福州市质检答案.docx

福州市质检答案.docx

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福州市质检答案

2018—2019学年度福州市九年级质量检测

数学试题答案及评分标准

评分说明：

1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要

考查内容比照评分参考制定相应的评分细则．

2．对于计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容

和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一

半；如果后继部分的解答有较严重的错误，就不再给分．

3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．

4．只给整数分数．选择题和填空题不给中间分．

一、选择题：

每小题4分，满分40分．

1．A2．B3．D4．B5．C

6．D7．C8．B9．C10．B

二、填空题：

每小题4分，满分24分．

11．m（m2）（m2）12．正方体

13．甲14．4

15．4316．51

2

注：

12题答案不唯一，能够正确给出一种符合题意的几何体即可给分，如：

某个面是正方

形的长方体，底面直径和高相等的圆柱，等．

三、解答题：

本题共9小题，共86分．解答应写出文字说明、证明过程和演算步骤．

17．解：

原式3331

····································································6分

3

311·············································································7分

3．················································································8分

18．证明：

∵∠1∠2，

∴∠ACB∠ACD．····································3分

B

在△ABC和△ADC中，

BD

，

ACBACD

，

A

C

1

2

ACAC，

D

∴△ABC≌△ADC（AAS），·······················································6分

∴CBCD．··········································································8分

注：

在全等的获得过程中，∠B＝∠D，AC＝AC，△ABC≌△ADC，各有1分．

数学试题答案及评分参考第1页（共6页）

19．解：

原式

2

x1x2x1

2

xx

·······························································1分

2

x1x

2

x（x1）

····································································3分

x

，·············································································5分

x

1

当x31时，原式31

311

····················································6分

31

3

33

3

．

···················································8分

20．解：

B

O

CDA

···········································3分如图，⊙O就是所求作的圆．·······················································4分

证明：

连接OD．

∵BD平分∠ABC，

∴∠CBD∠ABD．···························································5分

∵OBOD，

∴∠OBD∠ODB，

∴∠CBD∠ODB，··························································6分

∴OD∥BC，

∴∠ODA∠ACB

又∠ACB90°，

∴∠ODA90°，

即OD⊥AC．····································································7分

∵点D是半径OD的外端点，

∴AC与⊙O相切．····························································8分

注：

垂直平分线画对得1分，标注点O得1分，画出⊙O得1分；结论1分．

21．

（1）四边形ABB′A′是菱形．·································································1分

证明如下：

由平移得AA′∥BB′，AA′BB′，

∴四边形ABB′A′是平行四边形，∠AA′B∠A′BC．·············2分

∵BA′平分∠ABC，

∴∠ABA′∠A′BC，

∴∠AA′B∠A′BA，····················································3分

∴ABAA′，

∴□ABB′A′是菱形．·····················································4分

数学试题答案及评分参考第2页（共6页）

（2）解：

过点A作AF⊥BC于点F．

由

（1）得BB′BA6．

由平移得△A′B′C′≌△ABC，

∴BCBC4′′，

AA'

D

∴BC′10．····························5分

∵AC′⊥A′B′，

E

∴∠B′EC′90°，

∵AB∥A′B′，

B

FCB'C'

∴∠BAC′∠B′EC′90°．

在Rt△ABC′中，AC′

228

BCAB．

··································6分

∵S

△ABC′

11

ABACBCAF，

22

∴AF24

ABAC

BC5

，··························································7分

∴S

菱形ABB′A′144

BBAF，菱形ABB′A′144

5

∴菱形ABB′A′的面积是144

5

．··················································8分

22．

（1）是；························································································2分

（2）①85.5；336；············································································6分

②由表中数据可知，30名同学中，A等级的有10人，B等级的有11人，

C等级的有5人，D等级的有4人．

依题意得，154105511010

30

·········································8分

5.5．····································································9分

∴根据算得的样本数据提高的平均成绩，可以估计，强化训练后，全年

级学生的平均成绩约提高5.5分．············································10分

23．解：

（1）y27250.1（x2）0.1x2.2；········································4分

（2）依题意，得（0.1x2.2）x0.5101（x10）20.6，················7分

解得

x1x216．

································································9分

答：

x的值是16．·······························································10分

注：

（1）中的解析式未整理成一般式的扣1分．

24．

（1）①证明：

∵四边形ABCD是正方形，

∴∠ADC∠BCD90°，CA平分∠BCD．

G

∵EF⊥EB，

∴∠BEF90°．

证法一：

过点E作EN⊥BC于点N，···········1分

∴∠ENBENC90∠°．

D

M

F

C

H

∵四边形AEGD是平行四边形，

∴AD∥GE，

E

N

∴∠EMF∠ADC90°，

∴EM⊥CD，∠MEN90°，

AB

∴EMEN，·····················································2分

∵∠BEF90°，

∴∠MEF∠BEN，

数学试题答案及评分参考第3页（共6页）

∴△EFM≌△EBN，

∴EBEF．······················································3分

证明二：

过点E作EK⊥AC交CD延长线于点K，·················1分

∴∠KEC∠BEF90°，

∴∠BEC∠KEF，

G

∵∠BEF∠BCD180°，

DF

∴∠CBE∠CFE180°．

K

M

∵∠EFK∠CFE180°，

H

C

∴∠CBE∠KFE．

又∠ECK1

2

∠BCD45°，

E

∴∠K＝45°，

∴∠K∠ECK，

AB

∴ECEK，·······················································2分

∴△EBC≌△EFK，

∴EBEF．······················································3分

证明三：

连接BF，取BF中点O，连接OE，OC．················1分

∵∠BEF∠BCF90°，

G

∴OE1

BFOC，

2

∴点B，C，E，F都在

以O为圆心，

D

M

FC

H

OB为半径的⊙O上．

∵BEBE，

E

O

∴∠BFE∠BCA45°，

∴∠EBF45°∠BFE，

·······2分

A

B

∴EBEF．······················································3分

②GH⊥AC．··············································································4分

证明如下：

∵四边形ABCD是正方形，

四边形AEGD是平行四边形，

∴AEDG，EGADAB，AE∥DG，

∠DGE∠DAC∠DCA45°，

∴∠GDC∠ACD45°．

由

（1）可知，

··········································5分

G

∠GEF∠BEN，EFEB．

F

DC

M

∵EN∥AB，

H

∴∠ABE∠BEN∠GEF，

∴△EFG≌△BEA，·····················6分

∴GFAEDG，

∴∠GFD∠GDF45°，

E

N

∴∠CFH∠GFD45°，

AB

∴∠FHC90°，

∴GF⊥AC．···························································7分

（2）解：

过点B作BQ⊥BP，交直线AP于点Q，取AC中点O，

∴∠PBQ∠ABC90°．

数学试题答案及评分参考第4页（共6页）

∵AP⊥CG，

∴∠APC90°．

①当点E在线段AO上时，（或“当01

AEAC时”）

2

∠PBQ∠ABP∠ABC∠ABP，

即∠QBA∠PBC．································8分

G

P

∵∠ABC90°，

∴∠BCP∠BAP180°．

DC

M

∵∠BAP∠BAQ180°，

O

∴∠BAQ∠BCP．································9分

∵BABC，

E

∴△BAQ≌△BCP，·····························10分

∴BQBP10，AQCP，

A

B

在Rt△PBQ中，PQ

22

BPBQ102．

Q

∴PAPCPAAQPQ102．········································11分

②当点E在线段OC上时，（或“当1

2

ACAEAC时”）

G

∠PBQ∠QBC∠ABC∠QBC，

即∠QBA∠PBC．

∵∠ABC∠APC90°，∠AKB∠CKP，

∴∠BAQ∠BCP．······························12分

D

CM

∵BABC，

∴△BAQ≌△BCP，

∴BQBP10，AQCP，

Q

在Rt△PBQ中，PQ

22

BPBQ102．

E

OK

P

AB

∴PAPCPAAQPQ102．···········13分

综上所述，当点E在线段AO上时，PAPC102；

当点E在线段OC上时，PA－PC102．

25．

（1）B（m，0），C（0，5

2

m）；··························································2分

解：

（2）设点E，F的坐标分别为（a，

a），（a，

2

a

2

），······················3分

代入

1（5）（）121（5）5

yxxmxmxm，

2222

得

11（5）

5a

2

amam

，①

2222

11（5）a

5

2

amam，②

2222

········································4分

由①②，得（m5）aa．

∵a0，

∴m6，···········································································5分

∴抛物线的解析式为

1115

2

yxx．

22

································6分

数学试题答案及评分参考第5页（共6页）

（3）依题意得A（5，0），C（0，5

m），

2

由m0，设过A，C两点的一次函数解析式是ykxb，

将A，C代入，得

5kb0

，

解得

5

bm．

2

1

km

2

，

5

bm，

2

∴过A，C两点的一次函数解析式是15

ymxm．

22

设点P（t，0），则5剟tm（m0），

··················7分

∴M（t，

1t1（m5）tm），N（t，15

25

mtm）．

22222

①当5剟t0时，

∴MN

11（5）

（1）

255

tmtmmtm

22222

125

tt．····························································8分

22

∵10

2

，∴该二次函数图象开口向下，

又对称轴是直线5

t，

2

∴当5

t时，MN的长最大，

2

此时MN

1（5）5（5）25

2

22228

．·······························9分

②当0t,m时，

∴MN

15[11（5）5]

2

mtmtmtm

22222

125

tt．

22

··········10分

∵10

，∴该二次函数图象开口向上，

2

又对称轴是直线5

t，

2

∴当0t,m时，MN的长随t的增大而增大，

∴当tm时，MN的长最大，此时MN

125

mm．

22

·············11分

∵线段MN长的最大值为25

8

，

∴

12525

mm,，

228

·····························································12分

整理得

550

2

（m）,，

24

由图象可得552552

剟m．

22

∵m0，

∴m的取值范围是0552

m,．·······································13分

2

数学试题答案及评分参考第6页（共6页）