：原式=4?3?=

1919?1 ··············································································· 1+1+1+1分 2203 A 4 D 5 B 6 D 7 B 8 C 9 A 10 D B C 11 y 12 3 13 11014 10 15 1 16 75 的不扣分，第16题的答案写成

1410的不扣分)

·········································································································· 5分

(注：运算的第一步正确一项给1分) 18.解：

xx?1?3x?1?3 ····························································································· 1分

去分母，得：x?3?3x?3····················································································· 3分 解得：x?3 ············································································································ 5分

经检验：x?3是原方程的根， ∴原方程的根是：x?3. ······················································································· 5分 (注：不检验扣1分;只要验根，用其他方式书写不扣分.) 19.解：(1)20;(2)3600;(3)1250.

(注：每小题答对给2分)

20.(1)证明：?四边形ABCD是正方形，BE?BF ?AB?CB，?ABC??EBF?90° ································1分 A ??ABC??EBC??EBF??E B 即?ABE??CBF ···························································2分 又BE?BF ······································································3分

F ?△ABE≌△CB ··························································4分 (2)解：?BE?BF，?EBF?90°

D

E

B G F 图9 C C

??BEF?45° ······································································5分 又?EBG??ABC??ABE?40°·········································6分 ??EGC??EBG??BEF?85° ········································8分 (注：其它方法酌情给分) 21.(1)证明：连结OC ············································1分 由DC是切线得OC?DC 又AD?DC ?AD∥OC······················································ 2分 D E A B ∴?DAC??ACO 又由OA?OC得?BAC??ACO························ 3分 - 5 - O 图10

C ??DAC??BA ··············································· 4分

即AC平分?BAD

(2)解：方法一：?AB为直径

∴?ACB?90° ··················································· 5分 又??BAC??BEC

?BC?AB·sin?BAC?AB·sin?BEC?6························································· 6分

?AC?AB?BC22···················································································· 7分 ?8 ·

245又??DAC??BAC??BEC且AD?DC

?CD?AC·sin?DAC?AC·sin?BEC?····················································· 8分

方法一：?AB为直径 ??ACB?90° ····································································································· 5分 又??BAC??BEC

?BC?AB·sin?BAC?AB·sin?BEC?6························································· 6分

?AC?AB?BC22···················································································· 7分 ?8 ·

又??DAC??BAC，?D??ACB?90° ?△ADC∽△ACB

DCCB?ACAB，即245DC6?810

解得DC? ······································································································· 8分

(注：其它方法酌情给分)

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C??DAC??BA··············;即AC平分?BAD;(2)解：方法一：?AB为直径;∴?ACB?90°···············;?BC?AB·sin?BAC?AB·sin?BE;?AC?AB?BC22·············;245又??DAC??BAC??BEC且AD?D;?CD?AC·sin?DAC?AC·sin?BE;方法一：

C ??DAC??BA ··············································· 4分

即AC平分?BAD

(2)解：方法一：?AB为直径

∴?ACB?90° ··················································· 5分 又??BAC??BEC

?BC?AB·sin?BAC?AB·sin?BEC?6························································· 6分

?AC?AB?BC22···················································································· 7分 ?8 ·

245又??DAC??BAC??BEC且AD?DC

?CD?AC·sin?DAC?AC·sin?BEC?····················································· 8分

方法一：?AB为直径 ??ACB?90° ····································································································· 5分 又??BAC??BEC

?BC?AB·sin?BAC?AB·sin?BEC?6························································· 6分

?AC?AB?BC22···················································································· 7分 ?8 ·

又??DAC??BAC，?D??ACB?90° ?△ADC∽△ACB

DCCB?ACAB，即245DC6?810

解得DC? ······································································································· 8分

(注：其它方法酌情给分)

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