南京市初中数学测试卷(含详细答案)

一、选择题

1．5的相反数是（ ▲ ）．

A．

1 5236 B．1 523 C．5

D．5

32．下列运算正确的是（ ▲ ）．

A．a·aa B．aa C．aba6b3 D．a6a2a3

823．为迎接2014年青奥会，在未来两到三年时间内，一条长53公里，总面积约11000亩的鸀色长廊将串起南京的观音门、仙鹤门、沧波门等8座老城门遗址．数据11000用科学记数法可表示为（ ▲ ）． A．1110 B．1.110 C．1.110 D．0.1110

x＋1＞0，4．如图，不等式组 的解集在数轴上表示正确的是（ ▲ ）．

x－1≤0

3455

0 1 -1 0 1 -1 0 1 B． D． C．

5．如图，在12网格的两个格点上任意摆放黑、白两个棋子，且两棋子不在同一条格线上．其中恰好如

-1 -1 图示位置摆放的概率是（ ▲ ）． A．

0 1 A．

1 6 B．

11 C． 912 D．

1 186．如图，在扇形纸片AOB中，OA =10，AOB=36，OB在桌面内的直线l上．现将此扇形沿l按顺时针方向旋转（旋转过程中无滑动），当OA落在l上时，停止旋转．则点O所经过的路线长为（ ▲ ）． A． 12

（第5题图）

B．11 C．10 D．10555 A O B （第6题图）

l

二、填空题

7．数据3，5，5，1，1，1，1的众数是 ▲ ． 8．分解因式x26x9的结果是 ▲ ．

A 9．如图，已知AB∥CD，AEF80°，则DCF为 ▲ °．

F C A

E

（第9题图）

D B

C O D B 2

4

（第11题图）

（第12题图）

10．观察：a11，a213111111． ，a3，a4，…，则an ▲ （n为正整数）

35577911．如图，AB是⊙O直径，且AB=4cm，弦CD⊥AB，∠COB=45°，则CD为 ▲ cm． 12．如图，是水平放置的长方体，它的底面边长为2和4，左视图的面积为6，则该长方体的体积为 ▲ ． 13．当分式

1与3的值相等时，x的值为 ▲ ．

x2xk的图象都经过点A（1，1）．则在第一象限内，当y1y2x14．如图，正比例函数y1x和反比例函数y2时，x的取值范围是 ▲ ．

O （第14题图） y A(1，1) x A E F B D H G C

（第15题图）

15．如图，在梯形ABCD中，AD∥BC，点E、F、G、H是两腰上的点，AE=EF=FB，CG=GH=HD，且四边形EFGH

的面积为6cm2，则梯形ABCD的面积为 ▲ cm2．

16．一张矩形纸片经过折叠得到一个三角形（如图），则矩形的长与宽的比为 ▲ ．

（第16题图）

三、解答题

17．（5分）计算：(13)0218．

a2b2b2＋22，其中a＝－2，b＝1． 18．（5分）先化简，再求值：

a＋ba－b

19．（6分）如图，在△ABC中，AB=AC，AD⊥BC，垂足为D，AE∥BC, DE∥AB.

证明：（1）AE=DC；（2）四边形ADCE为矩形．

20．（6分）某区为了解全区2800名九年级学生英语口语考试成绩的情况，从中随机抽取了部分学生的成

B

D

（第19题图）

A E

C

绩（满分24分，得分均为整数），制成下表： 分数段（x分） 人 数 （1）填空：

①本次抽样调查共抽取了 ▲ 名学生； ②学生成绩的中位数落在 ▲ 分数段；

③若用扇形统计图表示统计结果，则分数段为x≤16的人数所对应扇形的圆心角为 ▲ °； （2）如果将21分以上（含21分）定为优秀，请估计该区九年级考生成绩为优秀的人数．

21．（6分）某初级中学准备随机选出七、八、九三个年级各1名学生担任领操员．现已知这三个年级分别选送

一男、一女共6名学生为备选人.

（1）请你利用树状图或表格列出所有可能的选法； （2）求选出“两男一女”三名领操员的概率．

x≤16 10 17≤x≤18 15 19≤x≤20 35 21≤x≤22 112 23≤x≤24 128 22．（6分）受国际原油价格持续上涨影响，某市对出租车的收费标准进行调整．

车费y/元 16 15

调整前： 14

13 调整方案： 12 加收1元燃油附调整后： 11 ．

加费，其它收费标准10

9 保持不变． 0 1 2 3 4 5 6 路程x/km

（第22题图）

（1）调整前出租车的起步价为 ▲ 元，超过3km收费 ▲ 元/km；

（2）求调整后的车费y（元）与行驶路程x（km）（x>3）之间的函数关系式，并在图中画出其函数图象．

23．(8分) 现有一张宽为12cm练习纸，相邻两条格线间的距离均为．调皮的小聪在纸的左上角用印章印

出一个矩形卡通图案，图案的顶点恰好在四条格线上（如图），测得∠α＝32°．

（1）求矩形图案的面积；

（2）若小聪在第一个图案的右边以同样的方式继续盖印（如图），最多能印几个完整的图案

（参考数据：sin32°≈，cos32°≈，tan32°≈）

α …… 12cm （第23题图）

24．（8分）某手机专营店代理销售A、B两种型号手机．手机的进价、售价如下表：

（1）第一季度：用36000元购进 A、B两种型号的手机，全部售完后获利6300元，求第一季度购进A、

B两种型号手机的数量；

（2）第二季度：计划购进A、B两种型号手机共34部，且不超出第一季度的购机总费用，则A型号手

机最多能购多少部

25. （8分）如图，在△ABC中，AB=AC，点O为底边上的中点，以点O为圆心，1为半径的半圆与边AB

相切于点D．

A （1）判断直线AC与⊙O的位置关系，并说明理由；

（2）当∠A＝60°时，求图中阴影部分的面积．

26．（9分）已知二次函数yx2xm的图象与x轴相交于A、B两点（A左B右），与y轴相交于

点C，顶点为D． （1）求m的取值范围；

（2）当点A的坐标为(3,0)，求点B的坐标； （3）当BC⊥CD时，求m的值．

2型 号 进 价 售 价 A 1200元/部 1380元/部 B 1000元/部 1200元/部 D B O （第25题图）

C

27．（9分）

操作：小明准备制作棱长为1cm的正方体纸盒，现选用一些废弃的圆形纸片进行如下设计： C 说明：

B

方案一图形中的圆过点A、B、C；

方案二直角三角形的两直角边与

展开图左下角的正方形边重合，斜A

边经过两个正方形的顶点．

方案二 方案一

纸片被利用的面积

纸片利用率= ×100%

纸片的总面积

发现：（1）方案一中的点A、B恰好为该圆一直径的两个端点．

你认为小明的这个发现是否正确，请说明理由． （2）小明通过计算，发现方案一中纸片的利用率仅约为%．

请帮忙计算方案二的利用率，并写出求解过程．

探究：（3）小明感觉上面两个方案的利用率均偏低，又进行了新的设计（方案三），请直接写出方案

三的利用率．

说明：

方案三中的每条边均过其中两个

正方形的顶点．

方案三

28．（12分）如图，在Rt△ABC中，∠C=90°，AC=BC=4cm，点D为AC边上一点，且AD=3cm，动点E从

点A出发，以1cm/s的速度沿线段AB向终点B运动，运动时间为x s．作∠DEF=45°，与边BC相交于点F．设BF长为ycm． （1）当x= ▲ s时，DE⊥AB；

（2）求在点E运动过程中，y与x之间的函数关系式及点F运动路线的长； （3）当△BEF为等腰三角形时，求x的值．

C D F A E B （第28题图）

C D A B （第28题备用图）

参考答案

一、选择题（每小题2分，共计12分）

题号 答案 二、填空题

7．1 8．x3 9．100 10．

21 C 2 C 3 B 4 B 5 C 6 A 11 2n12n1 11．22

223

12．24 13．3 14．x＞1 15．18 16．2︰3 （或 或3 ）

3 三、解答题 17．（本题5分）

解：原式＝1－2+32 ················································································································ 3分

＝－1+32 ·················································································································· 5分 18．（本题5分）

(a2b)(ab)2b2解：原式 ·········································································· 2分

(ab)(ab)(ab)(ab)a2ab ········································································································· 3分 (ab)(ab)a ························································································································ 4分 ab－22

= 3 ·········································································· 5分

－2－1

A E

当a=－2，b=1时，原式=

19．（本题6分）

证明：

（1）在△ABC中，∵AB=AC，AD⊥BC， ∴BD=DC ········································································································································ 1分 ∵AE∥BC, DE∥AB， B C D ∴四边形ABDE为平行四边形 ···································································································· 2分 ∴BD=AE， ···································································································································· 3分 ∵BD=DC ∴AE = DC． ··································································································································· 4分 （2）

解法一：∵AE∥BC，AE = DC， ∴四边形ADCE为平行四边形． ································································································ 5分 又∵AD⊥BC， ∴∠ADC=90°，

∴四边形ADCE为矩形． ············································································································ 6分 解法二：

∵AE∥BC，AE = DC，

∴四边形ADCE为平行四边形 ···································································································· 5分 又∵四边形ABDE为平行四边形 ∴AB=DE．∵AB=AC，∴DE=AC． ∴四边形ADCE为矩形． ············································································································ 6分 20．（本题6分） 解法一：（1）用表格列出所有可能结果： 七年级 八年级 九年级 结果 男 男 男 （男，男，男） 男 男 女 （男，男，女） 男 女 男 （男，女，男） 男 女 女 （男，女，女） 女 女 女 （女，女，女）

女 女 男 （女，女，男）

女 男 女 （女，男，女）

女 男 男 （女，男，男） ······················································································································································· 3分

（2）从上表可知：共有8种结果，且每种结果都是等可能的，其中“两男一女”的结果有3种．所以，P（两男一女）=3

8 ． ······································································································· 6分 解法二：（1）用树状图列出所有可能结果： 七年级 八年级 九年级 结果 男

男 （男，男，男） 男

女

（男，男，女） 男 （男，女，男） 女 （男，女，女） 开始

男 男 （女，男，男）

女

（女，男，女）

女

女

男 （女，女，男）

女 （女，女，女） ······················································································································································· 3分

（2）从上图可知：共有8种结果，且每种结果都是等可能的，其中“两男一女”的结果有3种．所以，P（两男一女）=3

8 ． ······································································································· 6分 21．（本题6分）

（1）①300 ··································································································································· 1分

②21≤x≤22 ······················································································································· 3分 ③12 ····································································································································· 4分 （2）2800×112+128

300 =2240（人） ··························································································· 5分 答：该区所有学生中口语成绩为满分的人数约为2240人． ················································· 6分 22．（本题6分） 解：（1）9；； ······························································································································· 2分

5分5分

（2）y=10+（x－3）=+ ·········································································································· 5分

调整后的图像如图： 16 15 14 13 12 11 10 9 车费y/元 调整前： 调整后： 0 1 2 3 4 5 6路程 x/km

······················································································································································· 6分

23．（本题8分）

（1）如图，在Rt△BCE中，

CECE0.8∵sinα=BC ，∴BC = sinα = = ························································································· 2分

0.5∵矩形ABCD中，∴∠BCD=90°，∴∠BCE+∠FCD=90°， 又∵在Rt△BCE中，∴∠EBC+∠BCE=90°，∴∠FCD=32°．

FC1.6FC在Rt△FCD中，∵cos∠FCD=CD ，∴CD===2 ···················································· 4分

cos320.8∴橡皮的长和宽分别为2cm和． E Bα C

……

G H F D

A 12cm AD

（2）如图，在Rt△ADH中，易求得∠DAH=32°．∵cos∠DAH=AH ， ∴AH=

AD1.6==2 ················································································································ 5分

cos320.8GH

在Rt△CGH中，∠GCH=32°．∵tan∠GCH=CG ，

∴GH=CG tan32°= × = ················································································································ 7分 又∵6×2+＞12，5×2+＜12，3×4+，∴最多能摆放5块橡皮． ··············································· 8分 24．（本题8分）

（1）解：设该专营店第一季度购进A、B两种型号手机的数量分别为x部和y部． ········ 1分

1200x+1000 y＝36000，由题意可知：  ··················································································· 3分

180x+200y＝6300x＝15，

解得：

y＝18

答：该专营店本次购进A、B两种型号手机的数分别为15部和18部． ····························· 4分

（2）解：设第二季度购进A型号手机a部． ········································································· 5分 由题意可知：1200a+1000(34－a)≤36000，············································································· 6分 解得：a≤10 ································································································································· 7分 不等式的最大整数解为10

答：第二季度最多能购A型号手机10部． ············································································· 8分

A

25．（本题8分）

解：（1）直线AC与⊙O相切． ································································································· 1分 理由是：

连接OD,过点O作OE⊥AC，垂足为点E． ∵⊙O与边AB相切于点D，

B C O ·∴OD⊥AB． ································································································································· 2分

D E ∵AB=AC，点O为底边上的中点，

∴AO平分∠BAC ··························································································································· 3分 又∵OD⊥AB，OE⊥AC

∴OD= OE ······································································································································· 4分 ∴OE是⊙O的半径．

又∵OE⊥AC，∴直线AC与⊙O相切．····················································································· 5分 （2）∵AO平分∠BAC，且∠BAC=60°， ∴∠OAD=∠OAE=30°， ∴∠AOD=∠AOE=60°，

ODOD

在Rt△OAD中，∵tan∠OAD = AD ，∴AD= tan∠OAD =3，同理可得AE=3

11

∴S四边形ADOE =2 ×OD×AD×2=2 ×1×3×2=3 ······························································· 6分 120π×121又∵S扇形形ODE= 360 =3 π ····································································································· 7分 1

∴S阴影= S四边形ADOE －S扇形形ODE=3 －3 π． ·············································································· 8分 26．（本题9分）

解：（1）∵二次函数yx2xm的图象与x轴相交于A、B两点

∴b2－4ac＞0，∴4+4m＞0， ····································································································· 2分 解得：m＞－1 ······························································································································ 3分 （2）解法一：

b2∵二次函数yx2xm的图象的对称轴为直线x＝－2a ＝1 ···································· 4分 ∴根据抛物线的对称性得点B的坐标为（5，0） ··································································· 6分 解法二：

把x=－3，y=0代入yx2xm中得m=15 ··································································· 4分

22∴二次函数的表达式为yx2x15

令y=0得x2x150 ······································································································ 5分 解得x1=－3，x2=5

y ∴点B的坐标为（5，0） ··········································································································· 6分 D E

C （3）如图，过D作DE⊥y轴，垂足为E．∴∠DEC＝∠COB＝90°， 当BC⊥CD时，∠DCE +∠BCO＝90°，

∵∠DEC＝90°，∴∠DCE +∠EDC＝90°，∴∠EDC＝∠BCO．

B x A O ECED

∴△DEC∽△COB，∴OB ＝OC ．····························································································· 7分 11

由题意得：OE＝m+1，OC＝m，DE＝1，∴EC＝1．∴ OB ＝m ．

∴OB＝m，∴B的坐标为（m，0）． ························································································· 8分 将（m，0）代入yx2xm得：－m 2+2 m + m＝0．

解得：m1＝0（舍去）， m2＝3． ······························································································ 9分 27．（本题9分） 发现：（1）小明的这个发现正确． ··························································································· 1分 理由：解法一：如图一：连接AC、BC、AB，∵AC＝BC＝5 ，AB＝10

∴AC2+BC2＝AB2 ∴∠BAC＝90°， ······················································· 2分 ∴AB为该圆的直径． ·············································································· 3分

解法二：如图二：连接AC、BC、AB．易证△AMC≌△BNC，∴∠ACM＝∠CBN．

又∵∠BCN＋∠CBN＝90°，∴∠BCN＋∠ACM＝90°，即∠BAC＝90°， ······· 2分 ∴AB为该圆的直径． ······················································································ 3分

A E N D F

H

M

C B 图一 图二 图三

（2）如图三：易证△ADE≌△EHF，∴AD＝EH＝1． ····························································· 4分 ADDE12

∵DE∥BC，∴△ADE∽△ACB，∴AC ＝CB ∴4 ＝CB ，∴BC＝8． ···································· 5分 ∴S△ACB＝16． ······························································································································· 6分 展开图的面积6∴该方案纸片利用率＝ ×100%＝16 ×100%＝% ········································· 7分

纸板的总面积180

探究：（3）361 ··························································································································· 9分 28．（本题12分）

2223

解：（1）2

2 ···························································································································· 2分

（2）∵在△ABC中，∠C=90°，AC=BC=4．

∴∠A＝∠B＝45°，AB=42 ，∴∠ADE＋∠AED＝135°；

又∵∠DEF＝45°，∴∠BEF＋∠AED＝135°，∴∠ADE＝∠BEF；

∴△ADE∽△BEF··························································································································· 4分 ADAE∴BE ＝BF ， ∴

3x14

＝y ，∴y＝－3 x2+3

42 －x

2 x ················································································· 5分

14

∴y＝－3 x2+3 18

2 x＝－3 ( x－22 )2+3

8

∴当x＝22 时，y有最大值＝3 ···························································································· 6分 16

∴点F运动路程为3 cm ············································································································ 7分

C D F

A E B 第28题（1）（2）图

（3）这里有三种情况：

①如图，若EF＝BF，则∠B＝∠BEF；

又∵△ADE∽△BEF，∴∠A＝∠ADE＝45° 3

∴∠AED＝90°，∴AE＝DE＝2 2 ，

2 s；

C

D

F

A

第28题（3）①图

E B

3

∵动点E的速度为1cm/s ，∴此时x＝2 ②如图，若EF＝BE，则∠B＝∠EFB；

又∵△ADE∽△BEF，∴∠A＝∠AED＝45° ∴∠ADE＝90°，∴AE＝32 ， ∵动点E的速度为1cm/s ∴此时x＝32 s；

C

D

A

第28题（3）②图

C D

F

F

E B

A

第28题（3）③图

E B

③如图，若BF＝BE，则∠FEB＝∠EFB； 又∵△ADE∽△BEF，∴∠ADE＝∠AED ∴AE＝AD＝3，

∵动点E的速度为1cm/s ∴此时x＝3s；

3

综上所述，当△BEF为等腰三角形时，x的值为2

2 s或32 s或3s．

（注：求对一个结论得2分，求对两个结论得4分，求对三个结论得5分）