江苏省南京市秦淮区中考二模数学试题及参考答案.docx
- 文档编号:8038612
- 上传时间:2023-01-28
- 格式:DOCX
- 页数:17
- 大小:90.80KB
江苏省南京市秦淮区中考二模数学试题及参考答案.docx
《江苏省南京市秦淮区中考二模数学试题及参考答案.docx》由会员分享,可在线阅读,更多相关《江苏省南京市秦淮区中考二模数学试题及参考答案.docx(17页珍藏版)》请在冰豆网上搜索。
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+=
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 江苏省 南京市 秦淮 中考 数学试题 参考答案