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2012年中考真题分类阅读理解题

2012年中考真题分类阅读理解题
2012年中考真题分类阅读理解题

中考数学阅读理解题试题分类汇编

一、选择题

1、(2007四川眉山)为确保信息安全,信息需加密传翰,发送方将明文加密为密文传输给接收方,接收方收到密文后解密还原为明文.己知某种加密规则为:明文a 、b 对应的密文为2a -b 、2a +b .例如,明文1、2对应的密文是-3、4.当接收方收到密文是1、7时,解密得到的明文是( ). C

A .-1,1

B .1,3

C . 3,I

D .1,l

2、(2007湖南长沙)在密码学中,直接可以看到内容为明码,对明码进行某种处理后得到的内容为密码.有一种密码,将英文26个字母a b c ,,,…,z (不论大小写)依次对应1,2,3,…,26这26个自然数(见表格).当明码对应的序号x 为奇数时,密码对应的序号1

2

x y +=;当明码对应的序号x 为偶数时,密码对应的序号132x

y =

+. 字母 a

b

c

d

e

f

g

h

i

j

k

l

m

序号 1 2 3

4 5

6 7 8 9 10 11 12 13

字母 n

o

p

q

r s

t

u v w x y z

序号

14 15 16 17 18 19 20

21

22

23

24

25

26

按上述规定,将明码“love ”译成密码是( ) B A .gawq B .shxc C .sdri D .love

二、填空题

1、(2007四川德阳)阅读材料:设一元二次方程2

0ax bx c ++=的两根为1x ,2x ,则两根与方程系数之间有如下关系:12b x x a +=-

,12c

x x a

= .根据该材料填空: 已知1x ,2x 是方程2

630x x ++=的两实数根,则

21

12

x x x x +的值为______.10 2、(2007四川巴中)先阅读下列材料,然后解答问题:

从A B C ,,三张卡片中选两张,有三种不同选法,抽象成数学问题就是从3个元素中选取2个元素组合,记作2

332

C 321

?=

=?. 一般地,从m 个元素中选取n 个元素组合,记作:(1)(1)

C (1)321

n

m m m m n n n --+=-???

例:从7个元素中选5个元素,共有5

776543

C 2154321

????=

=????种不同的选法.

问题:从某学习小组10人中选取3人参加活动,不同的选法共有

种.120

3、(2007广东梅州)将4个数a b c d ,,,排成2行、2列,两边各加一条竖直线记成

a b c

d

,定义

a b

c

d

ad bc =-,上述记号就叫做2阶行列式.若

11

11

x x x x +--+

6=,则x =__________. 答:2±

三、解答题

1、(2007浙江临安)阅读下列题目的解题过程:

已知a 、b 、c 为?ABC 的三边,且满足a c b c a b 2

2

2

2

4

4

-=-,试判断?ABC 的形状. 解: a c b c a b A 222244

-=-()

2222

2

22

222

()()()()

()

ABC c a b a b

a b

B c a b

C ?∴-=+-∴=+∴是直角三角形

问:(1)上述解题过程,从哪一步开始出现错误?请写出该步的代号:________________; (2)错误的原因为:_______________________________________________________; (3)本题正确的结论为:____________.

解:(1) C ---2分 (2)没有考虑2

2

0a b -=---4分

(3)ABC ?是直角三角形或等腰三角形 ---6分

2、(2007云南双柏)阅读下列材料,并解决后面的问题.

材料:一般地,n 个相同的因数a 相乘:n

n a a a a 记为个

?.如23=8,此时,3叫做以2为底8的对数,记为()38log 8log 22=即.

一般地,若()0,10>≠>=b a a b a n

且,则n 叫做以a 为底b 的对数,记为

()813.log log 4==如即n b b a a ,则4叫做以3为底81的对数,记为)481log (81log 33=即.

问题:(1)计算以下各对数的值:(3分) =

==

64log 16log 4log 222 .

(2)观察(1)中三数4、16、64之间满足怎样的关系式?64log 16log 4log 222、、 之间又满足怎样

的关系式?(2分)

(3)由(2)的结果,你能归纳出一个一般性的结论吗?(2分)

()0,0,10l o g l o g >>≠>=

+N M a a N M a a 且

(4)根据幂的运算法则:m n m

n

a a a +=?以及对数的含义证明上述结论.(3分)

证明:

解:(1)24log 2= , 416log 2= ,664log 2=

(2)4×16=64 ,4log 2 + 16log 2 = 64log 2 (3)M a log + N a log = )(log MN a (4)证明:设M a log =b 1 , N a log =b 2 则M a

b =1

,N a b =2

∴2121b b b b a a a MN

+=?=

∴b 1+b 2=)(log MN a 即M a log + N a log = )(log MN a

3、(2007安徽芜湖)阅读以下材料,并解答以下问题.

“完成一件事有两类不同的方案,在第一类方案中有m 种不同的方法,在第二类方案中有n 种不同的方法.那么完成这件事共有N = m + n 种不同的方法,这是分类加法计数原理;完成一件事需要两个步骤,做第一步有m 种不同的方法,做第二步有n 种不同的方法.那么完成这件事共有N =m ×n 种不同的方法, 这就是分步乘法计数原理. ”如完成沿图1所示的街道从A 点出发向B 点行进这件事(规定必须向北走,或向东走), 会有多种不同的走法,其中从A 点出发到某些交叉点的走法数已在图2填出. (1) 根据以上原理和图2的提示, 算出从A 出发到达其余交叉点的走法数,将数字填入图2的空圆中,

并回答从A 点出发到B 点的走法共有多少种?

(2) 运用适当的原理和方法算出从A 点出发到达B 点,并禁止通过交叉点C 的走法有多少种?

(3) 现由于交叉点C 道路施工,禁止通行. 求如任选一种走法,从A 点出发能顺利开车到达B 点(无返回)概率是多少?

解:

解: (1)∵完成从A 点到B 点必须向北走,或向东走,

∴到达A 点以外的任意交叉点的走法数只能是与其相邻的南边交叉点和西边交叉点的数字之和. 故使用分类加法计数原理,由此算出从A 点到达其余各交叉点的走法数,填表如图1,

答:从A 点到B 点的走法共有35种. ……………………………………5分

(1) 方法一: 可先求从A 点到B 点,并经过交叉点C 的走法数,再用从A 点到B 点总走法数减去它,即得从A 点到B 点,但不经过交叉点C 的走法数.

完成从A 点出发经C 点到B 点这件事可分两步,先从A 点到C 点,再从C 点到B 点. 使用分类加法计数原理,算出从A 点到C 点的走法是3种,见图2;算出从C 点到B 点的走法为6种,见图3,再运用分步乘法计数原理,得到从A 点经C 点到B 点的走法有3×6=18种.

∴从A 点到B 点但不经过C 点的走法数为35-18=17种. ………………………10分

方法二:由于交叉点C 道路施工,禁止通行,故视为相邻道路不通,可删除与C 点紧相连的线段.运用分类加法计数原理,算出从A 点到B 点并禁止通过交叉点C 的走法有17种. 从A 点到各交叉点的走法数见图4.

∴从A 点到B 点并禁止经过C 点的走法数为35-18=17种.………10分 (3) P (顺利开车到达B 点)=

1735

. 答:任选一种走法,顺利开车到达B 点的概率是

17

35

. ………………12分

4、(2007江苏连云港)如图1,点C 将线段AB 分成两.

部分,如果AC BC

AB AC

=,那么称点C 为线段AB 的黄金分割点.

某研究小组在进行课题学习时,由黄金分割点联想到“黄金分割线”,类似地给出“黄金分割线”的定义:直线l 将一个面积为S 的图形分成两部分,这两部分的面积分别为1S ,2S ,如果

12

1

S S S S =

,那么称直线l 为该图形的黄金分割线.

(1)研究小组猜想:在ABC △中,若点D 为AB 边上的黄金分割点(如图2),则直线CD 是ABC △的黄金分割线.你认为对吗?为什么?

(2)请你说明:三角形的中线是否也是该三角形的黄金分割线?

(3)研究小组在进一步探究中发现:过点C 任作一条直线交AB 于点E ,再过点D 作直线DF CE ∥,交AC 于点F ,连接EF (如图3),则直线EF 也是ABC △的黄金分割线. 请你说明理由.

(4)如图4,点E 是ABCD

的边AB 的黄金分割点,过点E 作EF AD ∥,交DC 于点F ,显然直线EF 是ABCD

的黄金分割线.请你画一条ABCD 的黄金分割线,使它不经过ABCD

各边黄金分割点.

解:(1)直线CD 是ABC △的黄金分割线.理由如下: 设ABC △的边AB 上的高为h . 12A

D C

S A D h =

△,12BDC S BD h = △,12

ABC S AB h = △, 所以,

ADC ABC S AD S AB =△△,BDC ADC S BD

S AD

=

△△. ··············································· 2分 又因为点D 为边AB 的黄金分割点,所以有

AD BD AB AD =.因此ADC BDC ABC ADC

S S

S S =△△△△. 所以,直线CD 是ABC △的黄金分割线.············································ 4分 (2)因为三角形的中线将三角形分成面积相等的两部分,此时121

2

s s s ==

,即 12

1

s s s s ≠,所以三角形的中线不可能是该三角形的黄金分割线. ·

············ 6分 (3)因为DF CE ∥,所以DEC △和FCE △的公共边CE 上的高也相等, 所以有DEC FCE S S =△△. ································································ 7分 设直线EF 与CD 交于点G .所以DGE FGC S S =△△. 所以ADC FGC AFGD S S S =+△△四边形

DGE AEF AFGD S S S =+=△△四边形,BDC BEFC S S =△四边形. 又因为

ADC BDC ABC ADC S S S S =△△△△,所以BEFC

AEF ABC AEF

S S S S =

四边形△△△. ·························· 9分 因此,直线EF 也是ABC △的黄金分割线. ··································· 10分

(4)画法不惟一,现提供两种画法; ············································· 12分

画法一:如答图1,取EF 的中点G ,再过点G 作一条直线分别交AB ,DC 于M ,N 点,则直线MN 就是ABCD

的黄金分割线.

画法二:如答图2,在DF 上取一点N ,连接EN ,再过点F 作FM NE ∥交AB 于点M ,连接MN ,则直线MN 就是ABCD

的黄金分割线.

5、(2007浙江衢州)请阅读下列材料:

问题:如图(2),一圆柱的底面半径为5dm ,BC 是底面直径,求一只蚂蚁从A 点出发沿圆柱表面爬行到点C 的最短路线.小明设计了两条路线:

路线1:侧面展开图中的先端AC .如下图(2)所示:

F

C

B D E A N M G

(第4题答图1)

F

C

B

D E A N M (第4题答图2)

设路线1的长度为1l ,则2222222

12525)5(5ππ+=+=+==AC AB AC l 路线2:高线AB + 底面直径BC .如上图(1)所示: 设路线2的长度为2l ,则225)105()(2222=+=+=AC AB l

0)8(25200252252525222

221>-=-=-+=-πππl l

∴2

22

1l l > ∴21l l >

所以要选择路线2较短.

(1)小明对上述结论有些疑惑,于是他把条件改成:“圆柱的底面半径为1dm ,高AB 为

5dm ”继续按前面的路线进行计算.请你帮小明完成下面的计算: 路线1:==22

1AC l ___________________; 路线2:=+=222)(AC AB l __________ ∵222

1_____l l ∴ 21_____l l ( 填>或<)

所以应选择路线____________(填1或2)较短.

(2)请你帮小明继续研究:在一般情况下,当圆柱的底面半径为r ,高为h 时,应如何选择上面的两条路线才能使蚂蚁从点A 出发沿圆柱表面爬行到C 点的路线最短. 解:(1)22222221525l AC AB AC ππ==+=+=+

2222()(52)49l AB AC =+=+

= 2212l l < ∴12l l <

所以要选择路线1较短.

(2)2222221()l AC AB AC h r π==+=+

2222()(2)l AB AC h r =+=+

2212l l - =22()h r π+-2(2)h r +=2(44)r r r h π--=2[(4)4]r r h π--

当244h r π=

-时,2212l l =;当r >244h π-时,21l >2

2l ;当r <2

44

h π-时,21l <22l . 6、(2007甘肃白银等3市)阅读下边一元二次方程求根公式的两种推导方法: 方法一:教材中方法 方法二:

∵ ax 2+bx +c =0,

∴ 4a 2x 2+4abx +4ac =0,

配方可得: ∴ (2ax +b )2=b 2-4ac . 当 b 2-4ac ≥0时,

2

222

2,4(),244ax bx c o b b ac a x a a b b ac ++=-∴+=- 2

22,4ax bx c o b b ac ++=- 2

22

,4(),24ax bx c o b b ac a x a a ++=-∴+= 比较两个正数的大小,有时用它们的平方来比较更方便

2ax +b =±24b ac -,

∴ 2ax =-b ±24b ac -.

当 b 2

-4ac ≥0时, ∴ x =242b b ac

a

-±-.

请回答下列问题:

(1)两种方法有什么异同?你认为哪个方法好? (2)说说你有什么感想?

解:(1)都采用配方法.方法一是将二次项的系数化为1,方法二是将二次项系数变成一个平方式.方法一较好.

7、(2007江苏无锡)图1是由若干个小圆圈堆成的一个形如正三角形的图案,最上面一层有一个圆圈,以下各层均比上一层多一个圆圈,一共堆了n 层.将图1倒置后与原图1拼成图2的形状,这样我们可以算出图1中所有圆圈的个数为(1)

1232

n n n +++++=

图1 图2 图3

图4

如果图1中的圆圈共有12层,(1)我们自上往下,在每个圆圈中都按图3的方式填上一串连续的正整

数1234,,,, ,则最底层最左边这个圆圈中的数是 ;(2)我们自上往下,在每个圆圈中都按图4的方式填上一串连续的整数23-,22-,21-, ,求图4中所有圆圈中各数的绝对值之和. 解:(1)67. ······························································································· 2分 (2)图4中所有圆圈中共有12(121)

12312782

+++++=

= 个数, 其中23个负数,1个0,54个正数, ·································································· 4分

∴图4中所有圆圈中各数的绝对值之和|23||22||1|01254=-+-++-+++++

(12323)(12354)27614851761=+++++++++=+= . ··························· 6分

8、(2007鄂尔多斯)我们给出如下定义:若一个四边形中存在相邻两边的平方和等于一条对角线的平方,则称这个四边形为勾股四边形,这两条相邻的边称为这个四边形的勾股边.

(1)写出你所学过的特殊四边形中是勾股四边形的两种图形的名称_________,________;

(2)如图16(1),已知格点(小正方形的顶点)(00)O ,,(30)A ,,(04)B ,,请你画出以格点为顶点,OA OB ,为勾股边且对角线相等的勾股四边形OAMB ;

y B

(3)如图16(2),将ABC △绕顶点B 按顺时针方向旋转60

,得到DBE △,连结AD DC ,,

30DCB = ∠.

求证:2

2

2

DC BC AC +=,即四边形ABCD 是勾股四边形.

解:

(1)正方形、长方形、直角梯形.(任选两个均可) ············· 2分(填正确一个得1分) (2)答案如图所示.(34)M ,

或(43)M ,.(没有写出不扣分) ·························································· 2分(根据图形给分,一个图形正确得1分)

(3)证明:连结EC

ABC DBE △≌△ ······················································································· 5分 AC DE ∴=,BC BE = ················································································· 6分

60CBE = ∠ EC BC ∴=,60BCE = ∠ ······················································ 7分 30DCB = ∠ 90DCE ∴= ∠ 222DC EC DE ∴+= ······································· 8分 222DC BC AC ∴+=,即四边形ABCD 是勾股四边形 ·········································· 9分

A

B C

D E

60 图16(2)

y B O M

M

A x A

B

C D

E 60

【英语】中考英语阅读理解经典题型带答案

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