2012年中考模拟题
数 学 试 卷(六)
*考试时间120分钟 试卷满分150分
一、选择题(下列各题的备选答案中,只有一个答案是正确的,将正确答案的序号填在题后的括号内,每小题4分,共40分) 1.估算272-的值( )
A .在1到2之间
B .在2到3之间
C .在3到4之间
D .在4到5之间
2.把多项式2
288x x -+分解因式,结果正确的是( ) A .()2
24x -
B .()2
24x -
C .()2
22x -
D .()2
22x +
3.若m +n =3,则2
2
2426m mn n ++-的值为( ) A.12
B.6
C.3
D.0
4.二元一次方程组2,
0x y x y +=??
-=?
的解是(
)
A .0,2.x y =??=?
B .2,0.x y =??=?
C .1,1.x y =??=?
D .1,
1.x y =-??=-?
5. 如图所示的几何体的主视图是(
)
6.下列运算中,正确的是( )
A.x+x=2x
B. 2x -x=1
C.(x 3)3=x 6
D. x 8÷x 2=x 4
7.如图,点A 在双曲线6
y x
=
上,且OA =4,过A 作AC ⊥x 轴,垂足为C ,OA 的垂直平分线交OC 于B ,则△ABC 的周长为 ( ) A .27
B .5
C .47
D .22
A .
B .
C .
D .
图5
8.如图,正五边形FGHMN 是由正五边形ABCDE 经过位似变换得到的,若AB:FG=2:3,则下列结论正确的是(
)
A .2DE=3MN ,
B .3DE=2MN ,
C . 3∠A=2∠F
D .2∠A=3∠F
9.在下图4×4的正方形网格中,△MNP 绕某点旋转一定的角度,得到△M 1N 1P 1,则其旋转中心可能是(
)
A .点A
B .点B
C .点C
D .点D
10.如图, AD 是以等边三角形ABC 一边AB 为半径的四分之一圆周,
P 为 AD 上任意一点,若AC=5,则四边形ACBP 周长的最大值是(
)
A . 15
B . 20
C .15+52
D .15+55
二、填空题(共5小题,每题4分,满分20分.请将答案填入答题卡的相应位置) 11.分解因式:2
2x x -= 12.请写出一个比5小的整数
13. a 、b 为实数,且ab =1,设P =11a b a b +++,Q =11
11
a b +++,则P Q (填“>”、“<”或“=”).
14. 如图4所示,A 、B 、C 、D 是圆上的点,17040A ∠=∠=°,°,则C ∠= 度.
15.已知, A 、B 、C 、D 、E 是反比例函数16
y x
=
(x>0)图象上五个整数点(横、纵坐标均为整数),分别以这些点向横轴或纵轴作垂线段,由垂线段所在的正方形边长为半径作四分之一圆周的两条弧,组成如图5所示的五个橄榄形(阴影部分),
E
D
C
N
M
H
G
F B
A
C
B
A
D
P
A
B C
D M
N
P
P 1 M 1
N 1 A
B
C
D
1
(图4)
则这五个橄榄形的面积总和是 (用含π的代数式表示)
三、解答题(满分90分.请将答案填入答题卡的相应位置) 16.(每小题7分,共14分) (1)解不等式:5x –12≤2(4x -3)
(2)先化简,再求值。其中3=
x ,2=y
2
2
2)11(y
xy x xy
x y +--
17.(每小题8分,共16分)
(1)计算:8-(3-1)0+|-1|.
(2)整理一批图书,如果由一个人单独做要花60小时。现先由一部分人用一小时整理,随后增加15人和他们一起又做了两小时,恰好完成整理工作。假设每个人的工作效率相同,那么先安排整理的人员有多少人?
18.(满分10分)
在梯形ABCD 中,AB ∥CD ,∠A =90°, AB =2,BC =3,CD =1,E 是AD 中点. 求证:CE ⊥BE .
19.(满分12分)以下统计图描述了九年级(1)班学生在为期一个月的读书月活动中,三个阶段(上旬、中旬、下旬)日人均阅读时间的情况:
(1)从以上统计图可知,九年级(1)班共有学生 人; (2)图7-1中a 的值是 ;
(3)从图7-1、7-2中判断,在这次读书月活动中,该班学生每日阅读时间 (填“普遍增加了”或“普遍减少了”);
(4)通过这次读书月活动,如果该班学生初步形成了良好的每日阅读习惯,参照以上统计图的变化趋势,至读书月活动结束时,该班学生日人均阅读时间在0.5~1小时的人数比活动开展初期增加了 人。
A
C B D
E
20.(满分12分)
如图8,在边长为1的小正方形组成的网格中,ABC △的三个顶点均在格点上, 请按要求完成下列各题:
(1) 用签字笔...画AD ∥BC (D 为格点),连接CD ; (2) 线段CD 的长为 ;
(3) 请你在ACD △的三个内角中任选一个锐角..
,若你所选的锐角是 ,则它所对应的正弦函数值是 。
(4) 若E 为BC 中点,则tan ∠CAE 的值是
图8
如图, 四边形OABC 为直角梯形,A (4,0),B (3,4),C (0,4). 点M 从O 出发以每秒2个单位长度的速度向A 运动;点N 从B 同时出发,以每秒1个单位长度的速度向C 运动.其中一个动点到达终点时,另一个动点也随之停止运动.过点N 作NP 垂直x 轴于点P ,连结AC 交NP 于Q ,连结MQ .
(1)点 (填M 或N )能到达终点;
(2)求△AQM 的面积S 与运动时间t 的函数关系式,并写出自变量t 的取值范围,当t 为何值时,S 的值最大;
(3)是否存在点M ,使得△AQM 为直角三角形?若存在,求出点M 的坐标,若不存在,说明理由.
y
x
P Q
B
C
N M
O
A
如图,已知直线128
:33
l y x =
+与直线2:216l y x =-+相交于点C l l 12,、分别交x 轴A B 、两点.矩形DEFG 的顶点D E 、分别在直线12l l 、上,顶点F G 、都在x 轴上,且
点G 与点B 重合.
(1)求ABC △的面积;
(2)求矩形DEFG 的边DE 与EF 的长;
(3)若矩形DEFG 从原点出发,沿x 轴的反方向以每秒1个单位长度的速度平移,设移动时间为(012)t t ≤≤秒,矩形DEFG 与ABC △重叠部分的面积为S ,求S 关于t 的函数关系式,并写出相应的t 的取值范围.
A D
B E O
C F x
y
1
l 2l
(G )
2010年中考模拟题(六) 数学试题参考答案及评分标准
一、选择题(每小题4分,共40分)
1.C 2.C 3.A ; 4.C 5.D ;6.A 7.A 8.B 9.B 10.C
二、填空题(每小题4分,共20分)
11.x (x -2);12.答案不唯一,小于或等于2的整数均可,如:2,1等;13.=; 14.40;15.13π-26
三、解答题
16. (1)(本题满分7分)
解:5x –12≤8x -6. ················································································· 3分
3x -≤6. ·································································· 5分 x ≥-2 . ····································································· 7分
(2)解:原式=
2
)
(y x xy
xy y x -?- =
y x -1
……………………………………………………4分
将3=x ,2=y 代入,则
原式=232
31+=-……………………………………7分
17.
(1)解:08(31)1221122--+-=-+=
……………………8分
(2)解:设先安排整理的人员有x 人,依题意得,
2(15)16060
x x ++= ……………………4分
解得, x =10.
答:先安排整理的人员有10人.……………………8分
18.证明: 过点C 作CF ⊥AB ,垂足为F .……………… 1分
∵ 在梯形ABCD 中,AB ∥CD ,∠A=90°, ∴ ∠D =∠A =∠CFA =90°. ∴四边形AFCD 是矩形.
AD=CF, BF=AB-AF=1.……………………………… 3分 在Rt △BCF 中, CF 2=BC 2-BF 2=8,
∴ CF=22.
∴ AD=CF=22.……………………………………………………………… 5分 ∵ E 是AD 中点, ∴ DE=AE=
1
2
AD=2.…………………………………… 6分 在Rt △ABE 和 Rt △DEC 中, EB 2=AE 2+AB 2=6, EC 2= DE 2+CD 2=3, EB 2+ EC 2=9=BC 2.
∴ ∠CEB =90°.………………………………………………………9分 ∴ EB ⊥EC .………………………… 10分
(其他不同证法,参照以上标准评分)
19.(每小题各3分,共12分)
(1)50 (2)3 (3)普遍增加了
(4)15
20.(每小题3分,共12分)
(1)如图 (2)5
(3)∠CAD ,
55(或∠ADC ,5
5
2) (4)
2
1
21.解:(1)点 M ········································································································ 1分 (2)经过t 秒时,NB t =,2OM t = 则3CN t =-,42AM t =-
∵BCA ∠=MAQ ∠=45
∴ 3QN CN t ==- ∴ 1 PQ t =+ ······································································· 2分 ∴11
(42)(1)22
AMQ S AM PQ t t =
=-+ △ 22t t =-++ ······················································································································· 3分
∴2
219
224S t t t ??=-++=--+ ???
···················································································· 5分
∵02t ≤≤∴当1
2
t =
时,S 的值最大. ······································································ 6分 (3)存在. ·················································································································· 7分 设经过t 秒时,NB =t ,OM=2t 则3CN t =-,42AM t =-
∴BCA ∠=MAQ ∠=45
······················································································ 8分 ①若90AQM ∠= ,则PQ 是等腰Rt △MQA 底边MA 上的高 ∴PQ 是底边MA 的中线 ∴1
2
PQ AP MA == ∴1
1(42)2
t t +=- ∴12
t =
∴点M 的坐标为(1,0) ·························································································· 10分
②若90QMA ∠=
,此时QM 与QP 重合 ∴QM QP MA ==
∴142t t +=- ∴1t =
∴点M 的坐标为(2,0) ·························································································· 12分
22.(1)解:由
28
033
x +=,得4x A =-∴.点坐标为()40-,.
由2160x -+=,得8x B =∴.点坐标为()80,.
∴()8412AB =--=. ·························································································· 2分
由2833216y x y x ?=+???=-+?
,.解得56x y =??
=?,.∴C 点的坐标为()56,. ······································· 3分 ∴11
1263622
ABC C S AB y =
=??=△·. ······························································· 4分 (2)解:∵点D 在1l 上且28
88833
D B D x x y ==∴=?+=,.
∴D 点坐标为()88,. ·
···························································································· 5分 又∵点E 在2l 上且821684E D E E y y x x ==∴-+=∴=,..
∴E 点坐标为()48,. ····························································································· 6分 ∴8448OE EF =-==,. ················································································· 8分
(3)解法一:①当03t <≤时,如图1,矩形DEFG 与ABC △重叠部分为五边形
CHFGR (0t =时,为四边形CHFG ).过C 作CM AB ⊥于M ,则Rt Rt RGB CMB △∽△.
∴
BG RG BM CM =,即36t RG
=,∴2RG t =. Rt Rt AFH AMC △∽△,
∴()()112
36288223
ABC BRG AFH S S S S t t t t =--=-??--?-△△△.
即241644
333
S t t =-++. ········································································· 14分
A D
B E
O
R
F x
y
1
l 2l
M
(图3)
G C
A D
B E
O C
F x
y
1
l 2l
G (图1)
R
M A D B E
O C F x
y
1
l 2l
G (图2)
R
M