2017-2018宁德市九年数学答案
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
宁德市2017-2018学年度第一学期期末九年级质量检测
数学试题参考答案及评分标准
⑴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，可参照本答案的评分标准的精神进行评分．
⑵对解答题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的立意，可酌情给分． ⑶解答右端所注分数表示考生正确作完该步应得的累加分数． ⑷评分只给整数分，选择题和填空题均不给中间分． 一、选择题：（本大题有10小题，每小题4分，满分40分）
1．B 2．C 3．A 4．D 5．B 6．A 7．A 8．D 9．D 10．C 二、填空题：（本大题有6小题，每小题4分，满分24分）
11．3； 12．中心投影； 13．12； 14．322+=x y ； 15．20； 16．4
1． 三、解答题（本大题共9小题，共86分） 17．（本题满分8分）
解法1：∵DE ∥BC ，
∴AD AE
AB AC
=
． ·················································································4分 即
AC
3
155=
． ·················································································· 6分 ∴5AC=45． ···················································································· 7分 解得 AC=9． ·················································································· 8分 解法2：∵DE ∥BC ，
∴∠ADE=∠B ，∠AED=∠C ． ···························································· 2分 ∴△ADE ∽△ABC ． ·········································································· 3分 ∴AD AE
AB AC
=
． ·················································································4分 即
AC
3
155=
． ·················································································· 6分 ∴5AC=45． ···················································································· 7分 解得 AC=9． ·················································································· 8分
18．（本题满分8分）
解：∵3=x 是042=+-c x x 的根，
∴9120c -+=． ······························································ 2分 解得 3=c ． ···························································· 4分
将3=c 代入042
=+-c x x ，
得 2
430x x -+=． ·
································································· 5分 ∴()24443
x ±--⨯= （或)2（21x -=）． ········································· 6分
∴31=x ，12=x ． ·································································· 8分 ∴3=c ，另一个根为x =1．
19．（本题满分8分）
解：记这四张联纸分别为①，②，③，④，列表可得：
第二张
第一张
① ②
③
④
① （①，②） （①，③） （①，④）
② （②，①）
（②，③） （②，④）
③ （③，①） （③，②）
（③，④）
④
（④，①） （④，②） （④，③）
·····
由表可知，一共有12种结果，每种结果出现的可能性相同，其中能组成对联的结果有4种，分别是（①，④）, （②，③）, （③，②）, （④，①）， ················································· 7分
∴两张联纸恰好组成一幅对联的概率是1
3． ··············································· 8分
(若学生未画树状图或列表，但罗列出所有结果，同样可得分） 20．（本题满分9分）
解：（1）将A （5，2）代入反比例函数k
y x
=
， 得 25
k
=
． ······································· 2分 解得 10=k ． ································ 3分 ∴反比例函数表达式为x
y 10
=
． ·
······ 4分 （2）过点B 作BD ⊥AC 于点D ．
∵点A 的坐标为（5，2），AC ⊥y 轴，
∴AC=5． ······································· 5分 ∵△ABC 的面积为10， ∴
1
102
AC BD ⋅=． ·
····························· 6分 ∴BD =4．
∴n =4+2=6． ··································· 7分
A
O
y
x
B
C D