2014年学业水平考试数学试题参考答案

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2014年学业水平考试数学试题参考答案

一、选择题: 二、填空题:

16. 2014 17. 500 18. a b

a

- 19. 6 20. 6 21. ①②④ 三、解答题:

22.(1)解:原式=2+3-23+1-6 ……………………………………………2分 =-23 …………………………………………………………..3分 (2)解:方程两边都乘以最简公分母(x ﹣3)(x +1)得:

3(x +1)=5(x ﹣3), ………………………………………………4分 解得:x =9, ………………………………………………………….5分 检验:当x =9时,(x ﹣3)(x+1)=60≠0, ……………………….6分 ∴原分式方程的解为x =9. ………………………………………….7分

23.(1)证明:∵AC =BD ,

∴AC +CD =BD +CD ,

即AD =BC . ……………………………………1分 在△ADE 和△BCF 中,

AD =BC

∠A =∠B AE =BF

∴△ADE ≌△BCF (SAS ). ……………………………………2分

∴∠E =∠F . ……………………………………3分 (2)解:∵在Rt △ADB 中,∠BDA =45°,AB =3 ∴DA =3 …………1分

在Rt △ADC 中,∠CDA =60°∴tan60°=

CA

AD …………2分

CA= 33 ………………………………………3分 ∴BC=CA -BA=(33-3) 米 ………………………4分

24.解:设甲种商品应购进x 件,乙种商品应购进y 件. …………1分

根据题意,得 160

5101100.x y x y +=⎧⎨+=⎩

…………5分

解得:100

60.x y =⎧⎨=⎩

………………………………7分

答:甲种商品购进100件,乙种商品购进60件. …………8分

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A

B

D

C

C

D

B

D

D

B

B

D

A

A

C

25.解:列表得

1 2 3 1 2 3 4 2 3 4 5 3

4

5

6

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

共有9种等可能的结果,其中摸出的两个小球标号之和是2的占1种, 摸出的两个小球标号之和是3的占2种, 摸出的两个小球标号之和是4的占3种, 摸出的两个小球标号之和是5的占2种, 摸出的两个小球标号之和是6的占1种; 所以棋子走到E 点的可能性最大, ···························································· 7分

棋子走到E 点的概率=31

93=. ······························································· 8分

26.解:(1)90331802

ACB l ππ

=

⨯= …………………….2分 扇形OAB 的周长为362

π

+……………………….3分 (2)连结OC ,交DE 于M ,

∵四边形ODCE 是矩形 ∴OM =CM ,EM =DM ………………….4分 又∵DG =HE

∴EM -EH =DM -DG ,即HM =GM …………………….5分 ∴四边形OGCH 是平行四边形 ……………………………6分 (3)DG 不变; …………………………………………….7分

在矩形ODCE 中,DE =OC =3,∴DG =1 ………………..9分

27.解:(1)CF =EF ························································································· 1分

连接BF (如图①).∵△ABC ≌△DBE ∴BC =BE ,AC =DE

∵∠ACB =∠DEB =90° ∴∠BCF =∠BEF =90°

又∵BF =BF ,∴Rt △BFC ≌Rt △BFE . ∴CF =EF . ··········································································································2分 AF +EF =DE ·········································································································3分 ∵AF +EF =AF +CF =AC 又∵AC =DE ∴AF +EF =DE . ··································································································4分 (2)画出正确图形(可不加辅助线)如图② ·································································5分

AF +EF =DE 仍然成立. ······················································································6分 (3)不成立.此时AF ,EF 与DE 的关系为AF - EF =DE ······································7分

理由:连接BF (如图③),∵△ABC ≌△DBE ,∴BC =BE ,AC =DE , ∵∠ACB =∠DEB =90°,∴∠BCF =∠BEF =90°.

第2次

A O B

C

E

H G D M 第1次

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