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）
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