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河南省中考数学试题及答案解析
2010 年河南省初中学业水平暨高级中等学校招生考试试卷
数学
一、选择题(每小题 3 分,共 18 分)
下列各小题均有四个答案,其中只有一个是正确的.
1. -
1
2
的相反数是
11
(A)(B) -(C) 2(D) -2
22
2.我省 2009 年全年生产总值比 2008 年增长 10.7%,达到约 19 367 亿元.19 367 亿元用科
学记数法表示为
(A)1.9367 ⨯1011 元(B)1.9367 ⨯1012 元
(C)1.9367 ⨯1013 元(D)1.9367 ⨯1014 元
, , , ,,
3 .在某次体育测试中,九年级三班6 位同学的立定跳远成绩(单位:
m)分别为:
1.711.851.851.96 2.10 2.31.则这组数据的众数和极差分别是
(A)1.85 和 0.21(B)2.31 和 0.46(C)1.85 和 0.60(D)2.31 和 0.60
4.如图, △ABC 中, 点D、E 分别是 AB、AC 的中点,则下列
结论:
①BC = 2DE
②ADE ∽△ ABC ;③
中正确的有
AD AB
=
AE AC
.其
A
D E
(A)3 个(B)2 个(C)1 个(D)0 个
5.方程 x2 - 3 = 0 的根是
(A) x = 3(B) x = 3,x = -3
12
B C
(第 4 题)
(C) x =
3 (D) x = 3,x = - 3 y
1 2
B'
A'
6.如图,将 △ABC 绕点 C (0, 1) 旋转180°得到 △A'B'C' ,设点 A'
的坐标为 (a,b) ,则点 A 的坐标为A
(A)(-a, b)(B)(-a, b - 1)(C)(-a, b + 1)(D)(-a, b - 2)
二、填空题(每小题 3 分,共 27 分)
7.计算:
-1 + (-2)2 =.
O
C
B
(第 6 题)
x
8.若将三个数 - 3,7,11 表示在数轴上,其中
-2 -1 0 1 2 3 4 5
(第 8 题)
能被如图所示的墨迹覆盖的数是.
9.写出一个 y 随 x 的增大而增大的一次函数的解析式:
.
10.将一副直角三角板如图放置,使含 30°角的三角板的短直角边和含 45°角的三角板的一
1
条直角边重合,则 ∠1 的度数为.
D
m
O
1
C
(第 10 题)
B A
(第 11 题)
11.如图, AB 切 ⊙O 于点 A , BO 交 ⊙O 于点 C ,点 D 是 CmA 上异于点 C、A 的一点,
若 ∠ABO = 32°,则 ∠ADC 的度数是.
12.现有点数为 2,3,4,5 的四张扑克牌,背面朝上洗匀,然后从中任意抽取两张,这两
张牌上的数字之和为偶数的概率是.
13.如图是由大小相同的小正方体组成的简单几何体的主视图和左视图,那么组成这个几何
体的小正方体的个数最多为.
A
D C
E
主视图左视图
B
E C
A
D
B
(第 13 题)(第 14 题)
(第 15 题)
14.如图,矩形 ABCD 中,AB = 1,AD =
2 .以 AD 的长为半径的⊙A 交 BC 边于点 E ,
则图中阴影部分的面积为.
,,
15.如图,Rt△ ABC 中,∠C = 90° ∠ABC = 30° AB = 6 .点 D 在 AB 边上,点 E 是
BC 边上一点(不与点 B、C 重合),且 DA = DE ,则 AD 的取值范围是
三、解答题(本大题共 8 个小题,满分 75 分)
.
(
16. 8 分)已知 A =
1 1 x
,B = ,C = .将他们组合成 ( A - B) ÷ C 或 A - B ÷ C
x - 2 x2 - 4 x + 2
的形式,请你从中任选一种进行计算.先化简,再求值,其中 x = 3 .
....
2
17.9 分)如图,四边形 ABCD 是平行四边形,△AB'C 和 △ ABC
B'
关于 AC 所在的直线对称, AD 和 B'C 相交于点 O ,连结 BB' .A
(1)请直接写出图中所有的等腰三角形(不添加字母)
(2)求证:
△ AB'O ≌△CDO .
O
D
B
C
18.(9 分)“校园手机”现象越来越受到社会的关注.“五一”期间,小记者高凯随机调查
了城区若干名学生和家长对中学生带手机现象的看法,统计整理并制作了如下的统计图:
学生及家长对中学生带手机的态度统计图
人数
家长对中学生带手机
的态度统计图
280
学生
210
140
140
家长
赞成
反对
80
无所谓
70
40
30
30
20%
赞成
无所谓
反对 类别
图①图②
(1)求这次调查的家长人数,并补全图① ;
(2)求图 ② 中表示家长“赞成”的圆心角的度数;
(3)从这次接受调查的学生中,随机抽查一个,恰好是持“无所谓”态度的学生的概率是
多少?
3
C
19 .( 9 分 ) 如 图 , 在 梯 形 ABCD 中 , A D∥ B , E 是 BC 的 中 点 ,
AD = 5,BC = 12,CD = 4 2 , ∠C = 45°,点 P 是 BC 边上一动点,设 PB 的长为 x .
(1)当 x 的值为时,以点 P、A、D、E 为顶点的四边形为直角梯形.
(2)当 x 的值为时,以点 P、A、D、E 为顶点的四边形为平行四边形.
(3)当 P 在 BC 边上运动的过程中,以点 P、A、D、E 为顶点的四边形能否构成菱形?
试
说明理由.
A
D
B
P
E
C
(
20. 9 分)为鼓励学生参加体育锻炼,学校计划拿出不超过 1 600 元的资金再购买一批篮球
和排球.已知篮球和排球的单价比为3∶2 ,单价和为 80 元.
(1)篮球和排球的单价分别是多少元?
(2)若要求购买的篮球和排球的总数量是 36 个,且购买的篮球数量多于 25 个,有哪几种
购买方案?
4
21.(10 分)如图,直线 y = k x + b 与反比例函数
1
y = k 2 ( x > 0) 的图象交于 A(1, , B(a,
x
两点.
y
(1)求 k 、k 的值;
12
A
(2)直接写出 k x + b -
1
k
2 > 0 时 x 的取值范围;
x
B C
(3)如图,等腰梯形 OBCD 中, BC ∥ OD ,
OB = CD ,OD 边在 x 轴上,过点 C 作 CE ⊥ OD
于 E ,CE 和反比例函数的图象交于点 P .当梯形
OBCD 的面积为 12 时,请判断 PC 和 PE 的大小
关系,并说明理由.
P
O E D x
5
22.(10 分)
(1)操作发现
如图,矩形 ABCD 中, E 是 AD 的中点,将 △ ABE 沿 BE 折叠后
得到 △GBE ,且点G 在矩形 ABCD 内部.小明将 BG 延长交 DC
于点 F ,认为 GF = DF ,你同意吗?
说明理由.
A E
D
F
(2)问题解决
保持
(1)中的条件不变,若 DC = 2DF ,求
AD
AB
的值.
B G C
(3)类比探究
n
保持
(1)中的条件不变,若 DC = ·DF ,求
AD
AB
的值.
6
23. 11 分)在平面直角坐标系中,已知抛物线经过 A(-4, ,
B(0, 4) , C (2, 三点.
(1)求抛物线的解析式;
(2)若点 M 为第三象限内抛物线上一动点,点 M 的横坐标
为 m
AMB 的面积为 S .求S 关于 m 的函数关系式,并
求出 S 的最大值;A
(3)若点 P 是抛物线上的动点,点Q 是直线 y = - x 上的动
y
O
C x
点,判断有几个位置能使以点 P、Q、B、O 为顶点的四边
形为平行四边形,直接写出相应的点 Q 的坐标.
M
B
7
2010 年河南省初中学业水平暨高级中等学校招生考试
数学试题参考答案及评分标准
一、选择题(每小题 3 分,共 18 分)
题号
答案
1
A
2
B
3
C
4
A
5
D
6
D
二、填空题(每小题 3 分,共 27 分)
题号
7
8 9
10
11 12
13
14
15
答案
5
7
答案不唯一,
如 y = x 等
75° 29° 1
3
7
1 π
2 - - 2 ≤ AD < 3
2 4
三、解答题(本大题共 8 个小题,满分 75 分)
⎛ 12⎫x
-÷················································ 1 分
⎝ x - 2x2 - 4 ⎭x + 2
x + 2
⨯··················································································· 5 分
(x + 2)(x - 2)x
=
1
x - 2
. ······································································································ 7 分
当 x = 3 时,原式=
1
3 - 2
= 1. · ·········································································· 8 分
选二:
A - B ÷ C =
1 2 x
- ÷
x - 2 x2 - 4 x + 2
· ·························································· 1 分
2x + 2
-⨯········································································ 3 分
x - 2 (x + 2)(x - 2)x
=
=
1 2
- ··························································································· 4 分
x - 2 x( x - 2)
x - 2 1
= . ···························································································· 7 分
x( x - 2) x
当 x = 3 时,原式=
1
3
. ···················································································· 8 分
17.
(1) △ABB' , △AOC 和 △BB'C . ·························································· 3 分
(2)在ABCD 中, AB = DC,∠ABC = ∠D .
由轴对称知AB' = AB,∠ABC = ∠AB'C . · ···················································· 7 分
∴ AB' = CD,∠AB'O = ∠D .
在 △AB'O 和 △CDO 中,
⎧∠AB'O = ∠D,
⎪
⎩
8
AB'O ≌△CDO . ··················································································· 9 分
18.
(1)家长人数为80 ÷ 20% = 400 . ··························································· 3 分
(正确补全图 ① ). · ······················································································ 5 分
(2)表示家长“赞成”的圆心角的度数为
40
400
=
⨯ 360° 36︒ .· ···························· 7 分
(3)学生恰好持“无所谓”态度的概率是
30
140 + 30 + 30
= 0.15 . ························ 9 分
⎩ 48n + 32(36 - n)≤1 600.
19.
(1)3 或 8;(本空共 2 分,每答对一个给 1 分) ············································· 2 分
(2)1 或 11;(本空共 4 分,每答对一个给 2 分) · ··············································· 6 分
(3)由
(2)知,当 BP = 11 时,以点 P、A、D、E 为顶点的四边形是平行四边形.
∴ EP = AD = 5 . ·························································································· 7 分
过 D 作 DF ⊥ BC 于 F , 则 DF = FC = 4 ,∴ FP = 3 .
∴ DP = FP2 + DF 2 = 32 + 42 = 5 . ····························································· 8 分
∴ EP = DP ,故此时PDAE 是菱形.
即以点 P、A、D、E 为顶点的四边形能构成菱形.··············································· 9 分
2
20.
(1)设篮球的单价为 x 元,则排球的单价为x 元.依题意得
3
2
x +x = 80 . · ····························································································· 3 分
3
2
解得 x = 48 .∴x = 32.
3
即篮球和排球的单价分别是 48 元、32 元. ·························································· 4 分
(2)设购买的篮球数量为 n 个,则购买的排球数量为 (36 - n) 个.
⎧n > 25,
∴⎨·········································································· 6 分
9 8
解得 25 < n ≤ 28 . ························································································ 7 分
而 n 为整数,所以其取值为 26,27,28,对应的 36 - n 的值为10,,.所以共有三种购买
方案.
方案一:
购买篮球 26 个,排球 10 个;
方案二:
购买篮球 27 个,排球 9 个;
方案三:
购买篮球 28 个,排球 8 个. ································································· 9 分
21.
(1)由题意知k = 1⨯ 6 = 6 .····································································· 1 分
2
∴ 反比例函数的解析式为 y = 6
x
.
又 B(a, 在 y =
6
x
的图象上,∴ a = 2 .∴ B(2, .
63)
直线 y = k x + b 过 A(1,), B(2, 两点,
1
9
⎧k + b = 6,⎧k = -3,
1
2k + b = 3.⎩b = 9.
(2) x 的取值范围为1 < x < 2.········································································· 6 分
(3)当 S
梯形OBCD
= 12 , PC = PE . ································································· 7 分
B3
设点 P 的坐标为 (m,n) ,BC ∥OD,CE ⊥ OD,BO = CD,(2,),
3)
∴ C (m,,CE = 3,BC = m - 2,OD = m + 2 .
∴ S
梯形OBCD =
BC + OD m - 2 + m + 2
2 2
⨯ 3 .
31
∴
22
∴ PC = PE . ····························································································· 10 分
22.
(1)同意.连接 EF , ∠EGF = ∠D = 90°, EG = AE = ED,EF = EF .
∴∴
∴ Rt△EGF ≌ Rt△EDF .∴ GF = DF . · ······················································ 3 分
(2)由
(1)知, GF = DF .设 DF = x , BC = y ,则有 GF = x,AD = y
DC = 2DF, CF = x,DC = AB = BG = 2x. BF = BG + GF = 3x .
在 Rt△BCF 中, BC 2 + CF 2 = BF 2 ,即 y 2 + x 2 = (3x)2 .
y
== 2 . ······································································ 6 分
AB2 x
(3)由
(1)知, GF = DF .设 DF = x,BC = y ,则有 GF = x,AD = y.
n
DC = ·DF ,∴ DC = AB = BG = nx .
=
∴ CF = (n - 1)x,BF = BG + GF (n + 1)x .
[
在 Rt△BCF 中, BC 2 + CF 2 = BF 2 ,即 y 2 +(n - 1)x]2 = [(n + 1)x]2 .
y2 n ⎛2 ⎫
⎪ . · ···················································· 10 分
23.
(1)设抛物线的解析式为 y=ax2+bx+c(a≠0),则有
⎧16a - 4b + c = 0,⎧1
⎪⎪
⎩
⎪4a + 2b + c = 0.⎨b = 1,
⎪c = -4.
⎪
1 ⎩
∴抛物线的解析式 y=
(2)过点 M 作 MD⊥x 轴于点 D.设 M 点的坐标为(m,n).
10
则 AD=m+4,MD=﹣n,n= 1
2
m2+m-4 .
∴S =
AMD+S 梯形 DMBO-
ABO
11
( m+4) (﹣n)+(﹣n+4) (﹣m) -×4×4
222
= ﹣2n-2m-8
= ﹣2(
1
2
m2+m-4) -2m-8
= ﹣m2-4m (-4< m < 0)..............................6 分
∴S 最大值 = 4 …………………………………………………… 7 分
(3)满足题意的 Q 点的坐标有四个,分别是:
(-4 ,4 ),(4 ,-4),
(-2+ 2 5 ,2- 2 5 ),(-2- 2 5 ,2+ 2 5 )…………………………… 11 分
11