2011河北省初中毕业生升学文化课考试
参考答案:
1.C 2.B 3.D 4.D 5.D 6.A 7.C 8.C 9.B 10.B 11.A 12. B. 13. 14. 5 15. 1 16. 27° 17. 2 18. 3 19.
解:将x2,y
y
a中,得a
∴(a1)(a1)7a217a26
=269 20. 解:⑴如图1.
⑵ AA'CC'2
在Rt⊿OA'C'中,OA'OC'=2
,得A'C'
AC, ∴四边形AA'C'C的周长
=421. 解:⑴ P(得到负数)=
13
⑵用下表列举所有的可能结果:
从上表可知,一共有九种可能,其中两人得到的数相同的有三种, 因此 P(两人“不谋而合”)=
13
(注:画树状图正确也相应给分)
22. 解:⑴ 设乙单独整理x分钟完工,根据题意得:
2040
x
解得:x80.经检验x80是原方程的解.
2020
1
答:乙单独整理80分钟完工.
⑵ 设甲整理y分钟完工,根据题意得:
3080
y40
1,
解得:y 25
答:甲至少整理25分钟完工.
(注:以下解答也给分.设甲、乙分别整理y,z分钟,得∵z30,∴802y30,∴y25.)
23. 解:⑴证明:∵ 四边形ABCD是正方形 ,∴DCDA,DCEDAG90°. 又∵CE
AG,∴⊿DCE≌⊿DAG.∴EDCGDA,DEDG.又∵
z80
y40
1.∴z802y.
ADEEDC90,∴ADEGDA90,∴DEDG.
⑵如图2(注:图3或其它画法正确的相应给分)
⑶四边形CEFK是平行四边形. 证明:设CK,DE相交于M点.
∵四边形ABCD和四边形DEFG都是正方形,∴AB∥CD, AB=CD, EF=DG, EF∥DG, ∵BK=AG, ∴KG=AB=CD, ∴四边形CKGD为平行四边形. ∴CK=DG=EF, CK∥DG. ∴KMEGDEDEF90.∴KMEDEF180.∴CK∥EF, ∴四边形CEFK是平行四边形.
(注:由CK∥DG, EF∥DG得CK∥EF也可) ⑷
S正方形ABCDS正方形DEFG
=
n
22
n1
.
24. 解: ⑴ 60,100. ⑵依题意,得y汽=2402x
y汽=500x200. y火=2401.6x
240100
5x2280.
24060
5x200.
y火=396x2280.
若y汽 >y火,得500x200>396x2280, ∴x>20.
⑶上周货运量X(17201922222324)72120.
从平均数分析,建议预定火车费用较省.
从折线图走势分析,上周货运量周四(含周四)后大于20且呈上升趋势,建议预定火车费用较省.
25. 解:思考 90,2. 探究一 30,2.
探究二、⑴由已知得M与P的距离为4,∴当MPAB时,点P到AB的最大距离是4,从而点P到CD的最小距离为642.
P与AB相切,此时旋转角最大,当扇形MOP在AB,CD之间旋转到不能再转时,M
BMO的最大值为90°.
P与CD的切点时,a达到最大,即OPCD.此时,⑵如图4,由探究一可知,点P是M
延长PO交AB于点H,a最大值为OMHOHM3090120.
如图5,当点P在CD上且与AB距离最小时,MPCD,a达到最小,连接MP,作
OHMP于点H,由垂径定理,得MH3,在Rt⊿MOH中,MO=4, ∴sinMOH
MHOM
34
,∴MOH49,∵a2MOH,∴a最小为98.
∴a的取值范围是98a120
.
2
26. 解:⑴把x0,y0代入yxbxc,得c0.
22
再把xt,y0代入yxbx,得tbt0,∵t0,∴bt.
⑵①不变.
如图6,当x1时,y1t,故M(1,1t).
∵tanAMP1.∴AMP45
②SS四边形AMNP-SPAM
=SDPN+S梯形NDAM-SPAM =t-4)(4t-16)+=
232t
22
112
(4t16)(t1)3
12
(t1)(t1)
1522
t6 t6=
解
32
t
152181
,得t1
12
,t292
92
.
∵4t5,∴t1⑶
72t
113
2
舍去,∴t.
2011年邯郸市中考第二次模拟考试
数学参考答案及评分标准
一、选择题
5
二、填空题
13.x(xy)(xy); 14.a=b=c; 15.3; 16.4; 17.6; 18.三、解答题
19.解法一:①×2得:6x-2y=10 ③, „„„„„„„„„„„„„„
2分
②+③得:11x=33 , ∴ x=3.„„„„„„„„„„„„„„ 4分 把x=3代入①得: 9-y=5 . ∴y=4 „„„„„„„„„„ 6分 所以
x3y4
„„„„„„„„„„„„„„„„„„„„ 8分
解法二:由①得: y=3x-5③
把③代入②得:5x+2(3x-5)=23,
11x=33, ∴ x=3. „„„„„„„„„„„„„„„„„„„ 4分 把x=3代入③得:y=4. „„„„„„„„„„„„„„„„ 6分
所以
x3y4
. „„„„„„„„„„„„„„„„„„„„ 8分 4分
20.(1)作图略 „„„„„„„„„„„„„„„„„„„
(2)∵CF平分∠ACB ∴∠ACF=∠BCF
又∵DC=AC ∴CF是△ACD的中线
∴点F是AD的中点 „„„„„„„„„„„„„„„„„„„ 6分 ∵点E是AB的垂直平分线与AB的交点
∴点E是AB的中点 „„„„„„„„„„„„ 7分 ∴EF是△ABD中位线
1
∴EF BD=3 „„„„„„„„„„„„„
2
14
8分 3分
21.解:(1). „„„„„„„„„„„„„„„„
(2)正确画出树状图(或列表)
第1个开关
A
B
C
D
第2个开关„„„„„„„ BCDACDABDABC
12
7分
结果:任意闭合其中两个开关的情况共有12种,其中能使小灯泡发光的情况有6种 小灯泡发光的概率是
. „„„„„„„„„„„„
9分
22.解:⑴①当1≤x≤5时,设y
kx
,
把(1,200)代入,得k200,
即y
200x
; „„„„„„„„„„„„ 1分
②当x5时,y40,
所以当x>5时,
y4020(x5)20x60; „„„„„„„„„„„„
3分
⑵当y=200时,20x-60=200,
x=13, 13-5=8
所以治污改造工程顺利完工后经过8个月后,该企业利润达到200万元. „ 6分 ⑶对于y
200x
,当y=100时,x=2;
对于y=20x-60,当y=100时,x=8,
所以资金紧张的时间为:3,4,5,6,7月份,共5个月. „„„„„„„ 9分 (注:若学生按实际问题答共6个月同样给分) 23.解:(1)
12
(ab)c. „„„„„„„„„„„„
2分 6分 8分
(2)图略 „„„„„„„„„„„„„„„„„„„ 拓展:能,图略 „„„„„„„„„„„„„„„„„„„„„ 说明:分别取AB、BC的中点F、H,连接FH并延长 分别交AE、CD于点M、N,将△AMF与△CNH一起
拼接到△FBH位. „„„„„„„„„„„„„„„„„„„„„ 10分 24.(1)解:设DG为x, 由题意得:BG=1+x,CG=1-x, 由勾股定理得:BG有:1x2解得:x∴DG=
1414
2
2
BC
2
2
CG
2
,
11x,
.
2分
. „„„„„„„„„„„„„„„„„„„
(2)①证明:连接EG,
∵△FBE是由△ABE翻折得到的, ∴AE=FE, ∠EFB=∠EAB=90°, ∴∠EFG=∠EDG=90°. ∵AE=DE, ∴FE=DE.
∵EG=EG,
∴Rt△EFG≌Rt△EDG (HL) .
∴DG=FG. „„„„„„„„„„„„„„„„„„„ ②解:若G是CD的中点,则DG=CG=
12
,
2
2
在Rt△BCG中,BC
BG
2
CG
2
312
,
22
∴AD=
2
. „„„„„„„„„„„„„„
③解:由题意AB∥CD,∴∠ABG=∠CGB. ∵△FBE是由△ABE翻折得到的, ∴∠ABE=∠FBE=
1∠ABG,∴∠ABE=12
2
∠CGB.
∴若△ABE与△BCG相似,则必有∠ABE=∠CBG==30°. 在Rt△ABE中,AE=ABtan∠ABE=33
,
∴AD=2 AE=233
. „„„„„„„„„„„„„„„„„„„
25.(1)解:在Rt△ABE中,AB
AC
2
BC
2
3
2
4
2
5
. „„„„„ 过点O作OD⊥BC于点D,则OD∥AC,
55∴△ODB∽△ACB, ∴
ODAC
OBAB
, ∴
OD, ∴3
2
5
OD
32
,
∴点O到BC的距离为32
. „„„„„„„„„„„„„„„„„„„ (2)证明:过点O作OE⊥BC于点E, OF⊥AC于点F,
515∵△OEB∽△ACB, ∴
OE
OB ∴AC
AB
∴
OE3
8
5
,OE
158
.
∴直线BC与⊙O相切. „„„„„„„„„„„„„„„„„„„
此时,四边形OECF为矩形, ∴AF=AC-FC=3-158
=
98
,
∵OF⊥AC, ∴AP=2AF=94
. „„„„„„„„„„„„„„„„„„„
(3)
158
OA
52
; „„„„„„„„„„„„„„„„„„„
(4)过点O作OG⊥AC于点G, OH⊥BC于点H, 则四边形OGCH是矩形,且AP=2AG,
又∵CO平分∠ACB,∴OG=OH,∴矩形OGCH是正方形. „„„„„„„ 设正方形OGCH的边长为x,则AG=3-x, ∵OG∥BC,
5分
10分
1分
3分
5分
7分 9分
10分
∵△AOG∽△ABC, ∴∴3-x∴x
OGBC
BCAC
, ∴OG
34
x
,
127
34
x
,
,
247
∴AP=2AG=. „„„„„„„„„„„„„„„„„„„ 12分 2分
26.解:(1)对称轴是:直线x=1;点B的坐标是(3,0).„„„„„„„„ (2)由∠ACB=∠AOC=∠COB=90°得△AOC∽△COB, ∴
AOCO
COBO
, ∴CO=3,∴b=3.
当x1,y0时,a2a
30, ∴a
33
. „„„„„„„ ∴y
32
233
x
3
x3.
(3)点M的坐标是:(2,3),(1+7,-3)或(1-7,-3) „„„„(4)设点N的坐标为(m,n),则n
32
3
m
233
m3
,
过点N作ND⊥AB于点D,则有:
SBCNS梯形ODNCSBDNSOBC1
2(3n)m123-m)n33
2333
2m2
n323m
2
33
2
2
m32
(m
3932
)
2
8
„„„„„„„„„„„„ ∵
32
<0,
∴当m32
时,△BCN的面积最大,最大值是
938
,
点N的坐标为(32,534
) „„„„„„„„„„„„
4分
8分 10分
12分
2011年初中毕业生统练一数学试题参考答案
一、 选择题(每题2分)
17
二、填空题(每题3分)13、
17、 18
、 14 15、 16、 三、解答题:19、(本小题满分8分)解:原式=
x1(x2)(x2)x(x1)(x1)
-3分 2
x2x2x(x1)
=x+1----------5分
当x=-2时,原式=-2+1=-1------------8分 20、(本小题满分8分)解:(1)令y=0时,
12
x2=0,解得:x=4,∴A(4,0),OA=4
∵PC为△AOB的中位线,∴OC=2,即P点横坐标为2.---------2分
令X=0时,y=-2,∴B(0,-2),OB=2,即P点纵坐标为-1,∴P(2,-1)-----3分 (2)∵PQ∥y轴,∴Q点横坐标为2,-----4分 ∵SOQC
32
12
32
32
32
,∴
32
2CQ,∴CQ=
32
,∴Q(2,
3x
)------------6分
(3)将Q(2,)代入y
kx
中,=
k2
,k=3∴y=-----------8分
21、(本小题满分9分)解:(1)AB切⊙O于点B ∴OBAB,即B90°
OCD90° --------------1分 又DCOA,
在Rt△COD与Rt△BOD中ODOD,OBOC --------------------2分
Rt△COD≌Rt△BOD(HL) CDOBDO.---------------3分
(2)在Rt△ABO中,A30°,OB4OA8 ------------------4分
ACOAOC844 --------------5分
CD
在Rt△ACD中,tanA
AC
·tan30°又A30°
,AC4CDAC
3
--------------------6分
(3)由(2)知AC=OC=4,DC⊥OA,∴DC为OA的垂直平分线∴DO=DA,∠DOC=∠A=30° 由(1)知
,∴∠BOC=2∠DOC=60°,
∴S扇形OBC
604360
2
83
----------------7分
OBtan30
在Rt△AOB中,tan∠A=∴SAOB
12
OBAB12
,A30°,AC,OB=4,∴AB=
4483
=43
ABOB
∴SACD
12
ACCD
12
4
433
83383
--------------8分
∴S阴影SAOBS扇形OBCSACD83
833
1638
3
----------9分
22、(本小题满分9分)(1)300-------------1分(2)表:45;96;0.26 图:(略)(1+1+1+1+1)(3)
27100
----------7分(4)应增加文学类书籍-----------8分
∵96>81>78>45,∴最喜欢的书籍是文学类书籍-------------9分 23、(本小题满分10分)(1)
拓展应用(1)50 (2)
121003
S (2)
12
S (3)
12
S(1分+1分+1分)
(1分+1分)
(3)四边形DEBF的面积的值不随时间t的变化而变化;1;(1分+1分)
bv
AEAB
vta
bva
证明:∵AE=vt,AB=a ∴,∵BF=t,BC=b ∴
BFBC
a
b
t
vta
∴
AEAB
BFBC
8分 BFBC
∵△AED与△ABD同底,∴
SAEDSABD
AEAB
,∵△DBF与△DBC同底,∴
SDBFSDBC
∴
SAEDSABD
=
SDBFSDBC
,∵SABD=SDBC,∴SAED=SDBF,-----------------------9分
12
12
∴S四边形DEBFSABDSABCD
21-----------------------------10分
24、(本小题满分10分)(1)证明:过点M作MH⊥AB于H,MG⊥AD于G,连接AM
∵M是正方形ABCD的对称中心,∴M是正方形ABCD对角线的交点,
∴AM平分∠BAD,∴MH=MG
在正方形ABCD中,∠A=90°,∵∠MHA=∠MGA=90°∴∠HMG=90°,
在正方形QMNP,∠EMF=90°∴∠EMF=∠HMG.∴∠EMH=∠FMG,∵∠MHE=∠MGF, ∴△MHE≌△MGF,∴ME=MF.---------3分
(2) ME=MF。证明:过点M作MH⊥AB于H,MG⊥AD于G,连接AM,
∵M是菱形ABCD的对称中心,∴M是菱形ABCD对角线的交点,∴AM平分∠BAD,∴MH=MG,
∵BC∥AD,∴∠B+∠BAD=180°,∵∠M=∠B,∴∠M+∠BAD=180°
又∠MHA=∠MGF=90°,在四边形HMGA中,∠HMG+∠BAD=180°,∴∠EMF=∠HMG. ∴∠EMH=∠FMG,∵∠MHE=∠MGF,∴△MHE≌△MGF,∴ME=MF。----------6分
(3)ME=mMF.证明:过点M作MH⊥AB于H,MG⊥AD于G,
在矩形ABCD中,∠A=∠B=90°∴∠EMF=∠B=90°,
又∵∠MHA=∠MGA=90°,在四边形HMGA中,∴∠HMG=90°, ∴∠EMF=∠HMG,∴∠EMH=∠FMG.∵∠MHE=∠MGF, ∴△MHE∽△MGF,∴
MEMF
MHMG
,
又∵M是矩形ABCD的对称中心,∴M是矩形ABCD对角线的中点 ∴MG∥BC,∴MG=
12
BC.同理可得MH=
12
AB,
∵AB = mBC∴ME=mMF。-----------------9分
(4)平行四边形ABCD和平行四边形QMNP中,∠M=∠B,AB=mBD,
M是平行四边形ABCD的对称中心,MN交AB于F,AD交QM于E。
则ME=mMF.--------------10分 25、(本小题满分12分)
解:(1)y1(10a)x (1≤x≤200,x为正整数)
y210x0.05x
2
„„„„2分
(1≤x≤120,x为正整数) „„„„4分
(2)①∵3<a<8, ∴10-a>0,y1是x的正比例函数即y1随x的增大而增大 „5分
∴当x=200时,y1最大值=(10-a)×200=2000-200a(万美元) „„6分
②y2
0.05(x100)500
2
„„„„„„„„„„7分
„„„„8分
∵-0.05<0, ∴x=100时, y2最大值=500(万美元)
(3)由2000-200a>500,得a<7.5, ∴当3<a<7.5时,选择方案一;
„„„9分 „„„10分
由2000200a500,得 a7.5,
∴当a=7.5时,选择方案一或方案二均可; 由2000200a500,得 a7.5,
∴当7.5<a<8时,选择方案二.
26、(本小题满分12分)
„„„„„12分
------------2分
--------------4分
-------------------5分
(2)y
35
t3t6 -------------------------6分
2
---------------8分
-------------10分
③当沿AQ翻折时,PQ=AP,过P点作PH⊥AC于H,则点H必为AQ的中点, ∴Rt△AHP∽Rt△ACB,∴综上所述,当t
2521或t
APAB53
AHAC
5t5
t4
209
即,解得:t>2(不合题意应舍去)
时,所形成的四边形为菱形.-----------------------12分
2011年保定市初中毕业第二次模拟考试
参考答案
一、选择题(每小题2分,共24分)
13.a(ab)(ab); 14.1:3; 15.(6,0) 16.0
2xx1
.„„„„„„„„„„„„„„„„„„„„„„„2分
1
当x=时, 原式=-2.„„„„„„„„„„„„„„„„„„„„„„„4分
2
(2)作图略„„„„„„„„„„„„„„„„„„„„„„2分 AD221„„„„„„„„„„„„„„„„„„„„„„4分 20.(1)在Rt△ABD中,AD=ABsin45°=42
∴在Rt△ACD中,AC=
ADsin30
22
4,„„„„„„„„„„„2分
=2AD=8,
即新传送带AC的长度约为8米.„„„„„„„„„„„„„„„„„4分 (2)结论:货物MNQP不需挪走. „„„„„„„„„„„„„„„„„5分
解:在Rt△ABD中,BD=ABcos45°=42
在Rt△ACD中,CD=ACcos30°=8
∴CB=CD—BD=434
∵PC=PB—CB =5—(434)=9—43≈2.2>2
∴货物MNQP不需挪走. „„„„„„„„„„„„„„„„„„8分
21.(1)1020%=50(人); „„„„„„1分 (2)5024%=12(人); „„„„„„2分
补全频数分布直方图;„„„„„„3分 (3)表示户外活动时间1小时的扇形圆心角的度数
20 o o=360 =144; „„„„„„5分 50(4)户外活动的平均时间=
100.5201121.582
50
1.18(小时).
22
4
32
43
∵1.18>1 ,
∴平均活动时间符合上级要求;户外活动时间的众数和中位数均为1.„„9分
22.(1)将B(4,1)代入y
kx
得:1
k4
,∴k=4,∴y
4x
, „„„„„„„„„2分
将B(4,1)代入y=mx+5得:1=4m+5,∴m=-1,∴y=-x+5, „„„„„„„4分 (2)在y
4x
中,令x=1,解得y=4,∴A(1,4),∴S=
12
14=2, „„„„„6分
(3)作点A关于y轴的对称点N,则N(﹣1,4),
连接BN交y轴于点P,点P即为所求.
设直线BN的关系式为y=kx+b,
3k4kb15 ,∴y3x17,∴P(0, 17) „„„9分 由 得
17555kb4b5
23.(1)设矩形EFGH的宽为x,长为2x,则由△AEH∽△ABC,得:
EHBC92
AKAD
,即:
2x12
6x6
,解得:x3.
∴矩形EFGH的面积为3×6=18.„„„„„„„„„„„„„„„„„„4分(2);„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分
98
9
(3)(4)
;„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„6分 ,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 .„„„„„„„„„„„„„„„„„„„„„„„„„„„„„9分
2992
5
2
2n3
9
„„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分
24.(1) 猜想:BD+CE=DE.„„„„„„„„„„„„„„„„„„„„„„„„1分
证明:由已知条件可知:∠DAB+∠CAE=120°,∠ECA+∠CAE=120°,
∴∠DAB=∠ECA.
在△DAB和△ECA中,∠ADB=∠AEC=60°,∠DAB=∠ECA,AB=CA, ∴△DAB≌△ECA(AAS). ∴AD=CE,BD=AE.
∴BD+CE=AE+ AD=DE.„„„„„„„„„„„„„„„„„„„5分
(2) 猜想:CE-BD=DE.„„„„„„„„„„„„„„„„„„„„„„„„6分
证明:由已知条件可知:∠DAB+∠CAE=60°,∠ECA+∠CAE=60°,
∴∠DAB=∠ECA.
在△DAB和△ECA中,∠ADB=∠AEC=120°,∠DAB=∠ECA,AB=CA, ∴△DAB≌△ECA(AAS). ∴AD=CE,BD=AE.
∴CE-BD=AD-AE=DE.„„„„„„„„„„„„„„„„„„10分
25.(1)300, 250, 150; „„„„„„„„„„„„„„„„„„3分 (2)判断:y是x的一次函数.
设y=kx+b,∵x=10,y=300;x=11,y=250,∴
10kb30011kb250
,解得
k50b800
,
∴y=﹣50x+800,
经检验:x=13,y=150也适合上述关系式,∴y=﹣50x+800.„„„„„„„8分 (3)W=(x﹣8)y=(x﹣8)(﹣50x+800)=﹣50x2+1200x-6400 ∵a=﹣50
A12分 即当销售单价为12元时,每天可获得的利润最大,最大利润是80026.过点A作BC边上的高AM,垂足为M,交DE于N.
∵AB=10,sinB=
35
,∴AM= AB sinB= 6,
E
∵DE∥BC,△ADE∽△ABC, ADAB65DEBC
ANAM35
,即
t10
35
DE12
AN6
,
GMF
C
∴DE=t,AN=t,MN=6﹣t,
(备用图①)
(1)当正方形DEFG的边GF在BC上时,如图①,
DE=DG=MN,即∴当t=
103
65
t=6﹣
35
t,∴t=
103
,
时,正方形DEFG的边GF在BC上.„„„„„4分
(2) 当GF运动到△ABC外时,如图②, S△CEP+ S△BDQ= = S△ABC= 令
12
12
1212
PCPE
12
BQDQ
12
12
(PCBQ)MN 65t)(6
35t)
E
(BCDE)MN
12
(12
BCAM65t)(6
35
12636 1448,
(12t)
F (备用图②)
14
解得t1=15(舍去),t2=5,
∴当t=5时,△CEP与△BDQ的面积之和等于△ABC面积的
(3)分两种情况:
①当正方形DEFG在△ABC的内部时,如图14, S=DE=( 当t=
103
2
.„„„„8分
65
t)=
2
3625
t,此时t的范围是0≤t≤
2
103
,
B
图14
E C
时,S的最大值为16.
②当正方形DEFG的一部分在△ABC的外部时, 如图②,S=DE•MN=的范围是 ∵﹣
1825103
65
t(6﹣
35
t)=﹣
1825
t+
2
365
t,此时t
∵18>16,∴S的最大值为18.„„„„„„„„12分