2018-2019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．A 2．B 3．D 4．B 5．C

6．D 7．C 8．B 9．C 10．B

二、填空题：每小题4分，满分24分．

11．(2)(2)m m m +- 12．正方体

13．甲 14．4

15

． 16

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31=- ·························································································· 6分 311=+- ·

····································································································· 7分 3=． ·

··········································································································· 8分 18．证明：∵∠1=∠2，

∴∠ACB =∠ACD ． ·················································· 3分 在△ABC 和△ADC 中， B D ACB ACD AC AC ∠=∠⎧⎪∠=∠⎨⎪=⎩

，，， ∴△ABC ≌△ADC (AAS)，················································································· 6分 ∴CB =CD ． ······································································································ 8分 2 1 C A … B D

注：在全等的获得过程中，∠B ＝∠D ，AC ＝AC ，△ABC ≌△ADC ，各有1分．

19．解：原式22121x x x x x

--+=÷······················································································ 1分 221(1)

x x

x x -=⋅- ···························································································· 3分 1

x x =-， ······································································································ 5分 *

当1x

时，原式= ····································································· 6分

=

=． ······································································ 8分 20．解： ··························································· 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

∵OB =OD ，

∴∠OBD =∠ODB ，

∴∠CBD =∠ODB ， ················································································· 6分 ∴OD ∥BC ， ∴∠ODA =∠ACB

又∠ACB =90°，

;

∴∠ODA =90°，

即OD ⊥A C ． ···························································································· 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 ′B C ． ····················· 2分

∵BA ′平分∠ABC ，

∴∠ABA ′=∠A ′BC ，

|

∴∠AA ′B =∠A ′BA ， ········································································ 3分