2020年福州市九年级质量检测
数学试题答案及评分参考
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：共10小题，每小题4分，满分40分；在每小题给出的四个选项中，只有一项是符合题目要求的，
请在答题卡的相应位置填涂． 1．A 2．C 3．A 4．B 5．B 6．A 7．B 8．C 9．C 10．D
二、填空题：共6小题，每小题4分，满分24分，请在答题卡的相应位置作答． 11．1 14．4
12．1
415．18
13．15 16．9
4
三、解答题：共9小题，满分86分，请在答题卡的相应位置作答． 17．（本小题满分8分）
解：解不等式①，得x≤3． ······························································································ 3分
解不等式②，得x＞?1． ···························································································· 5分 ∴原不等式组的解集是?1＜x≤3， ··············································································· 6分 将该不等式组解集在数轴上表示如下： -5 -4 -3 -2 -1 0 1 2 3 4 5
······························································· 8分
18．（本小题满分8分）
证明：∵点E，F在BC上，BE?CF，
∴BE?EF?CF?EF， 即BF?CE． ········································································································· 3分
A D 在△ABF和△DCE中，
?AB?DC，? ??B??C，?BF?CE，?B E F C
∴△ABF≌△DCE， ······························································································· 6分 ∴∠A?∠D． ······································································································· 8分
九年级数学试题答案及评分参考第1页（共6页）
19．（本小题满分8分）
2解：原式?x?12?(x?1)?(x?1) ······················································································· 3分
(x?1)2(x?1)(x?1)?x?1? ·························································································· 4分 x?1x?122?x?1?x?1 ·································································································· 5分 x?1x?1?2． ··········································································································· 6分 x?12当x?3?1时，原式? ················································································· 7分 3?1?1?2 3···················································································· 8分 ?23． ·3 20．（本小题满分8分） 解：
画法一：
M A C D O B N D 画法二：
M A C O B N
······························································· 4分
如图，点C，D分别为（1），（2）所求作的点． ························································ 5分 （2）证明如下：由（1）得BC∥OA，BC?1OA，
2∴∠DBC?∠DAO，∠DCB?∠DOA， ∴△DBC∽△DAO， ············································································ 7分 ∴DC?BC?1， DOAO2∴OD?2CD． ····················································································· 8分
21．（本小题满分8分） 解：（1）由图1可得甲的速度是120?2=60m/min． ································································ 2分
由图2可知，当x?4时，甲，乙两人相遇，
3故(60?v乙)?4?200，
3解得v乙?90m/min． ···························································································· 4分 答：甲的速度是60 m/min，乙的速度是90 m/min．
（2）由图2可知：乙走完全程用了b min，甲走完全程用了a min，
∴b?200?20， ······························································································· 6分
909a?200?10． ································································································ 8分
603∴a的值为10，b的值为20．
39 22．（本小题满分10分） 解：（1）依题意得a?100．······························································································ 2分
这1000户家庭月均用水量的平均数为：
九年级数学试题答案及评分参考第2页（共6页）
x?2?40?6?100?10?180?14?280?18?220?22?100?26?60?30?20?14.72， ········· 6分
1000∴估计这1000户家庭月均用水量的平均数是14.72． （2）解法一：不合理．理由如下： ··············································································· 7分
由（1）可得14.72在12≤x＜16内，
∴这1000户家庭中月均用水量小于16 t的户数有
， ···························································· 8分 40?100?180?280?600（户）
∴这1000户家庭中月均用水量小于16 t的家庭所占的百分比是600?100%?60%，
1000∴月均用水量不超过14.72 t的户数小于60%． ············································· 9分 ∵该市政府希望70%的家庭的月均用水量不超过标准m， 而60%＜70%，
∴用14.72作为标准m不合理． ······························································· 10分
解法二：不合理．理由如下： ··············································································· 7分
∵该市政府希望70%的家庭的月均用水量不超过标准m， ∴数据中不超过m的频数应为700， ·························································· 8分 即有300户家庭的月均用水量超过m．
又20?60?100?160?300，20?60?100?220?380?300， ∴m应在16≤x＜20内． ·········································································· 9分 而14.72＜16，
∴用14.72作为标准m不合理． ······························································· 10分
23．（本小题满分10分）
（1）证明：连接OD，AD．
B ∵AB为⊙O直径，点D在⊙O上，
∴∠ADB?90°， ···························································································· 1分
F ∴∠ADC?90°．
O ∵E是AC的中点， D ∴DE＝AE，
∴∠EAD?∠EDA． ···················································································· 2分 A ·E ·C ·G ·∵OA?OD，
∴∠OAD?∠ODA． ······················································································· 3分 ∵∠OAD?∠EAD?∠BAC?90°， ∴∠ODA?∠EDA?90°， 即∠ODE?90°， ···························································································· 4分 ∴OD⊥DE．
∵D是半径OD的外端点， ∴DE是⊙O的切线． ····················································································· 5分
（2）解法一：过点F作FH⊥AB于点H，连接OF，
B ∴∠AHF?90°．
∵AB为⊙O直径，点F在⊙O上，
F H ∴∠AFB?90°， O D ∴∠BAF?∠ABF?90°．
∵∠BAC?90°，
∴∠G?∠ABF?90°， A E C G ∴∠G?∠BAF． ························································································· 6分 又∠AHF?∠GAB?90°， ∴△AFH∽△GBA， ···················································································· 7分 ∴AF?FH． ··························································································· 8分 GBBA由垂线段最短可得FH≤OF， ········································································ 9分 当且仅当点H，O重合时等号成立． ∵AC＜AB，
∴BD上存在点F使得FO⊥AB，此时点H，O重合， ∴AF?FH≤OF?1， ············································································ 10分 GBBABA2九年级数学试题答案及评分参考第3页（共6页）
即AF的最大值为1． GB2解法二：取GB中点M，连接AM．
∵∠BAG?90°，
B 1∴AM?GB． ··························································································· 6分 2F ∵AB为⊙O直径，点F在⊙O上， M O ∴∠AFB?90°， D ∴∠AFG?90°， ∴AF⊥GB． ··························································································· 7分 A ·E ·C ·G 由垂线段最短可得AF≤AM， ········································································ 8分 当且仅当点F，M重合时等号成立， 此时AF垂直平分GB， 即AG＝AB． ∵AC＜AB，
∴BD上存在点F使得F为GB中点， ∴AF≤1GB， ··························································································· 9分
2∴AF≤1， ···························································································· 10分 GB2即AF的最大值为1． GB2
24．（本小题满分12分）
（1）①证明：∵∠AED?45°，AE?DE，
∴∠EDA?180??45??67.5°． ······································································· 1分
2∵AB?AC，∠BAC?90°，
∴∠ACB?∠ABC?45°，∠DCA?22.5°， ························································· 2分 ∴∠DCB?22.5°， 即∠DCA?∠DCB， ∴CD平分∠ACB． ····················································································· 3分
A ②解：过点D作DF⊥BC于点F，
∴∠DFB?90°．
D ∵∠BAC?90°，
∴DA⊥CA． E 又CD平分∠ACB， B C F ∴AD?FD， ································································································· 4分 ∴AD?FD． DBDB在Rt△BFD中，∠ABC?45°， ∴sin∠DBF?FD?2， ················································································ 5分
2DB∴AD?2． ······························································································· 6分 DB2（2）证法一：过点A作AG⊥AE交CD的延长线于点G，连接BG，
∴∠GAE?90°．
又∠BAC?90°，∠AED?45°，
∴∠BAG?∠CAE，∠AGE?45°，∠AEC?135°， ·············································· 7分 ∴∠AGE?∠AEG， ∴AG?AE． ······························································································· 8分 ∵AB?AC，
∴△AGB≌△AEC， ···················································································· 9分 ∴∠AGB?∠AEC?135°，CE?BG， ∴∠BGE?90°． ························································································ 10分 ∵AE⊥BE，
九年级数学试题答案及评分参考第4页（共6页）
∴∠AEB?90°， ∴∠BEG?45°，
在Rt△BEG和Rt△AGE中，
A
G D
BE?GE?2GE，AE?GE?cos45??2GE， ·········································· 11分 E 2cos45?2GEB C
AE21??． ·在Rt△ABE中，tan△ABE?················································ 12分 BE2GE2（也可以将△AEB绕点A逆时针旋转90°至△AFC得到AE?2EF，CF?2EF） 2证法二：∵AE⊥BE，
∴∠AEB?90°，
∴∠BAE?∠ABE?90°． ∵∠AED?45°，
∴∠BED?45°，∠EAC?∠ECA?45°， ∴∠AEC?∠BEC?135°． ············································································ 7分
A ∵∠BAC?90°，
∴∠BAE?∠EAC?90°，
D
∴∠ABE?∠EAC．
E ∵∠ABC?45°，
∴∠ABE?∠EBC?45°，
B C
∴∠ECA?∠EBC， ····················································································· 8分 ∴△BEC∽△CEA， ∴BE?EC?BC． ···················································································· 9分 CEEACA在Rt△ABC中，BC?CA?2CA， ························································· 10分
cos45?∴BE?EC?2， CEEA∴BE?2CE，AE?2CE． ····································································· 11分
22CE?1． ·在Rt△ABE中，tan∠ABE?AE?2··············································· 12分
BE2CE2
25．（本小题满分14分） 解：（1）∵抛物线C的对称轴是y轴，
24k?k?0且k?0， ·∴?····················································································· 1分 2k∴4k?1?0，
2解得k?1， ······································································································ 3分
4∴抛物线C的解析式为y?1x2． ·········································································· 4分
4（2）点A在直线y??2上． ························································································ 5分
理由如下：∵过F（0，2）的直线与抛物线C交于P，Q两点，
∴直线PQ与x轴不垂直．
设直线PQ的解析式为y?tx?2，
将y?tx?2代入y?1x2，得x2?4tx?8?0，
4∴??16t2?32?0，
∴该方程有两个不相等的实数根x1，x2，
不妨设P（x1，y1），Q（x2，y2），
y∴直线OP的解析式为y?1x． ···························································· 6分
x1九年级数学试题答案及评分参考第5页（共6页）
设A（m，n）．
∵QA⊥x轴交直线OP于点A， ∴m?x2，
1x2?x12y∴n?1?x2?4····························································· 7分 ?1x1x2． ·
x1x14又方程x2?4tx?8?0的解为x?2t?2t2?2，
∴x1x2?(2t?2t2?2)(2t?2t2?2)?4t2?4(t2?2)??8， ∴1x1x2??2， 4即点A的纵坐标为?2， ········································································ 9分 ∴点A在直线y??2上．
（3）∵切线l不过抛物线C的顶点，
∴设切线l的解析式为y?ax?b（a?0）．
将y?ax?b代入y?1x2，得x2?4ax?4b?0， ······················································ 10分
4依题意得??0，
即(?4a)2?4?(?4b)?16a2?16b?0，
∴b??a2，
∴切线l的解析式为y?ax?a2． ·········································································· 11分
22a?2a当y?2时，x?，∴M（?2，2）． ·························································· 12分 aa22a?2a当y??2时，x?，∴N（?2，?2）． ······················································ 13分 aa∵F（0，2），
22a∴MF?(?2)2，
a22a由勾股定理得NF?(?2)2?(?2?2)2，
a22222a?2a)?[(?2)2?(?2?2)2] ∴MF?NF?(aa2222a?2a?2a?2a?(?)(??2)?16 aaaa22a??4?16 aa··············································································· 14分 ?8?16??8． ·
九年级数学试题答案及评分参考第6页（共6页）
2020年福州市九年级质量检测数学试题答案及评分参考 - 图文
