2013年南京市白下区中考一模数学试题及答案

白下区2012/2013学年度第二学期第一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铅笔将答题卷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答题卷相应位置上） ．．．．．．．1．如果a与－3互为相反数，那么a等于 A．3

B．－3

1C．

3

1D．－

3

2．2013年元宵节正值周末，观灯人数也创下历史新高．据统计，当天有520000游客在夫子庙地区观灯闹元宵，将520000用科学记数法表示为 A．0.52×105 A．a2＋a4

B．5.2×104 B．a8－a2

C．5.2×105 C．a2·a3

D．5.2×106 D．a7÷a

3．下列各式中，计算结果为a6的是

3

4．当x＜0时，函数y＝－的图象在

xA．第一象限

5．如图，在矩形ABCD内，以BC为一边作等边三角形EBC，连接AE、DE．若BC＝2，ED＝3，则AB的长为 A．22 C．2＋3

B．23 D．2＋3

B

C

（第5题）

B．第二象限 C．第三象限 D．第四象限

A

E D

6．把函数y＝2x2－4x的图象先沿x轴向右平移3个单位长度，再沿y轴向下平移2个单位长度得到新函数的图象，则新函数是 A．y＝2(x＋3)2－4(x＋3)－2 C．y＝2(x＋3)2－4(x＋3)＋2

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） ．．．．．．．

x7．若分式有意义，则x的取值范围是 ▲ ．

1－x8．计算(2＋1)(2－2)＝ ▲ ．

1

B．y＝2(x－3)2－4(x－3)－2 D．y＝2(x－3)2－4(x－3)＋2

9．我市市区3个PM2.5监测点连续两天测得的空气污染指数数据如下（主要污染物为可吸入颗粒物）：61，82，80，70，56，91，该组数据的中位数是 ▲ ．

10．一个等腰三角形的两边长分别是2和4，它的周长是 ▲ ． 11．如图，在平面直角坐标系中，□ABCD的顶点A、B、D的坐

标分别是（0，0）、（5，0）、（2，3），则点C的坐标是 ▲ ．

y D C B （A） O （第11题）

x 12．如图，F、G分别是正五边形ABCDE的边BC、CD上的点，CF＝DG，连接DF、EG．将

△DFC绕正五边形的中心按逆时针方向旋转到△EGD，旋转角为α（0°＜α＜180°），则∠α＝ ▲ °．

A A B F

D C G

（第12题）

B

（第13题）

C

B

（第14题）

C

E

D D A 13．如图，在梯形ABCD中，AD∥BC，BA＝BC，CA＝CD．若BC＝10 cm，CD＝6 cm，

则AD＝ ▲ cm．

14．在如图所示的正方形网格中，A、B、C都是小正方形的顶点，经过点A作射线CD，则

sin∠DAB的值等于 ▲ ． 15．课本上，公式(a－b)2＝a2－2ab＋b2是由公式(a＋b)2＝a2＋2ab＋b2推导得出的，该推导

过程的第一步是：(a－b)2＝ ▲ ． ．．．

16．“三角形的一个外角等于与它不相邻的两个内角的和”揭示了三角形的一个外角与它的

两个内角之间的数量关系，请探索并写出三角形没有公共顶点的两个外角与它的一个内角之间的数量关系： ▲ ．

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） 17．（6分）解方程x2－4x＋1＝0．

??2－x＞0，18．（6分）解不等式组?5x＋1并写出不等式组的整数解．

＋1≥x，?2?

a2－4a2＋2a2

19．（6分）计算(2－)÷．

a－4a＋4a－2a－2

2

20．（8分）为迎接2014年南京青奥会，某校组织了以“我为青奥加油”为主题的电子小报

制作比赛，评分结果只有60，70，80，90，100（单位：分）五种．现从中随机抽取部分电子小报，对其份数及成绩进行整理，制成如下两幅不完整的统计图．

抽取的电子小报份数统计图

份数 48 36 24 12 0

6 60

70

80

90

24 12 100 成绩/分

（第20题） 36 90分 30% 80分 70分 ▲ % 20% 抽取的电子小报成绩分布统计图

100分10% 60分5% 根据以上信息，解答下列问题：

（1）求本次抽取的电子小报的份数，并补全两幅统计图；

（2）已知该校收到参赛的电子小报共900份，请估计该校学生比赛成绩达到90分以上

（含90分）的电子小报有多少份？

21．（8分）

（1）如图，已知直线AB和直线外一点C．利用尺规，按下面的方法作图：

①取一点P，使点P与点C在直线AB的异侧．以C为圆心，CP的长为半径画弧，与直线AB交于点D、E；

1

②分别以D、E为圆心，大于DE的长为半径画弧，

2两弧交于点F（点F与点C在直线AB的异侧）；

③过C、F两点作直线．

（2）判断（1）中直线CF与直线AB的位置关系，并说明理由．

22．（8分）在歌唱比赛中，一位歌手分别转动如下的两个转盘（每个转盘都被分成3等份）

一次，根据指针指向的歌曲名演唱两首曲目．

（1）转动转盘①时，该转盘指针指向歌曲“3”的概率是 ▲ ；

（2）若允许该歌手替换他最不擅长的歌曲“3”，即指针指向歌曲“3”时，该歌手就

选择自己最擅长的歌曲“1”，求他演唱歌曲“1”和“4”的概率．

1 3 ① 2 4 6 ② 5 A

（第21题）

B

· C

（第22题）

3

23．（8分）如图，在四边形ABCD中，AB＝DC，E、F分别是AD、BC的中点，G、H分

别是对角线BD、AC的中点． （1）求证：四边形EGFH是菱形；

（2）若AB＝1，则当∠ABC＋∠DCB＝90°时，求四边形EGFH的面积．

A G E H D B F （第23题）

C

24．（8分）如图，D是⊙O的直径CA延长线上一点，点B在⊙O上，∠DBA＝∠C． （1）判断直线BD与⊙O的位置关系，并说明理由； （2）若AD＝AO＝1，求图中阴影部分的面积．

25．（8分）某批发商以每件50元的价格购进500件T恤．若以单价70元销售，预计可售

出200件．批发商的销售策略是：第一个月为增加销售量，降价销售，经过市场调查，单价每降低1元，可多售出10件，但最低单价高于购进的价格；第一个月结束后，将剩余的T恤一次性清仓销售，清仓时单价为40元．

（1）按照批发商的销售策略，销售完这批T恤可能亏本吗？请建立函数关系进行说明； （2）从增加销售量的角度看，第一个月批发商降价多少元时，销售完这批T恤获得的

利润为1000元？

D A O C B （第24题）

4

26．（10分）问题：如图，要在一个矩形木板ABCD上切割、拼接出一个圆形桌面，可在

该木板上切割出半径相等的半圆形O1和半圆形O2，其中O1、O2分别是AD、BC上的点，半圆O1分别与AB、BD 相切，半圆O2分别与CD、BD相切．若AB＝a m，BC＝b m，求最终拼接成的圆形桌面的半径（用含a、b的代数式表示）．

（1）请解决该问题；

（2）①下面方框中是小明简要的解答过程：

解：作O1E⊥BC，垂足为E，连接O1O2．

B O2

（第26题）

C

A O1 D 设半圆O1的半径为x m，则半圆O2的半径也为x m． 在Rt△O1EO2中，O1E2＋O2E2＝O1O22． 即O1E2＋(BC－BE－O2C)2＝O1O22． 所以a2＋(b－2x)2＝(2x)2． a2＋b2

解得x＝．

4b

a2＋b2 所以最终拼接成的圆形桌面的半径为m．

4b

老师说：“小明的解答是错误的！”请指出小明错误的原因．

②要使①中小明解得的答案是正确的，a、b需要满足什么数量关系？

B

E

A O1 O2

C D

5

27．（12分） 实际情境

王老师骑摩托车想尽快将甲、乙两位学生从学校送到同一个车站．由于摩托车后座只能坐1人，为了节约时间，王老师骑摩托车先带着乙出发，同时，甲步行出发． 已知甲、乙的步行速度都是5 km/h，摩托车的速度是45 km/h． 方案预设

（1）预设方案1：王老师将乙送到车站后，回去接甲，再将甲送到车站．图①中折线

A-B-C-D、线段AC分别表示王老师、甲在上述过程中，离车站的路程．．．．．．y（km）

与王老师所用时间x（h）之间的函数关系． ①学校与车站的距离为 ▲ km； ②求出点C的坐标，并说明它的实际意义； y/km

O 15 A C y/km 15 A C 10 10 5 5 B 1 3B ①

2 35 6D 1 x/h O D E ②

2 35 61 31 x/h （第27题）

（2）预设方案2：王老师骑摩托车行驶a h后，将乙放下，让乙步行去车站，与此同时，

5

王老师回去接甲并将甲送到车站，王老师骑摩托车一共行驶 h．图②中折线

6

A-B-C-D、线段AC、线段BE分别表示王老师、甲、乙在上述过程中，离车站的．．．．路程（km）与王老师所用时间x（h）．．y

之间的函数关系．求a的值．

优化方案

（3）请设计一种方案，使甲、乙两位学生．．

在出发50min内（不含50min）全部到达车站．

（要求：1.不需用文字写出方案，在图③中画出图象即可；2．写出你所画的

y/km 15 10 5 O 1 32 35 61 x/h （第27题③）

图象中y 与x的含义；3．不需算出甲、乙两位学生到达车站的具体时间！）

6

2012/2013学年度第二学期第一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 答案 1 A 2 C 3 D 4 B 5 C 6 B 二、填空题（每小题2分，共计20分）

7．x≠1 8．2 9．75 10．10 11．（7，3） 12．72 13．3.6 14．

2

15．[a＋(－b)]2（注：写a2＋2a·(－b)＋(－b)2也可） 2

16．表述方法不唯一，如：三角形没有公共顶点的两个外角之和等于与它们都不相邻的一个内角加上180°．

三、解答题（本大题共11小题，共计88分） 17．（本题6分）

解法一：移项，得x2－4x＝－1． ··················································································· 1分

配方，得 x2－4x＋4＝－1＋4， ········································································· 2分

(x－2)2 ＝3． ···················································································· 3分

由此可得 x－2 ＝±3． ················································································· 4分 x1＝2＋3，x2＝2－3． ·················································································· 6分

解法二：a＝1，b＝－4，c＝1．

b2－4ac＝(－4)2－4×1×1＝12＞0． ·································································· 2分 4±12

x＝ ··································································································· 4分

2＝2±3．

x1＝2＋3，x2＝2－3． ·················································································· 6分

18．（本题6分）

解：解不等式①，得x＜2． ·························································································· 2分

解不等式②，得x≥－1． ······················································································· 4分 所以不等式组的解集是－1≤x＜2． ·········································································· 5分 所以不等式组的整数解是－1，0，1． ······································································· 6分

7

19．（本题6分）

a2－4a2＋2a2

解：(2－)÷ a－4a＋4a－2a－2

(a＋2)(a－2)a(a＋2)2＝[]÷ ··········································································· 3分 2－ (a－2)a－2a－2

a－2a

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

a－2a(a＋2)1

＝． ··········································································································· 6分

a＋2

20．（本题8分）

解：（1）因为24÷20%＝120(份)，所以本次抽取了120份电子小报． ··································· 2分 条形统计图中80分的份数是42，扇形统计图中80分的百分比是35%． ···················· 4分 （2）900×(30%＋10%) ··························································································· 6分

＝360(份)． ··································································································· 7分 所以估计该校学生比赛成绩达到90分以上（含90分）的电子小报有360份．

····································································································································· 8分

21．（本题8分）

解：（1）作图正确． ·································································································· 3分

（2）CF⊥AB．

理由如下：连接CD、CE、FD、FE． ··································· 4分 由作图知CD＝CE，FD＝FE． ············································ 5分 A

方法一：所以点C、F都在线段DE的垂直平分线上．

········································································ 7分

D ·P F

E B C 所以CF是线段DE的垂直平分线，即CF⊥AB． ······································ 8分

方法二：因为CF＝CF，所以△CDF≌△CEF． ···················································· 6分 所以∠DCF＝∠ECF． ········································································· 7分 又因为CD＝CE，所以CF⊥AB． ·························································· 8分

22．（本题8分）

1解：（1）．············································································································· 2分

3（2）分别转动两个转盘一次，所有可能出现的结果有：（1，4）、（1，5）、（1，

6）、（2，4）、（2，5）、（2，6）、（3，4）、（3，5）、（3，6），共有9种，它们出现的可能性相同．由于指针指向歌曲“3”时，该歌手就选择自己最擅长的歌曲“1”，所以所有的结果中，该歌手演唱歌曲“1”和“4”（记为

2

事件A）的结果有2种，所以P（A）＝． ······························································· 8分

9

8

（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．）

23．（本题8分）

（1）证明：∵四边形ABCD中，E、F、G、H分别是AD、BC、BD、AC的中点，

1111

∴FG＝CD，HE＝CD，FH＝AB，GE＝AB． ·············································· 2分

2222∵AB＝CD，

∴FG＝FH＝HE＝EG． ················································································ 3分 ∴四边形EGFH是菱形． ·············································································· 4分 （其他方法参照给分）

（2）解：∵四边形ABCD中，G、F、H分别是BD、BC、AC的中点，

∴GF∥DC，HF∥AB． ···················································································· 5分 ∴∠GFB＝∠DCB，∠HFC＝∠ABC． ∴∠HFC＋∠GFB＝∠ABC＋∠DCB＝90°．

∴∠GFH＝90°． ·························································································· 6分 ∴菱形EGFH是正方形． ················································································· 7分 11

∵AB＝1，∴EG＝AB＝．

22

11

∴正方形EGFH的面积＝()2＝． ····································································· 8分

24

24．（本题8分）

解：（1）直线BD与⊙O相切．理由如下：

连接OB．

∵CA是⊙O的直径，

∴∠ABC＝90°． ······································· 1分

∵OB＝OC，

∴∠OBC＝∠C．

又∵∠DBA＝∠C，

∴∠DBA＋∠OBA＝∠OBC＋∠OBA＝∠ABC＝90°． ············································· 2分

∴OB⊥BD．

又∵直线BD经过半径OB的外端点B， ······························································ 3分

∴直线BD与⊙O相切． ·················································································· 4分

（2）∵∠DBO＝90°，AD＝AO＝1，

∴AB＝OA＝OB＝1．

∴△AOB是等边三角形．

9

（第24题）

D A O C B ∴∠AOB＝60°． ···························································································· 5分

60π×12π

∴S扇形OBA＝＝． ················································································· 6分

3606 ∵在Rt△DBO中，BD＝DO2－BO2＝3，

113

∴S?DBO＝OB·BD＝×1×3＝． ································································ 7分

222

∴S阴影＝S? DBO－S扇形OBA＝

25．（本题8分）

解：（1）解法一：设第一个月单价降低x元，批发商销售完这批T恤获得的总利润为y

元． ······························································································································· 1分 根据题意，得y＝(70－50－x)(200＋10x)＋(40－50)×[500－(200＋10x)]

＝－10x2＋100x＋1000． ································································ 4分 批发商销售这批T恤可能亏本，理由如下：（答案不唯一，以下方法供参考）

方法一：当x＝17（或18或19）时，y＜0． ·························································· 5分 方法二：当y＝0时，x＝55＋5（负根舍去）．

又因为当55＋5＜x＜20时，y随x的增大而减小，

所以当x＝17或18或19时，y＜0． ························································ 5分 解法二：设第一个月单价降低x元，当月出售T恤获得的利润为y1元，清仓剩

余T恤获得的利润为y2元． ··············································································· 1分 根据题意，得y1＝(70－50－x)(200＋10x)＝－10x2＋4000， ······································· 3分 y2＝(40－50)×[500－(200＋10x)]＝100x－3000． ···················································· 4分 批发商销售这批T恤可能亏本，理由如下：（答案不唯一，以下方法供参考）

方法一：当x＝17（或18或19）时，y1＋y2＜0． ··················································· 5分 方法二：当y1＋y2＝0时，x＝55＋5（负根舍去）．

又因为当55＋5＜x＜20时，y1＋y2随x的增大而减小，

所以当x＝17或18或19时，y1＋y2＜0． ·················································· 5分

（2）设第一个月单价降低x元时，销售完这批T恤获得的利润为1000元．

根据题意得－10x2＋100x＋1000＝1000． ································································ 6分 解这个方程，得x1＝0，x2＝10．

从增加销售量的角度看，取x＝10． ······································································ 7分 答：第一个月单价降低10元时，销售完这批T恤获得的利润为1000元． ····················· 8分

26．（本题10分）

3π

－． ····································································· 8分 26

10

解：（1）方法一：

如图，设半圆O2与BD 的切点为E，连接O2E，则O2E⊥BD． ······························· 1分 OA D 1 ∵半圆O2与CD 相切，且∠C＝90°， ∴O2E＝O2C，DC＝DE＝a．

在Rt△BEO2中，O2B2＝BE2＋O2E2． ·················· 2分 ∴(b－EO2)2＝(a2＋b2－a)2＋O2E2． ·················· 3分

aa2＋b2－a2aa2＋b2－a2 解得EO2＝，所以最终拼接成的圆形桌面的半径为bb

m． ············································································································ 4分 方法二：

如图，设半圆O2与BD 的切点为E，连接O2E，则O2E⊥BD． ······························· 1分 ∵半圆O2与CD 相切，且∠C＝90°， ∴O2E＝O2C．

∵∠EBO2＝∠CBD，∠BEO2＝∠BCD＝90°，

∴△BEO2∽△BCD． ······················································································· 2分 EO2BO2

∴＝．

CDBD

EO2b－EO2

∴＝22． ·························································································· 3分

aa＋b

aa2＋b2－a2aa2＋b2－a2 解得EO2＝，所以最终拼接成的圆形桌面的半径为bb

m． ····················································································································· 4分

（说明：求出最终拼接成的圆形桌面的半径为

ab

m不扣分．）

a2＋b2＋a

B O2

C E （2）①小明的错误是半圆O1与半圆O2不能保证外切，即“O1O2＝2x”是错误的．

····································································································································· 7分

②方法一：

要使小明解得的答案是正确的，就要半圆O1与半圆O2外切．

此时半圆O1与BD 的切点、半圆O2与BD的切点以及O1O2与BD的交点重合．

所以a2＋b2－a＝a． ··················································································· 8分

解得b＝3a． ·························································································· 10分 方法二：

aa2＋b2－a2a2＋b2令＝． ············································································· 8分

b4b4aa2＋b2－4a2＝a2＋b2． a2＋b2－4aa2＋b2＋4a2＝0．

11

即(a2＋b2－2a) 2＝0．

解得b＝3a． ·························································································· 10分

27．（本题12分）

解：方案预设

（1）预设方案1：

①15． ········································································································· 2分

②方法一：

设王老师把乙送到车站后，再经过m h与甲相遇．

1

(45＋5)m＝15－5×． ·················································································· 3分

34

解得m＝．

15

14333

＋＝，15－5×＝12，即C（，12）． ··················································· 5分

315555 方法二：

BC对应的函数关系式为y＝45x－15．

AC对应的函数关系式为y＝－5x＋15． ···························································· 3分

3

BC与AC的交点C的坐标为（，12）． ························································· 5分

5

3

点C的实际意义为王老师在出发 h后，在距离车站12 km处接到甲． ··················· 6分

5

（2）预设方案2：

方法一：

设王老师把乙放下后，再经过n h与甲相遇． (45＋5)n＝45a－5a．

4

解得n＝a． ································································································ 7分

55454

由于王老师骑摩托车一共行驶 h，可得方程15－5(a＋a)＝45×[－(a＋

6565

a)]． ······························································································································ 9分

5

解得a＝． ······························································································ 10分

16 方法二：

根据题意，得点B（a，15－45a）． ································································· 7分

9

求得点C（a，15－9a）． ············································································· 8分

5 所以CD对应的函数关系式为y＝－45x＋72a＋15． ············································· 9分

55

将（，0）代入，解得a＝．······································································ 10分

61615

（说明：未经过任何说明，直接判断点E坐标为（，0），从而解得a＝只

216

得1分）

12

优化方案

（3）本题答案不唯一，以下方法供参考． y/km A 15 C 10

5 B E D 251 O 1 x/h 363

······································································································· 11分 图中折线A-B-C-D、线段AC、线段BE分别表示王老师、甲、乙离车站的路程y（km）与王老师所用时间x（h）之间的函数图象．

······································································································· 12分

（说明：没有图象或图象错误不得分．）

13

本文来源：https://www.bwwdw.com/article/tg6.html