2006年安徽省中考数学试题(大纲) 数学参考答案及评分标准

2006年安徽省中考数学试题（大纲）

数学参考答案及评分标准

一、选择题（本题共10小题，每小题4分，共40分）

三、（本题共2小题，每小题8分，共16分）

16．解：去分母得：2

320x x -+=， ································································ 3分 (2)(1)0x x --=，

2x =或1x =． ················································································· 7分 经检验，原方程的解为12x =，21x =． ················································· 8分 17．解：PA ∵，PB 是O 的切线，OA ，OB 是半径，

90PAO PBO ∠=∠=

∴． ·································································· 3分 又360PAO PBO AOB P ∠+∠+∠+∠=

∵，70P ∠=

，

110AOB ∠=

∴． ············································································· 6分 AOB ∠∵是圆心角，ACB ∠是圆周角，

55ACB ∠= ∴． ··············································································· 8分 （注：其它解法参照赋分） 四、（本题共2小题，每小题10分，共20分） 18．解：∵六边形ABCDEF 是正六边形， 120B ∠= ∴． ··························· 2分

又∵点1A ，1B 分别为

AB ，BC 边的中点， 111cm BA BB

==∴，1130BA B ∠= ．

过点B 作11BM A B ⊥，垂足为

M ， 11

2

BM BA =

∴，1112A B A M =． ························································· 4分 又11cm BA =， 1

cm 2

BM =

∴，11A B =．

Ｍ

11211)22BA B S =

=△∴． ·

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

同理：11112)DC D FE F S S ==

△△

∴阴影部分的总面积23)S ==． ·

················································ 10分 （注：其它解法参照赋分）

18．（华东版）

解：由二视图得：圆柱的底面半径为1cm r =，圆柱的高为11cm h =，

圆锥的底面半径1cm r =，圆锥高2h =， ·············································· 2分 则圆柱的侧面积S

圆柱侧

122rh =π=π2(cm )，

圆柱的底面积2

cm S r 2

=π=π()， ······························································· 5分

又圆锥的母线2(cm)l =

==，

∴圆锥的侧面积S 圆锥侧2cm rl 2=π=π()． ·

···················································· 8分 ∴此工件的表面积为S S =表圆柱侧+S 圆锥侧+S =5cm 2π()． ·

·································· 10分 19．证明：（1）AM 平分90BAD

BAD =

，∠∠， 45BAE ∴= ∠．

BAE ∴△为等腰直角三角形，

又AB DC =， BE DC ∴=． ························································································· 4分 （2）由CM AM ⊥易得，MEC △为等腰直角三角形，

ME CM ∴=且45MEC MCE == ∠∠．

135BEM DCM ∴== ∠∠．

又BE DC =，

BEM DCM ∴△≌△． MBE MDC ∴=∠∠． ··········································································· 10分

（注：其它解法参照赋分） 五、（本题满分12分） 20．解：（1）由2211

(4)(2)222

y x x x =

-=--． ················································ 2分 ∴函数图象的顶点坐标为(22)-，

，对称轴为直线2x =． ································· 4分 （2）如右图． ······························································································· 7分

（3）从函数图象可以看出，从4月份开始新产品的销售累积利润盈利． ···················· 9分

（4）5x =时，2

1525 2.52y =

?-?=， 6x =时，21

62662

y =?-?=，

6 2.5 3.5-=．

∴这个公司第6个月所获的利润是3.5万元． ·

···· 12分

六、（本题满分12分） 21．解：（1）众数是0.1031； ·········································································· 2分

中位数是0.1016； ··················································································· 4分 平均数是0.1015． ··················································································· 6分 （2）若用平均数来估计总体，则10000千克柑橘中，被损坏的柑橘共有：

100000.10151015?=（千克），

∴可以出售的柑橘共有1000010158985-=（千克）

． ···································· 9分 而购柑橘款与利润共计100002500025000?+=（元），

∴出售价为：

25000

2.88985

≈（元）． ··························································· 12分 或解：若用中位数来估计总体，则10000千克柑橘中，被损坏的柑橘共有： 100000.10161016?=（千克），

∴可以出售的柑橘共有1000010168984-=（千克）

． ···································· 9分 而购柑橘款与利润共计25000（元），

∴出售价为：

25000

2.88984

≈（元）． ··························································· 12分 或解：若用众数来估计总体，则10000千克柑橘中，被损坏的柑橘共有： 100000.10311031?=（千克），

∴可以出售的柑橘共有1000010318969-=（千克）

． ···································· 9分 而购柑橘款与利润共计25000（元），

∴出售价为：

25000

2.88969

≈（元）． ··························································· 12分 七、（本题满分12分） 22．解：（1）A 点表示公交公司的该条公交路线的运营成本为1万元． ····················· 2分

B 点表示当乘客量为1.5万人时，公交公司的该条公交路线收支恰好平衡． ·········· 4分 （2）图③； ·································································································· 6分

图②； ·································································································· 8分 （3）把原射线略往上平移，再按逆时针旋转一个适当的角度即可得到，如图所示．

（若点A 沿y 轴平移至x 轴及x 轴上方不给分） ··········································· 12分

第20题答案图

第22题答案图

八、（本题满分13分） 23．（1）如图②，由题意CAC α'=∠，

要使AB DC ∥，须BAC ACD =∠∠，

30BAC ∴=

∠．

453015CAC BAC BAC α''==-=-= ∠∠∠，

即15α=

时，能使得AB DC ∥． ··················· 4分 （2）易得45α= 时，可得图③，

此时，若记DC 与AC BC '

'，分别交于点E F ，， 则共有两对相似三角形：

BFC ADC C FE ADE '，△∽△△∽△． ··········· 6分 下求BFC △与ADC △的相似比：

在图③中，设AB a =

，则易得AC =

．

则1):1)1:(2BC a BC AC a ===，

或(22． ······················································································· 8分 注：C FE '△与ADE △

的相似比为：:C F AD '=

2):2．

（3）解法一：

当045α<

≤时，总有EFC '△存在．

EFC BDC DBC CAC α'''=+= ，，∠∠∠∠ FEC C α'=+∠∠，

又180EFC FEC C '''+∠+=

∠∠，

180BDC DBC C C α''∴++++= ∠∠∠∠． ··········································· 11分

又4530C C '==

，∠∠，

105DBC CAC BDC ''∴++= ∠∠∠． ····················································· 13分

解法二：

在图②中，BD 分别交AC AC '，于点M N ，，

由于在AMN △中，180CAC AMN CAC ANM α''=++=

，∠∠∠∠，

180BDC C DBC C α''∴++++= ∠∠∠∠． 3045180BDC DBC α'∴++++= ∠∠．

105BDC DBC α'∴++= ∠∠． ······························································ 11分

Ａ图②

第23题答案图

在图③中，45CAC α'==

∠，

易得60DBC BDC '+=

∠∠，

也有105DBC CAC BDC ''++=

∠∠∠，

综上，当045a <

≤时，总有105DBC CAC BDC ''++=

∠∠∠． ·············· 13分