陕西省课改中考试题及答案(word版中考真题)

陕西省基础教育课程改革实验区 2007 年 初 中 毕 业 学 业 考 试

数 学 试 卷

第Ⅰ卷（选择题 共30分）

一、选择题（共10小题，每小题3分，计30分．每小题只有一个选项是符合题意的） 1．2的相反数为（ ） A．2

B．2

C．

12

D．

12

2．下面四个图形中，经过折叠能围成如图只有三个面上印有图案的正方体纸盒的是（ ）

（第2题图）

A． B． D．

3．不等式组

x

20，3x≥0

的解集是（ ）

A．

2≤x≤3 B．x2，或x≥3 C．2x3 D．2x≤3

4．将我省某日11个市、区的最高气温统计如下： 最高气温 10℃ 14℃ 21℃ 22℃ 23℃ 24℃ 25℃ 26℃

1 1 3 1 1 2 1 1 市、区个数

该天这11个市、区最高气温的平均数和众数分别是（ ） A．21℃，21℃ B．20℃，21℃ C．21℃，22℃ D．20℃，22℃

5．中国人民银行宣布，从2007年6月5日起，上调人民币存款利率，一年定期存款利率上调到3.06%．某人于2007年6月5日存入定期为1年的人民币5000元（到期后银行将扣除20%的利息锐）．设到期后银行应向储户支付现金x元，则所列方程正确的是（ ） A．x500050003.06%

B．x500020%5000(13.06%) C．x50003.06%20%5000(13.06%) D．x50003.06%20%50003.06% 6．如图，圆与圆之间不同的位置关系有（ ） A．2种 B．3种 C．4种 D．5种

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（第6题图）5/14/2013

7．如图，一次函数图象经过点A，且与正比例函数yx的 图象交于点B，则该一次函数的表达式为（ ） A．yx2 C．yx2

2

B．yx2 D．yx2

（第7题图）

8．抛物线yx4x7的顶点坐标是（ ） A．(2，11)

B．(2，7)

C．(2，11)

D．(2，3)

C

（第9题图）

9．如图，在矩形ABCD中，E为CD的中点，连接AE并 延长交BC的延长线于点F，则图中全等的直角三角形共有（ ）

A．3对 B．4对

C．5对 D．6对

10．如图，在等边△ABC中，AC9，点O在AC上， 且AO3，点P是AB上一动点，连结OP，将线段OP 绕点O逆时针旋转60得到线段OD．要使点D恰好落在



BC上，则AP的长是（ ）

A．4

B．5

C．6

D．8

（第10题图）

第Ⅱ卷（非选择题 共90分）

二、填空题（共6小题，每小题3分，计18分） 11．计算：(3xy)

2

12

xy 3

12．在△ABC的三个顶点A(2，3，)B(，4，5)C(，3中2，可能在反比例函数

y

kx

(k0)的图象上的点是



13．如图，ABC50，AD垂直平分线段BC于点D，ABC的 平分线BE交AD于点E，连结EC，则AEC的度数是 ．

Page 2

（第13题图）

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14．选作题（要求在（1）、（2）中任选一题作答） ．．．（1）用计算器计算：3sin38（结果保留三个有效数字）．

（2）小明在楼顶点A处测得对面大楼楼顶点C处的 仰角为52，楼底点D

处的俯角为13．若两座楼AB与







CD相距60米，则楼CD的高度约为

（结果保留三个有效数字）．











（第14题图）

（sin130.2250，cos130.9744，tan130.2309

，sin520.7880，cos520.6157

tan521.2799）

15．小说《达芬奇密码》中的一个故事里出现了一串神密排列的数，将这串令人费解的数按从小到大的顺序排列为：11，，2，3，5，8，…，则这列数的第8

个数是 ． 16．如图，要使输出值y大于100，则输入的最小正整数x是．

三、解答题（共9小题，计72分．解答应写出过程） 17．（本题满分5分） 设A

（第16题图）

xx1

，B

3x1

2

1，当x为何值时，A与B的值相等？

18．（本题满分6分）

如图，横、纵相邻格点间的距离均为1个单位．

（1）在格点中画出图形ABCD先向右平移6个单位，再向上平移2个单位后的图形； （2）请写出平移前后两图形应对点之间的距离．

5/14/2013

（第18题图）

19．（本题满分7分）

如图，在梯形ABCD中，AB∥DC，DA⊥AB，B45，



延长CD到点E，使DEDA，连接AE． （1）求证：AE∥BC；

（2）若AB3，CD1，求四边形ABCE的面积．

20．（本题满分8分）

（第19题图）

2006年，全国30个省区市在我省有投资项目，投资金额如下表：

省区市 广东 福建 北京 浙江 其它

124 67 66 47 119 金额（亿元）

根据表格中的信息解答下列问题：

（1）求2006年外省区市在陕投资总额； （2）补全图①中的条形统计图；

（3）2006年，外省区投资中有81亿元用于西安高新技术产业开发区，54亿元用于西安经济技术开发区，剩余资金用于我省其它地区．请在图②中画出外省区市在我省投资金额使用情况的扇形统计图（扇形统计图中的圆心角精确到1，百分比精确到1%）．

2006年外省区市在陕投资金额统计图 2006年外省区市

在陕投资金额使用情况统计图

东建京江它

图②

图①

21．（本题满分8分） （第20题图）

为了迎接暑期旅游，某旅行社推出了一种价格优惠方案：从现在开始，各条旅游线路的价格每人y（元）是原来价格每人x（元）的一次函数．现知道其中两条旅游线路原来旅游价格分别为每人2100元和2800元，而现在旅游的价格分别为每人1800元和2300元． （1）求y与x的函数关系式（不要求写出x的取值范围）； （2）王老师想参加该旅行社原价格为5600元的一条线路的 暑期旅游，请帮王老师算出这条线路的价格． 22．（本题满分8分） 在下列直角坐标系中， （1）请写出在ABCD内（不包括边界）横、纵坐标均为 ．整数的点，且和为零的点的坐标； （2）在ABCD内（不包括边界）任取一个横、纵坐标均为 ．整数的点，求该点的横、纵坐标之和为零的概率．

Page 4



（第225/14/2013 题图）

23．（本题满分8分）

如图，AB是半圆O的直径，过点O作弦AD的垂线交切线AC于点C，OC与半圆O交

C 于点E，连结BE，DE．

（1）求证：BEDC；

（2）若OA5，AD8，求AC的长．

E D

B O

24．（本题满分10分） 如图，在直角梯形OBCD中，OB8，BC1，CD10．

（1）求C，D两点的坐标；

（2）若线段OB上存在点

P，使PD

⊥PC，求过D，P，

三点的抛物线的表达式．

25．（本题满分12分）

（第24题图） 如图，O的半径均为R．

（1）请在图①中画出弦AB，CD，使图①为轴对称图形而不是中心对称图形；请在图②中．．画出弦

AB，CD，使图②仍为中心对称图形；

（2）如图③，在O中，ABCDm(0m2R)，且AB与CD交于点E，夹角为锐角．求四边形ACBD面积（用含m，的式子表示）； （3）若线段AB，CD是O的两条弦，且ABCD

，你认为在以点A，B，C，D

为顶点的四边形中，是否存在面积最大的四边形？请利用图④说明理由．

（第25题图①） （第25题图②） （第25题图③） （第25题图④）

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陕西省基础教育课程改革实验区2007年初中毕业学业考试

数 学 答案及评分标准

一、选择题

1．A 2．B 3．D 4．A 5．C 6．C 7．B 8．A 9．B 10．C 二、填空题

11．xy 12．B 13．115°（填115不扣分） 14．（1）0.433 （2）90.6 15．21 16．21 三、解答题

17．解：当AB时，

3

3

xx1



3x1

2

1．

xx1



3(x1)(x1)

···································································································· 1分 1．·

方程两边同时乘以(x1)(x1)，得

······························································································· 2分 x(x1)3(x1)(x1)． ·

x2x3x21．

·································································································································· 3分 x2． ·

检验：当x2时，(x1)(x1)30．

··································································································· 4分 ∴x2是分式方程的根． ·

因此，当x2时，AB． ································································································· 5分 18．解：（1）画图正确得4分．

（第18题答案图）

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5/14/2013

（2

）个单位． ············································································································· 6分 19．解：（1）证明：∵AB∥DC，DAAB，B45°，

································································································ 1分 ∴C135°，DADE．·

又∵DEDA，

······················································································································ 2分 ∴E45°． ·

·········································································································· 3分 ∴CE180°． ·

······················································································································ 4分 ∴AE∥BC． ·

（2）解：∵AE∥BC，CE∥AB，

························································································· 5分 ∴四边形ABCE是平行四边形． ·

∴CEAB3．

···························································································· 6分 ∴DADECECD2． ·

∴SABCECE·AD326． ························································································ 7分

20．解：（1）2006年外省区市在陕投资总额为：

． ··········································································· 2分 124676647119423（亿元）

（2）如图①所示． ················································································································· 5分 2006年外省区市在陕投资金额计图 2006年外省区市

在陕投资金额使用情况统计图

13%

西安高新技术 19%

省区 市

东建京江它

（第20题答案图①） （第20题答案图②）

（3）如图②所示． ················································································································· 8分 21．解：（1）设y与x的函数关系式为ykxb， ·························································· 1分

由题意，得

2100kb1800，

2800kb2300，

··························································································· 3分

5

k，

解之，得··············································································································· 5分 7 ·

b300．

∴y与x的函数关系式为y

57

·········································································· 6分 x300． ·

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（2）当x5600时，y

57

···················································· 7分 56003004300元． ·

········································································· 8分 ∴王老师旅游这条线路的价格是4300元． ·

22．解：（1）(11)····················································································· 3分 ，，，，，(00)(11)． ·（2）∵在ABCD内横、纵坐标均为整数的点有15个，

其中横、纵坐标和为零的点有3个， ···················································································· 6分



∴P

315



15

． ····················································································································· 8分

23．解：（1）证明：∵AC是O的切线，AB是O直径，

∴ABAC． 则1290°． ················································································································ 1分 又∵OCAD，

············································································································· 2分 ∴1C90°． ······················································································································· 3分 ∴C2． ·

而BED2，

················································································································ 4分 ∴BEDC． ·

（2）解：连接BD． C ∵AB是O直径， ∴ADB90°．

E

D

O

（第23题答案图）

B

∴BD

6． ··········································································· 5分

········································································································ 6分 ∴△OAC∽△BDA． ·

∴OA:BDAC:DA． 即5:6AC:8． ·················································································································· 7分

20

． ························································································································ 8分 ∴AC3

24．解：（1）过点C作CEOD于点E，则四边形OBCE为矩形．

∴CEOB8，OEBC1．

∴DE6．

∴ODDEOE7．

Page 8

（第24题答案图）

5/14/2013

···································································· 4分 1)D(0，7)． ∴C，D两点的坐标分别为C(8，，（2）∵PCPD，

∴1290°． 又1390°， ∴23．

∴Rt△POD∽Rt△CBP．∴PO:CBOD:BP．

即PO:17:(8PO)．

∴PO28PO70．

∴PO1，或PO7．

······················································································ 6分 0)． ·0)，或(7，∴点P的坐标为(1，①当点P的坐标为(1，0)时，

设经过D，P，C三点的抛物线表达式为yaxbxc，

2

25a，28

c7，

221

，则abc0， ∴b 2864a8bc1．

c7．



∴所求抛物线的表达式为：y

②当点P为(7，0)时，

2528

x2

22128

·························································· 9分 x7．

设经过D，P，C三点的抛物线表达式为yaxbxc，

2

1

a，4c7，

11

则49a7bc0， ∴b，

464a8bc1．

c7．



∴所求抛物线的表达式为：y

14

x2

114

··························································· 10分 x7． ·

（说明：求出一条抛物线表达式给3分，求出两条抛物线表达式给4分）

25．解：（1）答案不唯一，如图①、②（只要满足题意，画对一个图形给2分，画对两个给3分）

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················································································································································· 3分 （2）过点A，B分别作CD的垂线，垂足分别为M，N．

∵S△ACD

12

CD·AM

12

CD·AE·sin，

S△BCD

12

CD·BN

12

············································································ 5分 CD·BE·sin． ·

∴S四边形ACBDS△ACDS△BCD



1212

1212

CD·AE·sin

12

CD·BE·sin



CD·(AEBE·)sin CD·AB·sin

··········································· 7分 m2sin．

（第25题答案图③）

（3）存在．分两种情况说明如下： ···················································································· 8分 ①当AB与CD相交时，

由（2

）及ABCD

知S四边形ACBD

12

······················ 9分 AB·CD·sinR2sin． ·

②当AB与CD不相交时，如图④

∵ABCD，OCODOAOBR，

∴AOBCOD90°，

而S四边形ABCDSRt△AOBSRt△OCDS△AODS△BOC

RS△AODS△BOC

2

（第25题答案图④）

． ····································································································· 10分

延长BO交O于点E，连接EC，则132390°．

∴12．

∴△AOD≌△COE．

∴S△AODS△OCE．

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∴S△AODS△BOCS△OCES△BOCS△BCE．

过点C作CHBE，垂足为H， 则S△BCE12BE·CHR·CH．

··········································································· 11分 ∴当CHR时，S△BCE取最大值R2． ·

综合①、②可知，当1290°，即四边形ABCD

的正方形时， S四边形ABCDR2R22R2为最大值． ············································································· 12分

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