2014年哈尔滨市南岗区三模数学试卷及答案

2014年初中升学考试调研测试（三）
数学试卷参考答案与评分标准
一、选择题（每小题3分，共计30分）
二、（每小题3分，共计30分）
三、解答题（其中21~24题各6分，25~26题各8分，27~28题各10分，共计60分） 21．(本题满分6分) 解：4 (1)
)
1)(1(1)1)(1(11)1(1)111(2'
-=-+⨯+=-+⨯+-+=-÷
+-x x
x x x x x x x x x x x x 因为1..................................................................
32
1
2-460cos 24'=⨯=︒-=x 所以'
1..........................................................................................21311)111(2
=-=-=-÷+-x x x x
22．(本题满分6分) （1）
图①
..................................................................................................................................3'
题号 1 2 3 4 5 6 7 8 9 10 选项 C
D
B
C
A
B
B
C
B
A
题号 11 12
13
14
15 选项
2
1x ≥-
)3)(3(-+x x x
34x ≤＜
40
题号 16
17 18 19 20
选项
1=x
6
1 1或5
28
1255
A
D
C
B
（2）
图②
图②
画对一种情况..........................................................................................3' (本题满分6分) 23. 解：(1)240÷40%=600(人)．
600-180-60-240=120(人)
如图
………………………………………………………………………………………3'
(2)8000×
600
180
=2400(人)． 答：该居民区有8000人，估计爱吃A 粽的人有2400人．………………………………3'
24.(本题满分6分)
解：∵∠ADB=∠CBD-∠BAD=60°-30°=30°=∠BAD ∴BD=AB=72（米）……………………………'2
在Rt △BCD 中，∠DBC=60°， ∴BC=BD •cos60°=2172⨯
=36 CD=BD •sin60°=2
372⨯=336；………………………………'2
在Rt △BCE 中，∠EBC=45°，∴CE=BC=36； ………………………………'1 塔高DE=CD ﹣EC=36336- ················································································ '1 答：塔高DE 为)36336(-米． 25.（本题满分8分）
解：（1）∵E F ∥CD ∴∠EFB=∠DBF ∵弧BE=弧BE ∴∠EFB=∠BAC ···················································· '1
∴∠DBF=∠BAC 又∵∠CBE=∠DBF ∴∠CBE=∠BAC ······································································································ '1
∵AB 是直径 ∴∠AEB=90°
E
A C
B
A
E
C B
∴∠ABE+∠BAC=90°∴∠ABE+∠CBE=90°∴∠ABC=90° ····································· '1
∴AB CD ⊥ ∴CD 为⊙O 的切线 ········································································ '1
（2）∵CD ⊥AB ，E F ∥CD ，∴EF ⊥AB 又∵AB 是直径，∴EG=FG ． ···························································································· '1 连接E O ，设OG=x ，则BG=9-x ．
由勾股定理可知：2
2
2
2
2
OE OG BE BG EG -=-=，
即2222)9(69x x --=-，7=x ． ················································································· 2' 因此282
279
2
2
=-==EG EF ． ·
········································································· '1
26.（本题满分8分）
解：(1)设乙工程队每天完成x 米，则甲工程队每天完成2x 米.
1026000
6000+=x
x 解得x =300 ………………………… '2
经检验：x =300是原方程的解 ·························································································· '1
2x=2×300=600 ·············································································································· '1 答：略
（2）设两工程队合作施工a 天，
7600)500700(700600
)600300(6000≤++⨯+-a a
………………………… '2
∴4a ≤ ∴两工程队最多可合作施工4天………………………… '2
27.（本题满分10分） （1）依题意可得
2
2
4(8)4(8)4040a a c a a c ⎧=⨯-+⨯-+⎪⎨=⨯+⨯+⎪⎩ 解得184
a c ⎧=-⎪⎨⎪=⎩ 所
求抛物线的解析式为
211
4 (282)
y x x '=--+.
（2）解法一;如图1，可求D （4
3
-
，0）........................................................................................1'
x
y
L P Q
D
B
A C
O
图1
过点D 作DL ⊥AC ，垂足为点L.连接PC 、PQ.
∵∠DAL=∠CAO ∠ALD=∠AOC=90°∴△ADL ∽△ACO ∴20
34845
DL AL
== ∴DL=
534 AL=538 ∴CL=AL ==-53
8
53454 ∴∠DCL=45° ∴∠ACP=2∠DCL=90° ······································································· '1
由△CPO ∽△ACO 可得OP=2 PC=52 ∴CQ= PC=52 ∴t=PB=OB-OP=4-2=2 ·········································································································· '1
∴点Q 的运动速度为
25
52
=······················································································· '1 解法二：如图2，可求D （4
3
-，0）........................................................................................1'
连接DQ 、PQ. 设PQ 与CD 的交点为点E
则直线CD 垂直平分PQ ∴ PD=DQ ，PC=CQ ，∠PDE=∠EDQ ∴tan ∠EDQ=tan ∠PDE=
3OC
OD
= 令DE=n ，则PE=EQ=3n ，PD=10n.过点Q 作QF ⊥x 轴于F.
由△PDE ∽△PQF 得QF=3
105
a ，PF=9105a
∴DF=94
10101055
a a a -= 图2 ∵tan ∠QAF=tan ∠CAO=12OC OA = ∴3
10152
a
AF = ∴AF=6
105
a ∴AD=
64410102108553a a a +==- ∴a=103
∴AQ=525FQ = ·········································································································· '1
∴t=PB=BD-PD=103
10
344⨯-+=2 ·············································································· '1
∴点Q 的运动速度为25
52
=·
······················································································ '1
(3)如图3，由（2）可求 P(2，0) Q(-4，2) ∵2PN NQ NA =⨯ ∴
PN NA
NQ PN
= 又∵∠QNP=∠PNA ∴△NPQ ∽△NAP
∴∠NPQ=∠NAP ........................................................................................1'
∴tan ∠MPQ=tan ∠CAO=
12
∴
1
2
QM PM = 过点M 作直线l ∥x 轴，过点P 、Q 分别作PG ⊥l 、QH ⊥l ，垂足分别为点G 、H.
∵∠QMH+∠PMG=90°,∠MPG+∠PMG=90°∴∠QMH=∠MPG 又∵∠QHM=∠PGM=90°
∴△QMH ∽△MPG ∴1
2
QH HM QM MG PG PM === ∴MG=2QH PG=2HM ........................................................................................1'
设点M 的坐标为（m ，n ） ∴22(2)2(4)m n n m -=-⎧⎨
=+⎩ 解得2
4
m n =-⎧⎨=⎩
∴点M 的坐标为（-2，4）........................................................................................1'
211482y x x =--+ 当2x =-时，2119
(2)(2)44822
y =-⨯--⨯-+=≠
∴点M 不在（1）中的抛物线上．........................................................................................1'
图3
28.(本题满分10分) （1）如图1，
正方形ABCD 中，AB=BC=DC, ∠BCD=90°
∵BH ⊥CD ∴∠BHE=90°∴∠CBF+∠DEB=90°,
又∵∠CDE+∠DEB=90° 图1
∴∠CBF=∠CDE ················································································································ '1
∴△CBF ≌△CDE ············································································································ 2'
∴CF=CE ∵CD ∥AB ∴CE GE
BC AG
= ······························································································· '1 ∴
CF GE
AB AG
= ··············································································································· '1
（2） ○1当点F 在线段DC 上时 （如图2） 连接DQ ，连接QG 并延长交DE 于点N.
由△CQF ∽△AQB 得 CF QC
AB AQ
= ∵
CF GE AB AG = ∴GE QC
AG AQ
= ∴11GE QC AG AQ +=+ 即AE AC
AG AQ
=又∵∠QAG=∠CAE ∴△AQG ∽△ACE ∴∠AQG=∠ACE 图2
∴QG ∥CE △CQG 为等腰直角三角形 ································································· '1
∵BC=CD ∠BCQ=∠DCQ CQ=CQ ∴△CBQ ≌△CDQ ∴∠CBQ=∠CDQ
∵∠CBQ=∠CDE ∴∠CDQ=∠CDE 又∵DG=DG ∠DGQ=∠DGN=90° ∴△DQG ≌△DNG
∴QG=GN 又∵∠QHN=90°
∴GH=QG ∴∠QHG=∠HQG=∠HBC
∴∠CPQ=∠GHQ+∠CED =∠HBC +∠CED =90° ························································· '1 过点G 作GM ⊥GP 交CP 于点M ，设PC 与QG 的交点为O ∵∠PQG+∠POQ =∠MCG +∠COG =90° ∠POQ=∠COG ∴∠PQG=∠MCG 同理∠PGQ=∠MGC 又∵QG=CG ∴△GPQ ≌△GMC ∴PQ=CM 又∵2PM PG =
O M
N
G
Q
P
F
H
D A
B
C
E
∴2PC PQ PC CM PM PG -=-== ····································································· 2'
○
2当点F 在线段DC 延长线上时（如图3） 2PQ PC PG -=
············································································································································ '1
图3
（以上各解答题如有不同解法并且正确，请按相应步骤给分）。