初三数学答案

湛江市2012年初中毕业生学业调研测试

数学试卷 参考答案及评分意见

一、选择题（本大题共10小题，每小题4分，共40分）

1．C 2．D 3．B 4．D 5．B 6．C 7．A 8．A 9．D 10．D 二、填空题（本大题共5小题，每小题4分，共20分） 11．1.7×104 12．6 13．?12 14．3 15．35

三、解答题（本大题共10小题，共90分） 16．解：原式?3?12?12 ···································································································5分

?3 ·····································································································6分

17．解：原式＝a2?1?a2?2a·································································································5分

＝2a?1 ·············································································································6分

18．解：（1）将点A、B分别向右平移5个单位，

得到点A1、B1，顺次连接四点即可．（如图1） ····························································4分

（2）过点A、B1画一条直线或过点A1、B画一条直线即可．（如图2） ·································8分 y y 5 5 A1 A1 A 4 A 4 3 3 B -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1

图1 2 1 B 1B -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 图2

A D 2 1 B 119.证明：∵ AB∥DE，

∴?ABC??DEF. ……………………4分

在△ABC和△DEF中，

又∵?A??D，BC?EF，

∴△ABC≌△DEF． ……………………8分

20．解：（1）样本人数为

2050%?40B

E C F

（人）． ··············································································2分

骑自行车上学人数：40?30%?12（人），补全条形图如下：

【湛江市2012年初中毕业生学业调研测试 数学试卷 参考答案 第 1 页 共 4 页】

人数 步行 骑自 乘车 上学方式

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

行车

（2）该年级乘车上学的学生人数约为800?20%?160（人）． ··················································8分 21．解：（1）在Rt△BCD中， ∵?CBD?30?，BC?30， ∴CD?12BC?15．………………420 16 12 8 4 0 C 分 32A ?153B D （2）在Rt△BCD中，BD?BC?cos30??30?． ······················································6分

在△ABC中，∠CBD=30°，∠CAB=15°， ∴∠BCA=15°

∴AB=BC=15, ·············································································································8分 ∴AD=AB+BD=15?153?15(1?22．解：（1）设袋中有红球x个，则有

22?1?x?12·····································································9分 3)． ·

··································· 10分 3)米． ·

答：坡高15米，斜坡新起点A到点D的距离为15(1?， ·············································································································2分

解得x?1，所以口袋中有红球1个． ·············································································4分 （2）画树状图如下：

第一次 第二次

白 白

白 红 黄

白 红 黄

开 始

红

黄

白 白 红

白 白 黄

得分 2 1 3 2 1 3 1 1 2 3 3 2

······································································································································8分 由上述树状图可知：所有可能出现的结果共有12种．其中摸出两个球且得2分的有4种. 所以P(从中摸出两个球得2分)?412?13． ········································································· 10分

23. 解：(1) ∵AB是⊙O的直径 ∴?ACB?90°． ························································1分 ∵PA是⊙O的切线 ∴?OAP?90?． ······················· 2分

A ∵BC∥OP ∴?AOP??CBA． ······························· 3分 ∴△ABC∽△POA． ····················································· 5分 （2）∵AB是⊙O的直径，且AB=2．

∴OA=1． ······································································ 6分

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P ．O

B ∵在Rt?OAP中，PA=2, ∴OP?PA?OA?22···························································································7分 3． ·

∵由(1)可知△ABC∽△POA． ∴

BCOA?ABOP． ·············································································································9分

?2?13?233∴BC?AB?OAOP． ····················································································· 10分

24．解：（1）假设该厂现有原料能保证生产，且能生产A产品x件，则能生产B产品(100?x)件． ·······································································································································1分

根据题意，有 ??3x?2.5(100?x)≤263,?2x?3.5(100?x)≤314 ······································································4分

解得：24≤x≤26 ····························································································6分 由题意知，x应为整数，故x?24或x?25或x?26． ·····················································7分 此时对应的(100?x)分别为76、75、74．

即该厂现有原料能保证生产，可有三种生产方案：

生产A、B产品分别为24件，76件；25件，75件；26件，74件． ································8分 （2）生产A产品x件，则生产B产品(100?x)件．根据题意可得

y?120x?200(100?x)??80x?20000 ······································································· 10分 ∵?80?0，

∴y随x的增大而减小，从而当x?26，即生产A产品26件，B产品74件时，生产总成本最底，最低生产总成本为y??80?26?20000?17920元． ············································· 12分

25．解：（1）设所求抛物线的解析式为y?ax2?bx?c(a?0)····················································1分

y 1?B a???C 3?c?0F ?10??P 依题意，得?100a?10b?0，解得?b? 3H ???16a?4b?8?c?0??1103xO E D A x ∴ 所求抛物线的解析式为y??x2?3． ···································································3分

（2）作BE?OA于E，OE?BC?4，

在Rt△ABE中，AE?OA?OE?6，BE?OC?8， ∴AB?AE?BE22?10． ······························································································4分

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解法一：作OF?AB于F，DH?AB于H，

∵OA?BE?AB?OF， ································································································5分 ∴OF?∴S?12OA?BEAB?8，DH?12······································································6OF?4． 分

AP?DH?12?t?4?2t(0≤t≤10) ·································································8分

解法二：∵

S?APDS?ABDS20??APAB，S?ABD?12AD?BE?12?5?8?20························································6分

∴

t10，

∴S?2t（0≤t≤10） ·····························································································8分

（3）点P只能在AB或OC上才能满足题意，

S梯形COAB?12(BC?OA)?OC?12···························································9分 ?(4?10)?8?56 ·

（ⅰ）当点P在AB上时，设点P的坐标为(x,y)

由S?APD?得

12AD?y14S梯形COAB， 1285=?56，解得y?4，··············································································· 10分

由S?APD?12AP?DH?12t?4?14，得t?7．

285)?722此时，作BG?OA于G，由勾股定理得(AO?x)2?y2?AP2,即(10?x)2?(解得x?295，

，即在7秒时有点P1(2928,)55满足题意； ··················································· 11分

（ⅱ）当点P在OC上时，设点P的坐标为(0,y)．

由S?OPD?14S梯形COAB， 得1OD?y=1?56，解得y?28，

24528525此时t?10?4?(8?)?16． 即在t?16，在162525秒时，有点P2(0,285)285)满足题意；

综上，在7秒时有点P1(2928,)55秒时有点P2(0,

使PD将梯形COAB的面积分成1:3的两部分． ······················································ 12分

说明：以上各题，若还有其他解（证）法，请酌情给分．

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