5
分数
人数(人)
15 6分 0 20
10 8分 10分
第7题图
济南市2010年初三年级学业水平考试
数 学 试 题
注意事项:
1.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,满分120分.第Ⅰ卷1至2页,第Ⅱ卷3至8页.考试时间120分钟.
2.答第Ⅰ卷前,考生务必将自己的姓名、准考证号、考试科目用2B 铅笔涂写在答题卡上,并同时将考点、姓名、准考证号、座号填写在试卷规定的地方.
3.选择题选出答案后,用2B 铅笔把答题卡上对应题目的正确答案标号涂黑.如需改动,用橡皮擦干净后,再选涂其它答案,答案写在试卷上无效.
4.数学考试不允许使用计算器,考试结束后,应将本试卷和答题卡一并交回.
第Ⅰ卷(选择题 共48分)
一、选择题(本大题共12个小题,每小题4分,共48分.在每小题给出的四个选项中,只
有一项是符合题目要求的.) 1.2+(-2)的值是
A .-4
B .1
4- C .0 D .4
2.一组数据0、1、2、2、3、1、3、3的众数是 A .0 B .1 C .2 D .3
3.图中的几何体是由7个大小相同的小正方体组成的,该几何体的俯视图为
4.作为历史上第一个正式提出“低碳世博”理念的世博会,上海世博会从一开始就确定以“低碳、和谐、可持续发展的城市”为主题.如今在世博场馆和周边共运行着一千多辆新能源汽车,为目前世界上规模最大的新能源汽车示范运行,预计将减少温室气体排放约28400
吨.将28400吨用科学记数法表示为
A .0.284×105 吨
B .2.84×104吨
C .28.4×103吨
D .284×102吨 5.二元一次方程组42x y x y -=??+=?的解是
A .37x y =??=-?
B .1
1x y =??=?
C .7
3x y =??=?
D .3
1x y =??=-?
6.下列各选项的运算结果正确的是
A .236(2)8x x =
B .22523a b a b -=
C .6
2
3
x x x ÷= D .2
2
2
()a b a b -=- 7.在一次体育课上,体育老师对九年级一班的40名同学进行了立定跳远项目的测试,测试所得分数及相应的人数如图所示,则这次测试的平均分为
第4题图 A .
B .
C .
D . 第3题图
第10题图 A B
C
D
P
E
第12题图
A
B
C
D
M N
O 第9题图
⑴ 1+8=?
1+8+16=?
⑵ ⑶
1+8+16+24=?
第11题图
……
A .53分
B .354分
C .403
分 D .8分
8.一次函数21y x =-+的图象经过哪几个象限
A .一、二、三象限
B .一、二、四象限
C .一、三、四象限
D .二、三、四象限
9.如图所示,正方形ABCD 中,对角线AC 、BD 交于点O ,点M 、N 分别为OB 、OC 的中点,则cos ∠OMN 的值为 A .1
2
B
C
D .1
10.二次函数22y x x =--的图象如图所示,则函数值y <0时
x 的取值范围是
A .x <-1
B .x >2
C .-1<x <2
D .x <-1或x >2
11.观察下列图形及图形所对应的算式,根据你发现的规律计算1+8+16+24+……+8n (n 是正
整数)的结果为
A .2(21)n +
B .2(21)n -
C .2(2)n +
D .2n
12.如图所示,矩形ABCD 中,AB =4,BC
=,点E 是折线段A -D -C 上的一个动点(点E 与点A 不重合),点P 是点A 关于B E 的对称点.在点E 运动的过程中,使△PCB 为等腰三角形的点E 的位置共有
A .2个
B .3个
C .4个
D .5个 绝密★启用前
济南市2010年初三年级学业水平考试
数 学 试 题
注意事项:
1.第Ⅱ卷共6页.用蓝、黑色钢笔或圆珠笔直接答在考试卷上. 2.答卷前将密封线内的项目填写清楚.
第Ⅱ卷(非选择题 共72分) 二、填空题(本大题共5个小题,每小题3分,共15分.把答案填在题
中的横线上.)
A B C D
E
F
第14题图
第16题图
第17题图
13.分解因式:2
21x x ++= . 14.如图所示,△DEF 是
△ABC 沿水平方向向右平移后的对应
图形,若∠B =31°,∠C =79°,则∠D 的度
数是
度.
15.解方程2
3
123x x =
-+的结果是 . 16.如图所示,点A 是双曲线1
y x
=-
在第二象限的分支上的任意一点,点B 、C 、D 分别是点A 关于x 轴、原点、y 轴的对称点,则四边形ABCD 的面积是 .
17.如图所示,△ABC 的三个顶点的坐标分别为A (-1,3)、B (-2,-2)、C (4,-2),则
△ABC
7个小题,共57分.解答应写出文字说明、证明过程或演算步骤.)
18.(本小题满分7分)
⑴解不等式组:224x x x +>-??-?
≤
⑵如图所示,在梯形ABCD 中,BC ∥AD ,AB =DC ,点M 是AD 的中点. 求证:BM =CM .
B A
C D
M
第18题图
A
B
C
D
第19题图
第21题图
19.(本小题满分7分)
+0(3)-
⑵如图所示,△ABC 中,∠C =90°,∠B =30°,AD 是△ABC 的角平分线,若AC 求线段AD 的长.
20.(本小题满分8分)
如图所示,有一个可以自由转动的圆形转盘,被平均分成四个扇形,四个扇形内分别标有数字1、2、-3、-4.若将转盘转动两次,每一次停止转动后,指针指向的扇形内的数字分别记为a 、b (若指针恰好指在分界线上,则该次不计,重新转动一次,直至指针落在扇形内).
请你用列表法或树状图求a 与 b
的乘积等于2的概率.
21.(本小题满分8分)
如图所示,某幼儿园有一道长为16米的墙,计划用32米长的围栏靠墙围成一个面积为120平方米的矩形草坪ABCD .求该矩形草坪BC 边的长.
22.(本小题满分9分)
如图所示,菱形ABCD 的顶点A 、B 在x 轴上,点A 在点B 的左侧,点D 在y 轴的正半轴上,∠BAD =60°,点A 的坐标为(-2,0).
⑴求线段AD 所在直线的函数表达式.
第20题图
A
B
C
N M
P
A M
N
P 1 C
P 2
B A C
M
N
P 1 P 2 P 2009…… ……
B
第23题图2
第23题图1
第23题图3
第22题图
⑵动点P 从点A 出发,以每秒1个单位长度的速度,按照A →D →C →B →A 的顺序在菱形的边上匀速运动一周,设运动时间为t 秒.求t 为何值时,以点P 为圆心、以1为半径的圆与对角线AC 相切?
23.(本小题满分9分)
已知:△ABC 是任意三角形. ⑴如图1所示,点M 、P 、N 分别是边AB ⑵如图2所示,点M 、N
分别在边AB 、AC 上,且13AM AB =,1
3AN AC =,点P 1、P 2是边
BC 的三等分点,你认为∠MP 1N +∠MP 2N =∠A 是否正确?请说明你的理由. ⑶如图3所示,点M 、N 分别在边AB 、AC 上,且12010AM AB =,1
2010AN AC =
,点P 1、P 2、……、P 2009是边BC 的2010等分点,则∠MP 1N +∠MP 2N +……+∠MP 2009N =____________.
(请直接将该小问的答案写在横线上.)
24.(本小题满分9分)
如图所示,抛物线223y x x =-++与x 轴交于A 、B 两点,直线BD 的函数表达式为y =+称轴l 与直线BD 交于点C 、与x 轴交于点E .
⑴求A 、B 、C 三个点的坐标.
⑵点P 为线段AB 上的一个动点(与点A 、点B 不重合),以点A 为圆心、以AP 为半径的圆弧与线段AC 交于点M ,以点B 为圆心、以BP 为半径的圆弧与线段BC 交于点N ,分别连接AN 、BM 、MN .
①求证:AN =BM .
②在点P 运动的过程中,四边形AMNB 并求出该最大值或最小值.
济南市2010年初三年级学业水平考试
数学试题参考答案及评分标准
二、填空题
13. 2(1)x + 14. 70 15. 9x =- 16. 4 17. 三、解答题
18.(1)解:224
x x
x +-??-?>≤
解不等式①,得1x ->, ·································································· 1分 解不等式②,得2x ≥-, ··································································· 2分 ∴不等式组的解集为1x ->. ···································································· 3分 (2) 证明:∵BC ∥AD ,AB =DC ,
∴∠BAM =∠CDM , ··································································· 1分 ∵点M 是AD 的中点,
∴AM =DM , ··············································································· 2分
∴△ABM ≌△DCM , ···································
······························ 3分 ∴BM =CM . ·········································
······································· 4分 19.(1)解:原式0(3)+- ·
···············
··········································· 1分 2+1 ················································································ 2分 -1 ··················································································· 3分
(2)解:∵△ABC 中,∠C =90o,∠B =30o,
∴∠BAC =60o,
∵AD 是△ABC 的角平分线,
∴∠CAD =30o, ····
·········
································································ 1分 ∴在Rt △ADC 中,cos30AC
AD =
?
················································· 2分
·················
······························· 3分
=2 . ························································ 4分
20.解:a 与① ②
·················································································································· 6分 总共有16种结果,每种结果出现的可能性相同,其中ab =2的结果有2种, ··························································································································· 7分
∴a 与 b 的乘积等于2的概率是1
8. ·························································· 8分
21.解:设BC 边的长为x 米,根据题意得 ··················································· 1分
321202x
x -= ,
················································································· 4分 解得:121220x x ==,, ··········································································· 6分
∵20>16,
∴220x =不合题意,舍去, ·················································
··············
·· 7分
答:该矩形草坪BC 边的长为12米. ··············································· 8分 22. 解:⑴∵点A 的坐标为(-2,0)
,∠BAD =60°,∠AOD =90°,
∴OD =OA ·tan60°=
,
∴点D 的坐标为(
0,), ···········
·····
·········································· 1分 设直线AD 的函数表达式为y kx b =+, 20
k b b -+=???
=??,解得k b ?=??=??
∴直线AD 的函数表达式为y +. ······································· 3分 ⑵∵四边形ABCD 是菱形, ∴∠DCB =∠BAD =60°, ∴∠1=∠2=∠3=∠4=30°, AD =DC =CB =BA =4, ········································································· 5分 如图所示:
①点P 在AD 上与AC 相切时, AP 1=2r =2, ∴t 1=2. ·································································································· 6分
②点P 在DC 上与AC 相切时, CP 2=2r =2, ∴AD +DP 2=6, ∴t 2=6. ······································· 7分 ③点P 在BC 上与AC 相切时, CP 3=2r =2,
D
C
M
N
l
y
F
A
B C M N P 1
第23题图
P
2
1 2 ∴AD +DC +CP 3=10, ∴t 3=10. ······································· 8分 ④点P 在AB 上与AC 相切时, AP 4=2r =2,
∴AD +DC +CB +BP 4=14, ∴t 4=14,
∴当t =2、6、10、14时,以点P 为圆心、以1为半径的圆与对角线AC 相切. ··························································· 9分
23. ⑴证明:∵点M 、P 、N 分别是AB 、BC 、CA 的中点, ∴线段MP 、PN 是△ABC 的中位线,
∴MP ∥AN ,PN ∥AM , ··················· 1分
∴四边形AMPN 是平行四边形, ····· 2分
∴∠MPN =∠A . ···························· 3分
⑵∠MP 1N +∠MP 2N =∠A 正确. ··················· 4分 如图所示,连接MN , ··························· 5分 ∵13AM AN AB AC ==,∠A =∠A ,
∴△AMN ∽△ABC ,
∴∠AMN =∠B ,1
3
MN BC =,
∴MN ∥BC ,MN =1
3
BC , ······················· 6分
∵点P 1、P 2是边BC 的三等分点,
∴MN 与BP 1平行且相等,MN 与P 1P 2平行且相等,MN 与P 2C 平行且相等, ∴四边形MBP 1N 、MP 1P 2N 、MP 2CN 都是平行四边形, ∴MB ∥NP 1,MP 1∥NP 2,MP 2∥AC , ······································································ 7分 ∴∠MP 1N =∠1,∠MP 2N =∠2,∠BMP 2=∠A , ∴∠MP 1N +∠MP 2N =∠1+∠2=∠BMP 2=∠A . ···································································· 8分 ⑶∠A . ···················································· 9分
24.解:⑴令2230x x -++=,
解得:121,3x x =-=, ∴A (-1,0),B (3,0) ······························· 2分 ∵223y x x =-++=2(1)4x --+, ∴抛物线的对称轴为直线x =1,
将x =1
代入y =+y
∴C (1,. ··································· 3分
⑵①在Rt △ACE 中,tan ∠CAE =
CE
AE
= ∴∠CAE =60o,
由抛物线的对称性可知l 是线段AB 的垂直平分线, ∴AC=BC ,
∴△ABC 为等边三角形, ·································································· 4分 ∴AB = BC =AC = 4,∠ABC=∠ACB = 60o, 又∵AM=AP ,BN=BP , ∴BN = CM ,
∴△ABN ≌△BCM , ∴AN =BM . ··························································································· 5分 ②四边形AMNB 的面积有最小值. ··················································· 6分 设AP=m ,四边形AMNB 的面积为S ,
由①可知AB = BC= 4,BN = CM=BP ,S △ABC ×42=, ∴CM=BN= BP=4-m ,CN=m , 过M 作MF ⊥BC ,垂足为F ,
则MF =MC ?sin60o)m -,
∴S △CMN =12CN MF =1
2
m )m -=2+, ·
························ 7分 ∴S =S △AB C -S △CMN
=-(2
)
22)m -+ ·
············································································ 8分
∴m =2时,S 取得最小值 ·························································· 9分