乌鲁木齐中考数学试题及答案
2010年新疆乌鲁木齐市初中毕业生学业水平测试数学试卷
一、选择题（本大题共10小题，每小题4分，共40分）每题的选项中只有一项符合题目要求. 1.在0，2-，1，2-这四个数中负整数是 A.2- B. 0 C.22- D. 1
2.如图1是由五个相同的小正方体组成的几何体，则它的左视图是
3.“十二五”期间，新疆将建成横贯东西、沟通天山的“十”字形高速公路主骨架，全疆高 速公路总里程突破4 000km ，交通运输条件得到全面改善，将4 000用科学记数法可以表 示为
A.24010?
B. 3410?
C. 4
0.410? D. 4410?
4.阳光公司销售一种进价为21元的电子产品，按标价的九折销售，仍可获得20%，则这种 电子产品的标价为
A. 26元
B. 27元
C. 28元
D. 29元 5.已知整式2
5
2
x x -
的值为6，则2256x x -+的值为 A. 9 B. 12 C. 18 D. 24
6.如图2，在平面直角坐标系中，点A B C 、、的坐标为 （1，4）、（5，4）、（1、2-），则ABC △外接圆的圆心 坐标是
A.（2，3）
B.（3，2）
C.（1，3）
D.（3，1）
7.有若干张面积分别为2
2
a b ab 、、的正方形和长方形纸片，阳阳从中抽取了1张面积为2
a
的正方形纸片，4张面积为ab 的长方形纸片，若他想拼成一个大正方形，则还需要抽取 面积为2
b 的正方形纸片
A. 2张
B.4张
C.6张
D.8张
8.某校九年级（2）班50名同学为玉树灾区献爱心捐款情况如下表：
捐款（元） 10 15 30 40 50 60
人数
3
6
11 11 13
6
则该班捐款金额的众数和中位数分别是
A. 13，11
B. 50，35
C. 50，40
D. 40，50
9.如图3，四边形OABC 为菱形，点A B 、在以点O 为圆心的DE 上，若312OA =∠=∠，，则扇形ODE 的面积为
A.
3π2 B. 2π C.5
π2
D. 3π 10.将边长为3cm 的正三角形各边三等分，以这六个分点为顶点构 成一个正六边形，则这个正六边形的面积为 A.
332cm 2 B.334cm 2 C.338
cm 2
D.33cm 2 图2 A
D
O E
C
B
图3
二、填空题（本大题共5小题，每小题4分，共20分）把答案直接填在答题卡的相应位置处. 11.计算：18322-+=_____________.
12.如图4，AB 是O ⊙的直径，C D 、为O ⊙上的两点， 若35CDB ∠=°，则ABC ∠的度数为__________. 13.在数轴上，点A B 、对应的数分别为2，
5
1
x x -+，且A B 、 两点关于原点对称，则x 的值为___________.
14.已知点1(1)A y -，，2(1)B y ，，3(2)C y ，在反比例函数(0)k y k x
=
<的图象上，则 123y y y 、、的大小关系为_________（用“>”或“<”连接）.
15.暑假期间，瑞瑞打算参观上海世博会.她要从中国馆、澳大利亚馆、德国馆、英国馆、日
本馆和瑞士馆中预约两个馆重点参观，想用抽签的方式来作决定，于是她做了分别写有以上馆名的六张卡片，从中任意抽取两张来确定预约的场馆，则他恰好抽中中国馆、澳大利亚馆的概率是___________.
三、解答题（本大题Ⅰ—Ⅴ，共9小题，共90分）解答时对应在答题卡的相应位置处写出文字说明、证明过程或演算过程.
Ⅰ.（本题满分15分，第16题6分，第17题9分）
16.解不等式组1
(4)223(1) 5.
x x x ?+?，
17．先化简，再求值：
2111
1211
a a a a a a ++-÷
+-+-，其中 2.a = 四．（本题满分30分，第18题8分，第19题、20题，每题11分）
18.如图5，在平行四边形ABCD 中，BE 平分ABC ∠交AD 于点E ，DF 平分∠ADC 交 BC 于点F . 求证：（1）ABE CDF △≌；
（2）若BD EF ⊥，则判断四边形EBFD 是什么特殊四边形，请证明你的结论.
19.如图6，在平面直角坐标系中，直线4:43
l y x =-
+分别交x 轴、y 轴于点A B 、，将
AOB △绕点O 顺时针旋转90°后得到A OB ''△. （1）求直线A B ''的解析式；
（2）若直线A B ''与直线l 相交于点C ，求A BC '△的面积.
C
O
图4
B
D
A
F
D 图5
E C A
B 图6
C
A y x O
l
A '
B '
20.某过街天桥的截面图为梯形，如图7所示，其中天桥斜面CD 的坡度为1:3i =
（1:3i =是指铅直高度DE 与水平宽度CE 的比），CD 的长为10m ，天桥另一斜面AB 坡角ABG ∠=45°.
（1）写出过街天桥斜面AB 的坡度； （2）求DE 的长；
（3）若决定对该过街天桥进行改建，使AB 斜面的坡度变缓，将其45°坡角改为30°， 方便过路群众，改建后斜面为AF .试计算此改建需占路面的宽度FB 的长（结果精确0.01）
Ⅲ.（本题满分23分，第21题11分，第22题12分）
21.2010年5月中央召开了新疆工作座谈会，为实现新疆跨越式发展和长治久安，作出了重 要战略决策部署．为此我市抓住机遇，加快发展，决定今年投入5亿元用于城市基础设施 维护和建设，以后逐年增加，计划到2012年当年用于城市基础设施维护与建设资金达到 8.45亿元.
（1）求从2010年至2012年我市每年投入城市基础设施维护和建设资金的年平均增长率； （2）若2010年至2012年我市每年投入城市基础设施维护和建设资金的年平均增长率相同， 预计我市这三年用于城市基础设施维护和建设资金共多少亿元？
22.2010年6月4日，乌鲁木齐市政府通报了首府2009年环境质量公报，其中空气质量级别分布统计图如图8所示，请根据统计图解答以下问题：
（1）写出2009年乌鲁木齐市全年三级轻度污染天数：
（2）求出空气质量为二级所对应扇形圆心角的度数（结果保留到个位）；
（3）若到2012年，首府空气质量良好（二级及二级以上）的天数与全年天数（2012年是闰年，全年有366天）之比超过85%，求2012年空气质量良好的天数要比2009年至少增加多少天？
Ⅳ.（本题满分10分）
23.已知二次函数2
(0)y ax bx c a =++≠的图象经过(00)(1)O M ，，
，1和()(0)N n n ≠，0 三点.
（1）若该函数图象顶点恰为点M ，写出此时n 的值及y 的最大值；
（2）当2n =-时，确定这个二次函数的解析式，并判断此时y 是否有最大值； （3）由（1）、（2）可知，n 的取值变化，会影响该函数图象的开口方向.请你求出n 满足 什么条件时，y 有最小值？
F A
B G D E C
图7
图8
Ⅴ.（本题满分12分）
24.如图9，边长为5的正方形OABC 的顶点O 在坐标原点处，点A C 、分别在x 轴、y 轴 的正半轴上，点E 是OA 边上的点（不与点A 重合），EF CE ⊥，且与正方形外角平分 线AC 交于点P .
（1）当点E 坐标为(30)，时，试证明CE EP =；
（2）如果将上述条件“点E 坐标为（3，0）”改为“点E 坐标为（t ，0）（0t >）”，结论 CE EP =是否仍然成立，请说明理由；
（3）在y 轴上是否存在点M ，使得四边形BMEP 是平行四边形？若存在，用t 表示点M 的坐标；若不存在，说明理由.
数学试题参考答案及评分标准
一、选择题（本大题共10小题，每小题4分，共40分）
题号 1 2 3 4 5 6 7 8 9 10 答案 A
D
B
C
C
D
B
C
D
A
二、填空题（本大题共5小题，每小题4分，共20分）
11.0 12.55° 13. 1 14. 231y y y <<或132y y y >> 15.
115
三、解答题（本大题1-V 题，共9小题，共90分） 16.解：由（1）得：440x x +<<， ··························································································· 2′
由（2）得：3351x x x -+><-， ·················································································· 4′ ∴不等式组的解集是：1x <- ··························································································· 6′ 17.解：原式=
()2111
11
1a a a a a +--++-·
······························································································· 3′ =
1111a a -+- ············································································································· 4′ =22
1
a -- ···················································································································· 7′
当2a =时，原式=()
2
2
221
-=-- ····································································· 9′ 18. 证明：（1）∵四边形ABCD 是平行四边形，∴A C AB CD ABC ADC ∠=∠=∠=∠，，
∵BE 平分ABC ∠，DF 平分ADC ∠，
∴ABE CDF ∠=∠ ································ 2′ ∴()ABE CDF ASA △≌△ ·················································································· 4′ （2）由ABE CDF △≌△，得AE CF = ····································································· 5′ 在平行四边形ABCD 中，AD BC AD BC =∥，
∴DE BF DE BF =∥，
∴四边形EBFD 是平行四边形 ··············································································· 6′ 若BD EF ⊥，则四边形EBFD 是菱形 ··································································· 8′
19.解：（1）由直线l ：4
43
y x =-
+分别交x 轴、y 轴于点A B 、，
可知；()()3004A B ，，，
∵AOB △绕点O 顺时针旋转90°而得到A OB ''△ ∴AOB A OB ''△≌△
故()()0340A B ''-，，， ······································································································· 2′ 设直线A B ''的解析式为y kx b =+（0k k b ≠，，为常数）
∴有340b k b =-??+=?解之得：343
k b ?
=???=-?
∴直线A B ''的解析式为3
34
y x =- ·················································································· 5′ （2）由题意得：
33444
3y x y x ?
=-???
?=-+??解之得：84251225x y ?=????=-??
∴84122525C ??- ???， ··················································· 9′ 又7A B '=
∴184294
722525
A C
B S =
??=
△′ ···························································································· 11′ 20.解：（1）在Rt AGB △中，45ABG ∠=° ∴AG BG =
∴AB 的坡度=1AG
BG
= ······································································································· 2′ （2）在Rt DEC △中，∵3
tan 3
DE C EC ∠==∴30C ∠=° 又∵10CD = ∴()1
5m 2
DE CD =
= ······································································· 5′ （3）由（1）知，5AG BG ==，在Rt AFG △中，30AFG ∠=°
tan AG
AFG FG
∠=
，即3535FB =+ ········································································· 7′ 解得535 3.66FB =-≈ ························································································ 10′
答：改建后需占路面宽度约为3.66m. ···································································· 11′
21.解：（1）设从2010至2012年我市每年投入城市基础设施维护和建设资金的年平均增长率为x ，
由题意得：()2
518.45x += ······························································································ 3′
解得，1230% 2.3x x ==-，（不合题意舍去） ····························································· 6′
答：从2010至2012年我市每年投入城市基础设施维护和建设资金的年平均增长率为30%. ··························································································································································· 7′
（2）这三年共投资()5518.45x +++
()5510.38.4519.95=+++=（亿元） ·
······························································ 10′ 答：预计我市这三年用于城市建设基础设施维护和建设资金共19.95亿元 ··············· 11′ 22． 解：（1）21.6%36578.8479?=≈（天） ······································································ 2′
（2）()19.0% 2.7% 3.9%21.6%360-+++?????°
226.08=°
226≈° ·
··················································································································· 5′ （3）设到2012年首府空气质量良好的天数比2009年增加了x 天，由题意得：
()
9.0%36562.8%36585%365
x +?+?>···························································· 8′
49.03x > ·
··········································································································· 10′ 由题意知x 应为正整数，∴50x ≥ ································································ 11′
答：2012年首府空气质量良好的天数比2009年首府空气质量良好的天数至少增加50天. ················································································································ 12′
23.解：（1）由二次函数图象的对称性可知2n =；y 的最大值为1. ··································· 2′
（2）由题意得：1420a b a b +=??-=?，解这个方程组得：13
23a b ?
=????=??
故这个二次函数的解析式为212
33
y x x =+ ···························································· 5′ ∵
1
03
> ∴y 没有最大值. ·················································································· 6′ （3）由题意，得21
0a b an bn +=??+=?
，整理得：()210an a n +-= ·································· 8′
∵0n ≠ ∴10an a +-=故()11n a -=，而1n ≠ 若y 有最小值，则需0a > ∴10n -> 即1n <
∴1n <时，y 有最小值. ························································································· 10′
24.解：（1）过点P 作PH x ⊥轴，垂足为H
∴2190∠=∠=° ∵EF CE ⊥ ∴34∠=∠ ∴COE EHP △∽△
A
R
H
O
M C
y B
G
P
F
x
∴
CO EH
OE HP
=
················································· 2′ 由题意知:5CO = 3OE = 2EH EA AH HP =+=+ ∴523HP HP += 得3HP = ∴5EH = ·························································································································· 3′ 在Rt COE △和Rt EHP △中
∴2234CE CO OE =+= 2234EP EH PH =+=
故CE EP = ······················································································································· 5′ （2）CE EP =仍成立.
同理.COE EHP △∽△ ∴
CO EH
OE HP
=
········································································ 6′ 由题意知:5CO = OE t = 5EH t HP =-+ ∴55t HP t HP
-+= 整理得()()55t HP t t -=- ∵点E 不与点A 重合 ∴50t -≠ ∴HP t = 5EH = ∴在Rt COE △和Rt EHP △中
225CE t =+ 225EP t =+ ∴CE EP = ·
·························································· 5′ （3）y 轴上存在点M ，使得四边形BMEP 是平行四边形. ············································· 9′
过点B 作BM EP ∥交y 轴于点M ∴590CEP ∠=∠=° ∴64∠=∠
在BCM △和COE △中
64BC OC
BCM COE ∠=∠??
=??∠=∠?
∴BCM COE △≌△ ∴BM CE = 而CE EP = ∴BM EP =
由于BM EP ∥ ∴四边形BMEP 是平行四边形. ················································ 11′ 故BCM COE △≌△可得CM OE t == ∴5OM CO CM t =-=-
故点M 的坐标为()05t -， ···························································································· 12′