二○○八年绥化市初中毕业学业考试
数 学 试 卷
考生注意:
1.考试时间120分钟
2.全卷共三道大题,总分120分
一、填空题(每空3分,满分33分)
1.在抗震救灾过程中,共产党员充分发挥了先锋模范作用,截止5月28日17时,全国党员已缴纳特殊党费26.84亿元,用科学记数法表示为 元(结果保留两个有效数字). 2
.函数y =
中,自变量x 的取值范围是 . 3.如图,BAC ABD ∠=∠,请你添加一个条件: ,使OC OD =(只添一个即可). 4.如图,小明想用图中所示的扇形纸片围成一个圆锥,已知扇形的半径为5cm ,弧长是6πcm ,那么围成的圆锥的高度是 cm .
5.如图,某商场正在热销2008年北京奥运会的纪念品,小华买了一盒福娃和一枚奥运徽章,已知一盒福娃的价格比一枚奥运徽章的价格贵120元,则一盒福娃价格是 元.
6.有一个正十二面体,12个面上分别写有1~12这12个整数,投掷这个正十二面体一次,向上一面的数字是3的倍数或4的倍数的概率是 .
7.在半径为5cm 的圆中,两条平行弦的长度分别为6cm 和8cm ,则这两条弦之间的距离为 . 8.一幅图案.在某个顶点处由三个边长相等的正多边形镶嵌而成.其中的两个分别是正方形和正六边形,则第三个正多边形的边数是 .
9.如图,矩形ABCD 中,3AB =cm ,6AD =cm ,点E 为AB 边上的任意一点,四边形EFGB 也是
矩形,且2EF BE =,则
S = 2
cm .
10.三角形的每条边的长都是方程2
680x x -+=的根,则三角形的周长是 .
D O
C B A 第3题图 O B A 第4题图 5cm
A
D
C
E
F G
B 第9题图 第6题图
一共花了170元 第5题图
11.如图,菱形111AB C D 的边长为1,160B ∠= ;
作2
11
A D
B
C ⊥于点2
D ,以2AD 为一边,做第二个菱形222AB C D ,使
260B ∠= ;作322AD B C ⊥于点3D ,以3AD 为一边做第三个
菱形333AB C D ,使360B ∠=
; 依此类推,这样做的第n 个菱形n n n AB C D 的边n AD 的长是 . 二、选择题(每题3分,满分27分)
12.下列各运算中,错误的个数是( )
①01333-+=-
=③235(2)8a a = ④844
a a a -÷=- A .1
B .2
C .3
D .4
13.用电器的输出功率P 与通过的电流I 、用电器的电阻R 之间的关系是2
P I R =,下面说法正确的是( )
A .P 为定值,I 与R 成反比例
B .P 为定值,2
I 与R 成反比例 C .P 为定值,I 与R 成正比例
D .P 为定值,2I 与R 成正比例
14.为紧急安置100名地震灾民,需要同时搭建可容纳6人和4人的两种帐篷,则搭建方案共有( ) A .8种 B .9种 C .16种 D .17种 15.对于抛物线2
1
(5)33
y x =--+,下列说法正确的是( )
A .开口向下,顶点坐标(53),
B .开口向上,顶点坐标(53),
C .开口向下,顶点坐标(53)-,
D .开口向上,顶点坐标(53)-,
16.下列图案中是中心对称图形的是( )
17.关于x 的分式方程
15
m
x =-,下列说法正确的是( ) A .方程的解是5x m =+ B .5m >-时,方程的解是正数 C .5m <-时,方程的解为负数 D .无法确定
18.5月23日8时40分,哈尔滨铁路局一列满载着2400吨“爱心”大米的专列向四川灾区进发,途中除3次因更换车头等原因必须停车外,一路快速行驶,经过80小时到达成都.描述上述过程的大致图象是( )
1
D B
3
第11题图
A
C 2
B 2
C 3
D 3 B 1
D 2
C 1 A . B . C .
D .
第16题图
第18题图
19
.已知5个正数12345a a a a a ,,,,的平均数是a ,且12345a a a a a >>>>,则数据
123450a a a a a ,,,,,的平均数和中位数是( )
A .3a a ,
B .3
4
2
a a a +, C .
23562
a a a +, D .
34
562
a a a +,
20.如图,将ABC △沿DE 折叠,使点A 与BC 边的中点F 重合,下列结论中:①EF AB ∥且
1
2
EF AB =
;②BAF CAF ∠=∠; ③1
2
ADFE S AF DE = 四边形;
④2BDF FEC BAC ∠+∠=∠,正确的个数是( ) A .1 B .2 C .3
D .4
三、解答题(满分60分) 21.(本小题满分5分)
先化简:22
42
26926
a a a a a --÷++++,再任选一个你喜欢的数代入求值. 22.(本小题满分6分)
如图,方格纸中每个小正方形的边长都是单位1.
(1)平移已知直角三角形,使直角顶点与点O 重合,画出平移后的三角形. (2)将平移后的三角形绕点O 逆时针旋转90
,画出旋转后的图形.
(3)在方格纸中任作一条直线作为对称轴,画出(1)和(2)所画图形的轴对称图形,得到一个美丽的图案.
第20题图
t
t
B.
C .
D .
23.(本小题满分6分)
有一底角为60
的直角梯形,上底长为10cm ,与底垂直的腰长为10cm ,以上底或与底垂直的腰为一边作三角形,使三角形的另一边长为15cm ,第三个顶点落在下底上.请计算所作的三角形的面积. 24.(本小题满分7分)
A B C ,,三名大学生竞选系学生会主席,他们的笔试成绩和口试成绩(单位:分)分别用了两种方式进行了统计,如表一和图一: 表一
(1)请将表一和图一中的空缺部分补充完整.
(2)竞选的最后一个程序是由本系的300名学生进行投票,三位候选人的得票情况如图二(没有弃权票,每名学生只能推荐一个),请计算每人的得票数.
(3)若每票计1分,系里将笔试、口试、
得票三项测试得分按4:3:3
的比例确定个人成绩,请计算三位候选人的最后成绩,并根据成绩判
断谁能当选.
25.(本小题满分8分)
武警战士乘一冲锋舟从A 地逆流而上,前往C 地营救受困群众,途经B 地时,由所携带的救生艇将B 地受困群众运回A 地,冲锋舟继续前进,到C 地接到群众后立刻返回A 地,途中曾与救生艇相遇.冲锋舟和救生艇距A 地的距离y
(千米)和冲锋舟出发后所用时间x
(分)之间的函数图象如图所示.假设营救群众的时间忽略不计,水流速度和冲锋舟在静水中的速度不变. (1)请直接写出冲锋舟从A 地到C 地所用的时间.
图二 100
95 90 85
80
75
70 分数/分 图一
竞选人 A B C
(2)求水流的速度.
(3)冲锋舟将C 地群众安全送到A 地后,又立即去接应救生艇.已知救生艇与A 地的距离y (千米)和冲锋舟出发后所用时间x (分)之间的函数关系式为1
1112
y x =-+,假设群众上下船的时间不计,求冲锋舟在距离A 地多远处与救生艇第二次相遇?
26.(本小题满分8分)
已知:正方形ABCD 中,45MAN ∠=
,MAN ∠绕点A 顺时针旋转,它的两边分别交CB DC ,(或它们的延长线)于点M N ,.
当MAN ∠绕点A 旋转到BM DN =时(如图1),易证BM DN MN +=.
(1)当M A N ∠绕点A 旋转到BM DN ≠时(如图2),线段BM DN ,和MN 之间有怎样的数量关系?
写出猜想,并加以证明.
(2)当MAN ∠绕点A 旋转到如图3的位置时,线段BM DN ,和MN 之间又有怎样的数量关系?请直接写出你的猜想.
27.(本小题满分10分)
某工厂计划为震区生产A B ,两种型号的学生桌椅500套,以解决1250名学生的学习问题,一套A 型桌椅(一桌两椅)需木料3
0.5m ,一套B 型桌椅(一桌三椅)需木料3
0.7m ,工厂现有库存木料3
302m . (1)有多少种生产方案?
B B M B
C N C
N M C N M 图1 图2 图3
A A A D D D x (分)
(2)现要把生产的全部桌椅运往震区,已知每套A 型桌椅的生产成本为100元,运费2元;每套B 型桌椅的生产成本为120元,运费4元,求总费用y (元)与生产A 型桌椅x (套)之间的关系式,并确定总费用最少的方案和最少的总费用.(总费用=生产成本+运费)
(3)按(2)的方案计算,有没有剩余木料?如果有,请直接写出用剩余木料再生产以上两种型号的桌椅,最多还可以为多少名学生提供桌椅;如果没有,请说明理由. 28.(本小题满分10分)
如图,在平面直角坐标系中,点(30)C -,,点A B ,分别在x 轴,y
轴的正半轴上,且满足10OA -=.
(1)求点A ,点B 的坐标.
(2)若点P 从C 点出发,以每秒1个单位的速度沿射线CB 运动,连结AP .设ABP △的面积为S ,点P 的运动时间为t 秒,求S 与t 的函数关系式,并写出自变量的取值范围.
(3)在(2)的条件下,是否存在点P ,使以点A B P ,,为顶点的三角形与AOB △相似?若存在,请直接写出点P 的坐标;若不存在,请说明理由.
二○○八年黑龙江省绥化市初中毕业学业考试
数学试卷参考答案及评分标准
x
一、填空题,每空3分,满分33分(多答案题全对得3分,否则不得分) 1.9
2.710?
2.3x ≤且1x ≠
3.C D ∠=∠或ABC BAD ∠=∠或AC BD =或OAD OBC ∠=∠ 4.4
5.145
6.
12
7.1cm 或7cm 8.12 9.9
10.6或10或12
11
.1
n -??
二、选择题,每题3分,满分27分.
12.C 13.B 14.A 15.A 16.B 17.C 18.D 19.D 20.B
三、解答题,满分60分.
21.解:22
42
26926a a a a a --÷++++ 2(2)(2)2(3)
2(3)2
a a a a a +-+=
++- ······················································································· (1分)
2426
33a a a a ++=-
+++ ······································································································· (2分) 23
a =+···························································································································· (3分) n 取3-和2以外的任何数,计算正确都可给分. ·
······················································ (5分) 22.平移正确,给2分;旋转正确,给2分;轴对称正确,给2分,计6分.
23.解:当15BE =cm 时,ABE △的面积是2
50cm ;
当15CF =cm 时,BCF △的面积是2
75cm ;
当15BE =cm 时,BCE △
的面积是2
cm .
B
(每种情况,图给1分,计算结果正确1分,共6分) 24.解:(1)90;补充后的图如下(每项1分,计2分)
(2)A :30035105?=% B :30040120?=% C :3002575?=%
(方法对1分,计算结果全部正确1分,计2分)
(3)A :
8549031053
92.5433
?+?+?=++(分)
B :
9548031203
98433
?+?+?=++(分) C :
904853753
84433
?+?+?=++(分) B 当选
(方法对1分,计算结果全部正确1分,判断正确1分,计3分) 25.解:(1)24分钟 ······································································································ (1分) (2)设水流速度为a 千米/分,冲锋舟速度为b 千米/分,根据题意得
24()20
(4424)()20
b a a b -=??
-+=? ·································································································· (3分) 解得1121112
a b ?=????=??
答:水流速度是
1
12
千米/分. ························································································ (4分) (3)如图,因为冲锋舟和水流的速度不变,所以设线段a 所在直线的函数解析式为
x (分)
95 90 85 80 75
70
分数/分
竞选人
A B C
5
6
y x b =
+ ····················································································································· (5分) 把(440),代入,得110
3
b =-
∴线段a 所在直线的函数解析式为5110
63y x =-
························································ (6分) 由11112511063y x y x ?
=-+????=-??
求出20523?? ???,这一点的坐标 ·························································· (7分)
∴冲锋舟在距离A 地
20
3
千米处与救生艇第二次相遇. ·············································· (8分) 26.解:(1)BM DN MN +=成立. ········································································ (2分)
如图,把AND △绕点A 顺时针90
,得到ABE △,
则可证得E B M ,,三点共线(图形画正确) ······ (3
分) 证明过程中,
证得:EAM NAM ∠=∠ ····································· (4分)
证得:AEM ANM △≌△ ······························· (5分)
ME MN ∴= ME BE BM DN BM =+=+ DN BM MN ∴+= ······································································································· (6分) (2)DN BM MN -= ································································································· (8分) 27.解:(1)设生产A 型桌椅x 套,则生产B 型桌椅(500)x -套,由题意得
0.50.7(500)302
23(500)1250
x x x x +?-??
+?-?≤≥ ······················································································ (2分) 解得240250x ≤≤ ······································································································ (3分) 因为x 是整数,所以有11种生产方案. ······································································· (4分) (2)(1002)(1204)(500)2262000y x x x =+++?-=-+ ···································· (6分)
220-< ,y 随x 的增大而减少.
∴当250x =时,y 有最小值. ·
··················································································· (7分) ∴当生产A 型桌椅250套、B 型桌椅250套时,总费用最少.
此时min 222506200056500y =-?+=(元) ··························································· (8分) (3)有剩余木料,最多还可以解决8名同学的桌椅问题. ······································ (10分) 28.解:(1)10OA -=
230OB ∴-=,10OA -= ·························································································· (1分)
E A C N D
OB ∴=,1OA =
点A ,点B 分别在x 轴,y 轴的正半轴上
(10)(0A B ∴,, ········································································································ (2分) (2)求得90ABC ∠=
································································································· (3分)
(0(t t S t t ?=?->?? ≤
(每个解析式各1分,两个取值范围共1分) ····························································· (6分) (3)1(30)P -,
;21P ?- ?
;31P ? ?
;4(3P (每个1分,计4分) ······································································································································· (10分) 注:本卷中所有题目,若由其它方法得出正确结论,酌情给分.