2012山东滨州中考数学
滨州市2012年初中学生学业考试
数学试题
(满分120分,考试时间120分钟)
一、选择题:本大题共12小题,在每小题的四个选项中只有一个是正确的,请把正确的选项选出来,并将其字母标号填写在答题栏内(每小题选对得3分,选错、不选或选出的
超过一个均记0分,满分36分)
31((2012山东滨州,1,3分)等于 ,2
A(-6 B(6 C(-8 D(8
【答案】C
2((2012山东滨州,2,3分)以下问题,不适合用全面调查的是
A(了解全班同学每周体育锻炼的时间
B(鞋厂检查生产的鞋底能承受的弯折次数
C(学校招聘老师,对应聘人员面试
D(黄河三角洲中学调查全校753名学生的身高
【答案】B
3((2012山东滨州,3,3分)借助一副三角尺,你能画出下面哪个度数的角
A(65? B(75? C(85? D(95?
【答案】B
4((2012山东滨州,4,3分)一个三角形的三个内角的度数之比为2:3:7,则这个三角形一定是
A(等腰三角形 B(直角三角形 C(锐角三角形 D(钝角三角形
【答案】D
2x,1,x,1,5((2012山东滨州,5,3分)不等式的解集是 ,x,8,4x,1,
A( B( C( D(空集 x,3x,22,x,3
【答案】A
6((2012山东滨州,6,3分)某几何体的三视图如图所示,则这个几何体是
主视图左视图俯视图
第6题图
A(圆柱 B(正方体 C(球 D(圆锥
【答案】D
7((2012山东滨州,7,3分)李明同学早上骑自行车上学,中途因道路施工步行一段路,到学校共用时15分种钟(他骑自行车的平均速度是250米/分钟,步行的平均速度是80米/1
分钟(他家离学校的距离是2900米(如果他骑自行车和步行的时间分别为x,y分钟,列出的方程组是
1,x,y,15,,x,y,A( B( ,,480x,250y,2900,,250x,80y,2900,
1,x,y,15,,x,y, D( C(,,4250x,80y,2900,,80x,250y,2900,
【答案】D
8((2012山东滨州,8,3分)直线y=x-1不经过
A(第一象限 B(第二象限 C(第三象限 D(第四象限
【答案】B
2y,,3x,x,49((2012山东滨州,9,3分)抛物线与坐标轴的交点的个数是
A(3 B(2 C(1 D(0
【答案】A
10((2012山东滨州,10,3分)把?ABC三边的长度都扩大为原来的3倍,则锐角A的正弦函数值
1A(不变 B(缩小为原来的 C(扩大为原来的3倍 D(不能确定 3
【答案】A
11((2012山东滨州,11,3分)若菱形的周长为8cm,高为1cm,则菱形两邻角的度数比为
A(3:1 B(4:1 C(5:1 D(6:1
【答案】C
23201212((2012山东滨州,12,3分)求的值,可令 1,2,2,2,,,,,2
2320122320132013=,则2S=,因此,1,2,2,2,,,,,22,2,2,,,,,22S,S,2,1S
232012仿照以上推理,计算出的值为 1,2,2,2,,,,,2
20132012,,515120122013A( B( C( D( 5,15,144
【答案】C
二、填空题:本大题共6个小题,每小题填对最后结果得4分,满分24分( 13((2012山东滨州,13,4分)下
是晨光中学男子篮球队队员的年龄统计:
年龄 13 14 15 16
人数 1 5 5 1 他们的平均年龄是 (
【答案】14.5(
52y,x,8x,2y,,14((2012山东滨州,14,4分)下列函数:?y,2x,1;?;?;x2
13a?;?;?中,y是x的反比例函数的有 (填序号)( y,y,y,32xxx
【答案】??(
6的算15((2012山东滨州,15,4分)根据你学习的数学知识,写出一个运算结果为a式 (
【答案】答案不唯一,只要合理就得满分.
16((2012山东滨州,16,4分)如图,在?ABC中,AB=AD=DC,?BAD=20?,则?C= ?(
AA
E
DF
BCBDC
(第16题图) (第18题图)
【答案】40(
17((2012山东滨州,17,4分)方程的根是 ( x(x,2),x
【答案】0,3(
18((2012山东滨州,18,4分)如图,锐角三角形ABC的边AB,AC上的高线EC,BF相交于点D,请写出图中的两对相似三角形 ((用相似符号连接)(
【答案】?BDE,?CDF,?ABF,?ACE.
三、解答题:本大题共7小题,满分60分(解答时请写出必要的演推过程( 19((2012山东滨州,19,6分)
20120,2计算:,2,(,1),(,,3),8,(,2)
1【答案】原式=2,1,1,22,„„„„„„„„„„„„„„„„„„„5分 4
13 =„„„„„„„„„„„„„„„„„„„„„„„„„6分 ,224
20((2012山东滨州,20,7分)
滨州市体育局要组织一次篮球赛,赛制为单循环形式(每两队之间都赛一场),
安排28场比赛,应邀请多少支球队参加比赛,学习以下解答过程,并完成填空( 解:设应邀请x支球队参赛,则每队共打 场比赛,比赛总场数用代数式表示为 (根据题意,可列出方程 (
整理,得 (
解这个方程,得 (
合乎实际意义的解为 (
答:应邀请 支球队参赛(
3
11【答案】;;;,;;8( x,8x,,7(x,1)x(x,1)x(x,1),28x,81222
21((2012山东滨州,21,8分)
如图,PA,PB是?O的切线,A,B为切点,AC是?O的直径,?P=50?,求?BAC的度数(
P
BA
O
C
【答案】解:?PA,PB分别切?O于A,B点,AC是?O的直径
?PA=PB,?PAC=90?„„„2分
??PAB=?PBA„„„„„„„„„„„„„„„„„„„„„„„3分
00180,500,65又??P=50?,??PAB=?PBA=„„„„„„„„„„„6分 2
??BAC=?PAC-?PAB=90?-65?=25?„„„„„„„„„„„„„„„„8分 22((2012山东滨州,22,8分)
在一个口袋中有4个完全相同的小球,把它们分别标上数字-1,0,1,2,随机的摸出一个小球记录数字然后放回,再随机的摸出一个小球记录数字(求下列事件的概率: (1)两次都是正数的概率P(A);
(2)两次的数字和等于0的概率P(B)(
【答案】
解:根据题意,可以用以下
表示所有不同的结果(
第一次 -1 0 1 2 第二次
(-1,-1) (0,-1) (1,-1) (2,-1) -1
(-1,0) (0,0) (1,0) (2,0) 0
(-1,1) (0,1) (1,1) (2,1) 1
(-1,2) (0,2) (1,2) (2,2) 2
„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„4分
(1)由上表可以看出,所有可能出现的结果共有16种,每种结果出现的可
41能性都相同,两个数字都是正数的结果有4种,所以P(A)=„„„6分 ,164
3(2)由上表可知,两个数字和为0的结果有3种,所以P(B)„„8分 1623((2012山东滨州,23,9分)
我们知道“连接三角形两边中点的线段叫做三角形的中位线”,“三角形的中位线平行于三角形的第三边,且等于第三边的一半”(类似地,我们连接梯形两腰中点的线段叫做梯形的中4
位线(如图,在梯形ABCD中,AD?BC,点E,F分别是AB,CD的中点,那么EF就是梯形ABCD的中位线(通过观察、测量,猜想EF和AD,BC有怎样的位置和数量关系,并证明你的结论(
AD
EF
CB
【答案】
1解:结论为:EF?AD?BC,EF=„„„„„„„„„„„4分 (AD,BC)2
证明:连接AF并延长交BC的延长线于点G(„„„„„„„„„„„„„„5分
?AD?BG,??DAF=?G,
AD
FE
GCB
在?ADF和?GCF中,
,DAF,,G,
,,DFA,,CFG ,
,DF,FG,
??ADF??GCF(„„„„„„„„„„„„„„„„„„„„„„„„„„6分
?AF=FG,AD=CG( „„„„„„„„„„„„„„„„„„„„„„„„„7分
1又?AE=EB,?EF?BG,EF=(„„„„„„„„„„„„8分 BG2
1即EF?AD?BC,EF=„„„„„„„„„„„„„„„„„„9分 (AD,BC)2
24((2012山东滨州,24,10分)
2y,ax,bx,c如图,在平面直角坐标系中,抛物线经过A(-2,-4),O(0,0),B(2,0)三点.
2y,ax,bx,c(1)求抛物线的解析式;
(2)若点M是抛物线对称轴上一点,求AM+OM的最小值(
5
【答案】
2y,ax,bx,c解:(1)把A(-2,-4),O(0,0),B(2,0)三点代入中,得
4a,2b,c,,4,
,4a,2b,c,0„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 ,
,c,0,
1解这个方程组,得,b=1,c=0. „„„„„„„„„„„„„„„„„4分 a,,2
12所以解析式为(„„„„„„„„„„„„„„„„„„„„„„5分 y,,x,x2
11122(2)由=(1),可得 y,,x,x,x,,222
抛物线的对称轴为x=1,并且对称垂直平分线段OB(„„„„„„„„„„„„6分
?OM=BM,
?OM+AM=BM+AM„„„„„„„„„„„„„„„„„„7分
连接AB交直线x=1于M,则此时OM+AM最小(„„„„„„„„„„„„8分
过A点作AN?x轴于点N,在Rt?ABN中,
2222AB=„„„„„„„„„„„„„„„„„„9分 AN,BN,4,4,42
因此OM+AM最小值为42„„„„„„„„„„„„„„„„„„„„„„10分
25((2012山东滨州,25,12分)
llll如图1,,,,是一组平行线,相邻2条平行线间的距离都是1个单位长度,正方3124
6
l形ABCD的4个顶点A,B,C,D都在这些平行线上,过点A作AF?于点F,交l于点32
l于点E,交于点G( H,过点C作CE?l32
(1)求证:?ADF??CBE;
(2)求正方形ABCD的面积;
h(3)如图2,如果四条平行线不等距,相邻的两条平行线间的距离依次为h,h,,试312
h用h,h,表示正方形ABCD的面积S( 312
AAll11
h1BEHBEHl2l2h2
Gl3GFDDFlh33
l4ClC4 图1 图2
【答案】
(1)证明:在Rt?AFD和Rt?CEB中,AD=BC,AF=CE, ?Rt?AFD? Rt?CEB„„„„„„„„„„„„„„„„„„„„„„„„3分 (2)解:?ABH+?CBE=90?,?ABH+?BAH=90?,
??CBE=?BAH(
又?AB=BC,?AHB=?CEB=90?,
??ABH??BCE„„„„„„„„„„„„„„„„„„„„„„„„„„6分 不难得出 ?ABH??BCE??CDG??DAF(„„„„„„„„„„„7分 S,4S,S? ,ABH正方形ABCD正方形EGFH
1= 4,,2,1,1,12
=5(„„„„„„„„„„„„„„„„„„„„„„„„„„8分 (3)解:由(1)知,Rt?AFD? Rt?CEB
h,h?„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„9分 13
由(2)知,?ABH??BCE??CDG??DAF(„„„„„„„„„„„„10分 S,4S,S? ,ABH正方形ABCD正方形EGFH
12 = 4,(h,h),h,h12122
22 =„„„„„„„„„„„„„„„„„„„„12分 2h,2hh,h1122
7