江苏省南京市秦淮区中考二模数学试题及参考答案.docx
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江苏省南京市秦淮区中考二模数学试题及参考答案
2015-2016学年度第二学期第二阶段学业质量监测试卷九年级数学
注意事项：
1.本试卷共6页．全卷满分120分．考试时间为120分钟．
2.答选择题必须用2B铅笔将答题卡上对应的答案标号涂黑．如需改动，请用橡皮擦干净后，再选涂其他答案．答非选择题必须用0.5毫米黑色墨水签字笔写在答题卷上的指定位置，在其他位置答题一律无效．
一、选择题（本大题共6小题，每小题2分，共12分．在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在学．校．发．的．答．题．卡．上．）
1．下列图案中，既是中心对称图形又是轴对称图形的是
A．B．C．D．
2．秦淮区将开展南部新城规划建设，在包括近10平方公里核心区及其外围的整个南部新城投入150000000
000元，10年后将其打造成南京“第二个河西”．将150000000000用科学记数法表示为
A．0.15×1012
3．下列计算正确的是
B．1.5×1011
C．1.5×1012
D．1.5×1013
A．a3＋a2＝a5
B．a6÷a3＝a2
C．（a2）3＝a8
D．a2·a3＝a5
4．若反比例函数yk3，2），则反比例函数y＝－k的图像在
＝的图像经过点（－
xx
A．一、二象限B．三、四象限C．一、三象限D．二、四象限
5．如图，数轴上的A、B、C三点所表示的数分别为a、b、c，AB＝BC，则下列关系正确的是
A
B
C
A．a＋c＝2b
a
b
（第5题）
c
B．b＞c
C．c－a＝2（a－b）
D．a＝c
6．记n边形（n＞3）的一个外角的度数为p，与该外角不相邻的（n－1）个内角的度数的和为q，则p与q的关系是
A．p＝qB．p＝q－（n－1）·180°C．p＝q－（n－2）·180°D．p＝q－（n－3）·180°
二、填空题（本大题共10小题，每小题2分，共20分.不需写出解答过程，请把答案直接填写在答．题．卷．相．
应．位．置．上）
7．4的算术平方根是．
8．函数y＝1＋x的自变量x的取值范围是．
9．不等式－3x＋1＞－8的正整数解是．
10．甲、乙两地5月下旬的日平均气温统计如下表（单位：
°C）：
甲地气温
24
30
28
24
22
26
27
26
29
24
乙地气温
24
26
25
26
24
27
28
26
28
26
甲
乙
则甲、乙两地这10天日平均气温的方差大小关系为：
S2S2．（填“＞”、“＜”或“＝”）
11．写出一个主视图、左视图和俯视图完全一样的几何体：
．
12．已知关于x的一元二次方程3（x－1）（x－m）＝0的两个根是1和2，则m的值是．
13．若解分式方程2x－a＝0时产生增根，则a＝．
x－44－x
14．如图，在△ABC中，BC的垂直平分线交它的外接圆于D、E两点．若∠B＝24°，∠C＝106°，
︵
则AD的度数为°．
15．如图，点B、C都在x轴上，AB⊥BC，垂足为B，M是AC的中点．若点A的坐标为（3，4），点M的坐标为（1，2），则点C的坐标为．
16．如图，在△ABC中，以B为圆心，BC为半径作弧，分别交AC、AB于点D、E，连接DE，若ED＝DC，
AE＝3，AD＝4，则S△ADE＝．
S△ABC
三、解答题（本大题共11小题，共88分．请在答题卷指定区域内作答，解答时应写出文字说明、证明过程
或演算步骤）
⎧x＋3y＝－1，
⎩
17．（6分）解方程组⎨3x－2y＝8．
181
1x2－2x＋1
．（6分）先化简，再求值：
（x－
＋＋2）÷x2－4，其中x＝3＋1．
19．（8分）如图，□ABCD中，E是AD的中点，连接BE并延长，交CD的延长线于点F．连接CE．
（1）求证：
△ABE≌△DFE；
（2）小丽在完成
（1）的证明后继续进行了探索：
当CE平分∠BCD时，她猜想△BCF是等腰三角形，请在下列框图中补全她的证明思路．
20．（8分）从3名男生和2名女生中随机抽取上海迪斯尼乐园志愿者．
（1）抽取1名，恰好是男生的概率是；
（2）抽取2名，求恰好是1名男生和1名女生的概率．
21．（8分）中学生使用手机的现象越来越受到社会的关注．某市记者随机调查了一些家长对这种现象的态度，并将调査结果绘制成图①和图②的统计图（不完整）．
家长对中学生使用手机三种态度分布统计图
A无所谓
B反对
C赞成
144
36
①
（第21题）
请根据图中提供的信息，解答下列问题：
（1）在图①中，C部分所占扇形的圆心角度数为°；
（2）将图②补充完整；
（3）根据抽样调查结果，请你估计该市10000名中学生家长中有多少名家长持反对态度？
22．（8分）下表给出了变量x与ax2、ax2＋bx＋c之间的部分对应关系（表格中的符号
“——”表示该项数据已经丢失）：
x
－1
0
1
ax2
——
——
1
ax2＋bx＋c
7
2
——
（1）求函数y＝ax2＋bx＋c的表达式；
（2）将函数y＝ax2＋bx＋c的图像向左平移1个单位长度，再向上平移2个单位长度，直．接．写出平移后图像的表达式.
23．（8分）如图，要利用一面长为25m的墙建羊圈，用100m的围栏围成总面积为400m2的三个大小相同的矩形羊圈，求羊圈的边AB、BC各多长？
A
D
24．（8分）2015年12月16日，南京大报恩寺遗址公园正式对外开放．某校数学兴趣小组想测量大报恩塔的高度．如图，成员小明利用测角仪在B处测得塔顶的仰角α＝63.5°，然后沿着正对该塔的方向前进了
13.1m到达E处，再次测得塔顶的仰角β＝71.6°．测角仪BD的高度为1.4m，那么该塔AC的高度是多少？
（参考数据：
sin63.5°≈0.90，cos63.5°≈0.45，tan63.5°≈2.00，sin71.6°≈0.95，cos71.6°≈0.30，tan71.6°≈3.00）
（第24题）
25．（8分）如图，在Rt△ABC中，∠C＝90°，AD是∠BAC的平分线，经过A、D两点的圆的圆心O恰好落在AB上，⊙O分别与AB、AC相交于点E、F．
（1）判断直线BC与⊙O的位置关系并证明；
（2）若⊙O的半径为2，AC＝3，求BD的长度．
26．（9分）“十·一”长假，小王与小叶相约分别驾车从南京出发，沿同一路线驶往距南京480km的甲地旅游．小王由于有事临时耽搁，比小叶迟出发1.25小时．而小
叶的汽车中途发生故障，等排除故障后，立即加速赶往甲地．若从小叶出发开始计时，图中的折线O-A-B-
D、线段EF分别表示小叶、小王两人与南京的距离
y1（km）、y2（km）与时间x（h）之间的函数关系．
（1）小叶在途中停留了h；
（2）求小叶的汽车在排除故障时与南京的距离；
（3）为了保证及时联络，小王、小叶在第一次相遇时约
定此后两车之间的距离不超过25km，试通过计算说明，他们实际的行驶过程是否符合约定？
27．（11分）如图，在矩形ABCD中，E是AD上一点．将矩形ABCD沿BE翻折，使得点F落在CD上．
（1）求证：
△DEF∽△CFB；
（2）若F恰是DC的中点，则AB与BC的数量关系是；
（3）在
（2）中，连接AF，G、M、N分别是AB、AF、BF上的点（都不与端点重合），若△GMN∽△ABF，
且△GMN的面积等于△ABF
面积的
1AG的值．
，求
2AB
2015-2016学年度第二学期第二阶段学业质量监测试卷
九年级数学参考答案及评分标准
说明：
本评分标准每题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，参照本评分标准的精神给分．
一、选择题（每小题2分，共计12分）
题号
1
2
3
4
5
6
答案
A
B
D
C
A
D
二、填空题（每小题2分，共计20分）
7．28．x≥－19．1和210．＞11．答案不唯一，如：
球、正方体等
12．213．－814．8215．（－1，0）16
9
．
23
三、解答题（本大题共11小题，共计88分）
17．（本题6分）
解法一：
由①，得x＝－3y－1③．···········································································1分将③代入②，得3（－3y－1）－2y＝8．·······························································3分解这个方程，得y＝－1．··············································································4分将y＝－1代入③，得x＝2．··········································································5分
⎧x＝2，
所以原方程组的解是⎨
⎩y＝－1．
········································································6分
解法二：
①×3，得3x＋9y＝－3③．··········································································1分
③－②，得11y＝－11．················································································3分解这个方程，得y＝－1．··············································································4分将y＝－1代入①，得x＝2．··········································································5分
⎧x＝2，
所以原方程组的解是⎨
⎩y＝－1．
········································································6分
18．（本题6分）
解：
（1
x－
＋1
＋
x2－2x＋1
）÷x2－4
＝2x÷（x－1）（x＋2）（x－2）（x＋2）（x－2）
···············································································2分
2x（x＋2）（x－2）
＝
（x＋2）（x－·
（x－1）2···············································································3分
2）
＝2x．·····································································································4分
（x－1）2
19．（本题8分）
（1）证明：
∵四边形ABCD是平行四边形，
∴AB∥CD．····························································································1分
∴∠ABE＝∠F，∠A＝∠FDE．···································································3分
∵E是AD的中点，
∴AE＝DE．····························································································4分
∴△ABE≌△DFE．···················································································5分
（2）BE＝FE，········································································································6分
CE平分∠BCD，·······························································································7分直径所对的圆周角是直角．··················································································8分
20．（本题8分）
解：
（1
3
）5．
··········································································································2分
（2）从3名男生和2名女生中随机抽取2名同学，所有可能出现的结果有：
（男1，女1）、（男1，女2）、（男1，男2）、（男1，男3）、（男2，女1）、（男2，女2）、（男2，男3）、（男3，女1）、
（男3，女2）、（女1，女2），共有10种，它们出现的可能性相同．所有的结果中，满足“恰好
是1名男生和1名女生”（记为事件A）的结果有6种，所以P（A）6＝3
··················8分
105
（说明：
通过枚举、画树状图或列表得出全部正确情况得4分；没有说明等可能性扣1分．）
21．（本题8分）
解：
（1）54．·········································································································2分
（2）60人，图略．····························································································5分
（3）10000×60%＝6000（人）．··············································································7分所以估计该市10000名中学生家长中有6000名家长持反对态度．····························8分
22．（本题8分）
解：
（1）因为当x＝1时，ax2＝1．
所以a＝1．·····································································································1分因为当x＝－1时，ax2＋bx＋c＝7；当x＝0时，ax2＋bx＋c＝2．
⎧1－b＋c＝7，
所以⎨
⎩c＝2．
····························································································3分
所以b＝－4．································································
································
···
4分
所以函数y＝ax2＋bx＋c的表达式为y＝x2－4x＋2．································
···················
5分
（2）y＝x2－2x＋1（或y＝（x－1）2）．································
································
····
8分
23．（本题8分）
解：
设AB＝xm，则BC＝（100－4x）m．·······································································2分由题意可知：
x（100－4x）＝400．···········································································4分化简得：
x2－25x＋100＝0．
解得x1＝20，x2＝5．·························································································6分
因为羊圈一面是长为25m的墙，所以100－4x≤25，解得
75
x≥4．
所以，x2＝5舍去．····························································································7分
BC＝100－4x＝20（m）．
答：
AB＝20m，BC＝20m．·················································································8分
A
24．（本题8分）
解：
延长DF，交AC于点G．·················································1分设AG＝xm．
由题意知：
DF＝13.1m，DB＝FE＝GC＝1.4m．
DG
在Rt△ADG中，tan∠ADG＝AG，
DαFβG
BEC
∴DG
AGxx
＝＝≈．···················································································3分
tanαtan63.5°2
FG
在Rt△AFG中，tan∠AFG＝AG，
∴FG＝AGx
＝
x．··················································································5分
tanβ
tan71.6°≈3
∵DF＝DG－FG，
∴x－x＝13.1．·································································································6分
23
解得x＝78.6．··································································································7分
∴AG＝78.6m．
∵AC＝AG＋GC，
∴AC＝78.6＋1.4＝80（m）．
答：
该塔AC的高度约80m．···············································································8分
25．（本题8分）
解：
（1）BC与⊙O相切．
证明：
连接OD．
∵AD是∠BAC的平分线，
∴∠BAD＝∠CAD．又∵OD＝OA，
∴∠OAD＝∠ODA．
∴∠CAD＝∠ODA．
∴OD∥AC．·····························································································1分
∴∠ODB＝∠C＝90°，即OD⊥BC．······························································2分
又∵BC过半径OD的外端点D，································
································
···
3分
∴BC与⊙O相切．································································
·····················
4分
（2）由
（1）知OD∥AC．
∴△BDO∽△BCA．·····················································································5分
∴BO＝DO．
BACA
∵⊙O的半径为2，∴DO＝OE＝2，AE＝4．
BE＋22
∴BE＋＝
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