DA甘肃省庆阳市中考真题

DA甘肃省庆阳市中考真题
DA甘肃省庆阳市中考真题

庆阳市2007年高中阶段招生考试

数学试卷参考答案及评分标准

一、选择题(本题有10道小题,每小题3分,共30分.每小题只有一个选项是正确的,不选、多选、错选均不得分)

1.B 2.A 3.A 4.A 5.C 6.C 7.C 8.A 9.B 10.B 二、填空题(本题共有10道小题,每小题3分,共30分) 11.6(0)

y x x =

> 12.1,

52

13.0 14.1

252

2??-- ??

?,

,直线12

x =-,上

15.最小值(2)1f =- 16.25;125;50 17.3000(1)x +,23000(1)y x =+,二 18.27.2元;20元;30元 19.

112

20.20

三、作图题(本题满分10分,每道小题5分)

21.①作A 点关于公路的对称点A '.②连结A B '与公路交于C .③连结AC BC ,,则C 就为机场的位置.

22.作图3分,写作法2分

作法1)选一点异于A 的一点O ;2)以O 为圆心,O A 为半径作O ;3)连结O A ,在O A 上选点P ,使A P a <;4)过P 作O A 的垂线l ;5)以A 为圆心,以线段a 长为半径作弧交l 于B C ,两点.

则O 和ABC △就是所求的图形. 作圆、作三角形、是轴对称图形共1分

四、解答题(在答题卷中作答,要有必要的解题步骤,按步骤给分.第23小题为8分,其余每小题均为9分,共80分) 23.3 24.(1)(4分)5或4- (2)(5分)解:由(1)得1(13)

2y x =

- (3) ················································· (1分)

代入(2),得21

138(13)32

x x x +-=-·

化简,得2430x x ++= ························································································ (1分) 解得11x =-或23x =- ··························································································· (1分) 代入(3),所求为1112

x y =-??

=?或2235

x y =-??

=? ·································································· (2分)

25.解:(1)ABE C D F △≌△,AD E C BF △≌△,ABD C D B △≌△.

(2)此题答案不唯一,选择其中一对证明即可. 如证明ABE C D F △≌△ ························································································· (1分)

证明:∵四边形ABC D 是平行四边形,

AB C D

=∴,AB C D ∥, ······················································································ (2分)

ABE C D F ∠=∠∴, ·

······························································································ (2分) AE BD ⊥∵,C F BD ⊥,

AEB C FD ∠=∠∴

··································································································· (2分) ABE C D F

∴△≌△. ····························································································· (2分)

26.解:把3x m y ==,代入3y x

=得,1m =,所以(13)M ,·

···································· (1分) 由一次函数y kx b =+经过点(20)(13)A M -,,,得203

k b k b -+=??

+=?, ·································· (2分)

所以解得12

k b =??

=?, ····································································································· (2分)

所以2y x =+ ········································································································· (1分)

(2)由23y x y x =+??

?=?

?

得1133x y =-??

=-?,;2213x y =??=?,································································· (2分)

所以(31)N --, ······································································································· (1分) 27.解:设A E x =,连结O D ,则90AD O ∠=°

又90ABC ∠=∵°,A A ∠=∠

AD O ABC

∴△∽△ ································································································ (1分)

AD O D AB

BC

=

316

62

AD x ==+,62

x AD +=

····················································································· (2分)

又2(6)AD x x =+∵

2

(6)

(6)4

x x x +=+∴ ······························································································· (2分)

即:24120x x +-=································································································ (2分)

26x x ==-∴,(舍)

即:2AE =,2(26)4AD =+= ··········································································· (2分)

A E

28.解:222210x x -+= 标准式为:2120

2

x x -+

= ··················································································· (1分)

2

10

2x ?

?-= ???∴ ····································································································· (2分)

1222

x x ==

∴ ······································································································· (2分)

2sin sin 2

A B ==

∵ ······························································································· (2分)

45A B ∠=∠=∴°···································································································· (2分)

29.解:4512

AF

=

,452AF = ········································································ (1分)

cot 303

45

BE ==°,453BE = ·············································································· (2分)

45225453

AB =++∴(米) ············································································· (2分)

又451sin 302

BC

==

°

90BC =(米) ······································································································ (2分)

BC

的坡度为1:3 ································································································· (2分)

30.解:90C ∠=∵°,4A C =,3B C = ································································ (1分)

2

2

22

435AB AC BC

=

+=

+=············································································· (2分)

作C D AB ⊥于D

112

2ABC S AC BC C D AB ==

△·· ·················································································· (2分)

43

125

5

AC BC C D AB

=

=

=··∴ ······················································································ (2分)

C

点到AB 的距离为

125

1245

R <<∴

时,C

与AB 相交 ············································································· (2分)

D C

F E B

30

31.解:如图作O C 交AB 于O ,则O C 为两个圆锥共同的底面的半径

2

2

2

2

345AB AC BC

=

+=

+=·

············································································ (1分)

AB O C AC BC

=··

125

O C =

∴ ············································································································· (2分)

以A C 为母线的圆锥侧面积21

12362π3π(cm )2

55=?=

··

·

············································· (2分)

以BC 为母线的圆锥侧面积2

1

12482π4π(cm )

2

55

=?=··

·············································· (2分)

∴表面积为

2

364884πππ(cm )

5

5

5

+

=

········································································· (2分)

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