2012答案福建省泉州市中考真题

2012年福建省泉州市初中毕业、升学考试

数学试题参考答案及评分标准

说明：

（一）考生的正确解法与“参考答案”不同时，可参照“参考答案及评分标准”的精神进行评分．

（二）如解答的某一步出现错误，这一错误没有改变后续部分的考查目的，可酌情给分．但原则上不超过后面应得的分数的二分之一；如属严重的概念性错误，就不给分．

（三）以下解答各行右端所注分数表示正确做完该步应得的累计分数．一、选择题（每小题3分，共21分）

1．B 2．C 3．B 4．A 5．D 6．D 7．C 二、填空题（每小题4分，共40分）

8．< 9．()5x x - 10．8

310? 11．4 12．7 13．1 14．3

15．80 16．2，30 17．（1）0；（2）

12或34或4

三、解答题（共89分）

18．（本小题9分）解：原式=6+431-- ···································································································· 8分 =6 ····································································································································· 9分 19．（本小题9分）解：原式=2

694x x x +++- ······················································································· 4分 =613x + ························································································································· 6分当2x =-时，原式()=62+13?- ················································································· 7分 =1 ····································································································································· 9分 20．（本小题9分）解：（1）P （白子）=

································································································ 4分（2）方法一：所有等可能的结果，画树状图如下：

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

P ∴（一黑一白）=

=122

···························································································· 9分方法二：所有等可能的结果，列表如下：

P ∴（一黑一白）=

61=122

···························································································· 9分 21．证明：在ABCD 中，AD BC AD BC =，∥ ··················································· 3分

=ADB CBD ∴∠∠ ·········································································································· 5分

AE BD CF BD ⊥⊥， 90AED CFB ∴∠=∠=° ······························································································ 6分 ADE CBF ∴△≌△ ······································································································ 7分 =DAE BCF ∴∠∠ ·········································································································· 9分

22．（本小题9分）解：（1）100分，90 条形统计图（略） ·········································································································· 6分（2）15

120020=9100

÷（名） ∴学校至少应安排9名高甲对兴趣小组的教师． ························································· 9分 23．（本小题9分）

解：（1）()14A --，、()41B --，，

平移后的直线A B ''（图略）； ························································································ 6分（2）C 点的坐标为()122C --，或()222C ，． ····························································· 9分 24．（本小题9分）

解：（1）90a =；4000100b =， ·················································································· 6分

（2）解法一：依题意及图象得：

()10090-504000*********x ?=+?

解得：200x =

答：200天后共节省燃料费40万元． ··········································································· 9分解法二：依题意，可得：

()400000

90-50+100=200100

÷（天）答：这200天后共节省燃料费40万元． ······································································ 9分 25．（本小题12分）解：（1）i ）点A 、B 、C 均在O ⊙上，

224590BOC A ∴∠=∠=?=°°． ·························· 3分 1OB OC ==，

BC ∴ ·

···························································· 5分注：也可延长BO 或过O 点作BC 的垂线构造直角三角形求得BC ．

ii ）证法一：如图，作直径CE ，则2E A CE R ∠=∠=， 90EBC ∴∠=°

sin sin 2BC

A E R

··················································· 8分证法二：如图，连结OB 、OC ，作OH BC ⊥于点H ，

则12A BOC BOH ∠=∠=∠，1

2BH BC =

2sin sin BC BH A BOH OB R

∴∠=∠==

=2BC

R ······························································ 8分

（2）解法一：如图，连结AP ，取AP 的中点K ，连结BK 、CK ，在Rt APC △中，1

CK AP AK PK =

==，同理得：BK AK PK == CK BK AK PK ∴==+，

∴点A 、B 、P 、C 都在K ⊙上． ·········································································· 10分 ∴

由（1）ii ）可知，sin 60BC

=° 2

sin 60AP ∴==°

故在整个滑动过程中，P 、A 两点间的距离保持不变． ··············································· 12 解法二：如图，连结AP ，并延长BP 交AN 于点Q ，

BP AM CP AN ⊥⊥，

cos BQ CQ

AQB AQ PQ

∴∠=

= AQP BQC ∠=∠ BCQ APQ ∴△∽△．．．

······································································································································· 10分

BC CQ AP PQ ∴

= c o s BC

AP AQB ∴=∠ 60QAB ∠=° 30AQB ∴∠=°

AP ∴=

（定值）故在整个滑动过程中，P 、A 两点间的距离保持不变． ·········································· 12分注：解法一中，由点A 、B 、P 、C 都在K ⊙上． ················································ 10分

可得=QAP QBC ∠∠ Q A P Q B ∴△∽△······················································ 11分

sin 60BQ BC

AQ AP

∴=

° s i n 63BC AP ∴==° ·················· 12分 26．（本小题14分）

解：（1）抛物线

y x h =

+ 经过点()01C ，，

014

h ∴?+=， ··········································································································· 2分解得：1h =． ················································································································ 3分

（2）操作、观察知：PQ x ∥轴，POQ △的面积最小，令（1）2

114

y x =

+中，2y =，解得：2x =±． ∴点()22P -，

、()22Q ， POQ ∴△的面积最小值为4． ··················································································· 6分．

（3）解法一：连结BQ ，若l 与x 轴不平行（如图），好PQ 与x 轴不平行依题意，设抛物线2

114

y x =

+上的点 2114P a a ??

+ ???

，、2114Q b b ??+ ???，

()0a b <<，

直线BC ：11y k x =+过点P ， ···················································································· 8分

211114a ak ∴+=+，得：11

k a =-，即1

y ax =+，

令0y =得：4

B x a

=-……（*） ················································································ 10分

同理：过点A 的直线l ：2y kx =+经过点P 、Q

124a ak ∴+=+…① 2

1124

b bk +=+…② ·································································································· 11分 b a ?-?①②得：()()221

24a b b a b a b a -+-=-， ············································ 12分

化简得：4

b a

∴点B 与Q 的横坐标相同，BQ y ∴∥轴，即BQ OA ∥，

又AQ 与OB 不平行，∴四边形AOBQ 是梯形． ····················································· 13分据抛物线的对称性可得()0a b >>结论相同．

故在直线l 旋转过程中：当l 与x 轴不平行时，四边形AOBQ 是梯形，当l 与x 轴平行时，四边形AOBQ 是正方形． ··························································································· 14分解法二：如图，作直线BM y ∥轴与直线PA 交于点M ，作1P P x ⊥轴于点1P ，

作1

M N P P ⊥于点N ，交y 轴于点H ，

易证：1BOC BPP △∽△，MHA MNP △∽△

OB OC HM HA

PB PP NM NP ∴

，． ························································································· 8分设()()110P x y B x ，，，0，()()0212P x y y y >，，，则

0011

x x x y =-…①，

0112

2x y x x y y -=--…②

又211114y x =

+，代入①得：20101114x x x x ??

+=- ???，即01

4x x =-……（*） ······ 10分由①②得：12

2y y y y -=

-， ························································································ 11分

11222111114

4114

x y y y x x +∴===+-，

（*）代入上式得：2

20114

y x =+， ············································································ 12分

∴点()02M x y ，也在抛物线21

y x =+上，即为点Q ，故QB y ∥轴，

又AQ 与OB 不平行，∴四边形AOBQ 为梯形． ····················································· 13分据抛物线的对称性可得()12y y <结论相同．

故在直线l 旋转过程中：当l 与x 轴不平行，四边形AOBQ 是梯形；当l 与x 轴平行时，四边形AOBQ 是正方形． ······························································································· 14分四、附加题（共10分） 1．5x = 2．130．

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