DA山西省太原市中考真题

山西省太原市 2008 年初中毕业生学业考试数学试卷参考答案

一、选择题题号答案 1 A 2 B 3 D 4 C 5 B 13． ( x  2)2 18．55° 6 B 7 C 8 A 9 D 10 D

二、填空题 11． x ≥ 2 16． 8π 三、解答题 12．

3 （或 0.3） 10

14． 3.16 10 19．15

15．5

17．（1，1）

20．（2）

21．解：解不等式 2 x  5 ≤ 3( x  2) ，得 x ≥ 1 ． ···················· 分 ··········· ········· ·········· ········· 2

2 ··········· ·········· ········· ·········· ··········· ········ x ，得 x  3 ． ······························4 分 3 所以，原不等式组的解集是 1 ≤ x  3 ． ·························· 分 ··········· ·········· ···· 5 ·········· ··········· ···· ， 22．解法一：这里 a  1 b  6，c  2 ． ························ 分 ··········· ·········· ··· ·········· ··········· ·· 1

解不等式 x  1  ··········· ·········· ··· ·········· ··········· ··· b2  4ac  (6)2  4 1 (2)  44  0 ，························2 分

x 

6  44 ． ··········· ··········· ·········· ······· 分 ··········· ·········· ··········· ······ 3 ·········· ··········· ··········· ······ 2 1

即 x  3  11 ． ······································· 4 分 ··········· ·········· ··········· ······· ·········· ··········· ··········· ······· 所以，方程的解为 x1  3  11 x2  3  11 ． ······················ 分 ··········· ·········· · ·········· ··········· 5 ，解法二：配方，得 ( x  3)2  11 ．······························ 分 ··········· ·········· ········· ·········· ··········· ········ 3 即 x  3  11 或 x  3   11 ． ······························ 4 分 ··········· ·········· ········· ·········· ··········· ········· 所以，方程的解为 x1  3  11 x2  3  11 ． ······················ 分 ··········· ·········· · ·········· ··········· 5 ， 23．解法一：设第二次捐款人数为 x 人，则第一次捐款人数为 ( x  50) 人． ······· 分

······· ······ 1

9000 12000  ． ··········· ··········· ········· 分 ··········· ·········· ········· 3 ·········· ··········· ········· x  50 x 解这个方程，得 x  200 ． ·································· 分 ··········· ·········· ··········· · 4 ·········· ··········· ··········· · 经检验， x  200 是所列方程的根． ····························· 分 ··········· ·········· ······· 5 ·········· ··········· ·······

根据题意，得答：该校第二次捐款人数为 200 人． ···························· 6 分 ··········· ·········· ······· ·········· ··········· ······· 解法二：人均捐款额为 (12000  9000)  50  60 （元） ················· 分． ··········· ······ ·········· ······ 3 第二次捐款人数为 12000  60  200 （人） ························ 分． ··········· ·········· ·· 5 ·········· ··········· ··

答：该校第二次捐款人数为 200 人． ···························· 6 分 ··········· ·········· ······· ·········· ··········· ······· 24．解：（1）如图， AD 即为所求． ··········· 分 ··········· ·········· 2 A （2） △ ABD ∽△CBA ，理由如下． ·········· 分 ·········· ········· 3  AD 平分 BAC，BAC  2C ， BAD  BCA ． ···················5 分 ··········· ········ ·········· ········· C B D 又 B  B ，△ ABD ∽△CBA ． ········· 分 ········· ········ 6 25．解：乙获胜的可能性大． ································ 分 ··········· ·········· ·········· 2 ·········· ··········· ·········· 进行一次游戏所有可能出现的结果如下表：························· 分 ··········· ·········· ··· 6 ·········· ··········· ··· 第二次第一次 J Q K1 K2 J （J，J）（Q，J）（K1，J）（K2，J） Q （J，Q）（Q，Q）（K1，Q）（K2，Q） K1 （J，K1）（Q，K1）（K1，K1）（K2，K1） K2 （J，K2）（Q，K2）（K1，K2）（K2，K2）

从上表可以看出，一次游戏可能出现的结果共有 16 种，而且每种结果出现的可能性相等，其中两次

取出的牌中都没有 K 的有（J，J）（J，Q）（Q，J）（Q，Q）等 4 种结果．，，， ··········· ··········· ·· 分 ··········· ·········· ··· ·········· ··········· ·· 7

4 1  ． 16 4 1 3  P （甲获胜）  ， P （乙获胜）  ． ························ 9 分 ··········· ·········· ··· ·········· ··········· ··· 4 4 1 3 ··········· ·········· ········ ·········· ··········· ·······   ， 乙获胜的可能性大．·····························10 分 4 4 k 26．解：设 f ，v 之间的关系式为 f  ( k  0) ． ····················· 分 ··········· ········· 1 ·········· ·········· v k  v  50 时， f  80， 80   ． ··········· ··········· ······· 分 ··········· ·········· ········ ·········· ··········· ······· 2 50 解，得 k  4000 ． ······································ 分 ··········· ·········· ··········· ······ ·········· ··········· ··········· ····· 3 4000 所以， f  ． ··········· ··········· ·········· ······ 分 ··········· ·········· ··········· ····· 4 ·········· ··········· ··········· ····· v 4000  40 （度） ··························· 分当 v  100 时， f  ． ··········· ·········· ······ ·········· ··········· ····· 5 100

 P （两次取出的牌中都没有 K） 

答：当车速为 100km/h 时视野为 40 度． ·························· 6 分 ··········· ·········· ····· ·········· ··········· ····· 27．解：（1） (15 1  60  2  65  3  35  4  20  5  5  6)  200

 600  200  3 （个/户） ························· 分．························· 2 ·········· ··········· ····

所以，这天这 200 户家庭平均每户丢弃 3 个塑料袋． ··················· 3 分 ··········· ········ ·········· ········· 100  3  365  109500 （万个） ··························· 分（2）． ··········· ·········· ····· 5 ·········· ··········· ····· 所以，我市所有家庭每年丢弃 109500 万个塑料袋．

···················· 分 ··········· ········· ·········· ········· 6 （3）如图，过点 C 作 CD  AB ，垂足为点 D ． ····················· 7 分 ··········· ·········· ·········· ··········· 在 Rt△BDC 中， BC  110，DBC  60 ，



由 sin 60 



CD  ，得 CD  110sin 60  55 3 ． ····················· 分 ··········· ·········· ·········· ·········· 8 BC

 AB  150 ， 1 1  S△ ABC  AB  150  55 3 ≈ 7000(km 2 ) ．················· 9 分 CD ··········· ······ ·········· ······· 2 2 109500  7000 ≈156000 （个/km2）.

答：我市每年平均每平方公里的土地上会增加 156000 个塑料袋．············ 分 ··········· · ··········· 10 28．解：（1） AFD  DCA （或相等） ························ 2 分． ··········· ·········· ··· ·········· ··········· ··· （2） AFD  DCA （或成立），理由如下： ······················ 3 分 ··········· ·········· · ·········· ··········· · 方法一：由 △ ABC ≌△DEF ，得 AB  DE，BC  EF （或 BF  EC ） A  D C F B EF A， D C  ， B E ． ABC  FBC  DEF  CBF ，ABF  DEC ． ·············· 分 ··········· ·· 4 ·········· ···

 AB  DE，  在 △ ABF 和 △DEC 中， ABF  DEC，  BF  EC， 

△ ABF ≌△DEC，BAF  EDC ． ························· 分 ··········· ·········· ···· ·········· ··········· ··· 5 BAC  BAF  EDF  EDC，FAC  CDF ．  AOD  FAC  AFD  CDF  DCA ， AFD  DCA ． ····································6 分 ··········· ·········· ··········· ···· ·········· ··········· ··········· ····  方法二：连接 AD ．同方法一 △ ABF ≌△DEC， AF  DC ． ············· 分 ··········· · 5 ·········· ·· △ ABC ≌△DEF ，得 FD  CA ．由

 AF  DC，  在 △ AFD ≌△DCA ，  FD  CA，  AD  DA， 

△ AFD ≌△DCA ， AFD  DCA ． ························· 分 ··········· ·········· ··· 6 ·········· ··········· ··· （3）如图， BO  AD ． ··································7

分 ··········· ·········· ··········· ·· ·········· ··········· ··········· ·· 方法一：由 △ ABC ≌△DEF ，点 B 与点 E 重合， A 得 BAC  BDF，BA  BD ．  点 B 在 AD 的垂直平分线上， G 且 BAD  BDA ． ··············· 8 分 ··········· ···· ·········· ····· F O C  OAD  BAD  BAC ， B(E) D ODA  BDA  BDF ， OAD  ODA ．  OA  OD ，点 O 在 AD 的垂直平分线上． ························ 分 ··········· ·········· ·· 9 ·········· ··········· ·· ····················· 10 ·········· ···········  直线 BO 是 AD 的垂直平分线， BO  AD ． ······················ 分方法二：延长 BO 交 AD 于点 G ，同方法一， OA  OD ． ················ 分 ··········· ····· ·········· ····· 8

 AB  DB，  在 △ ABO 和 △DBO 中，  BO  BO， OA  OD， 

△ ABO ≌△DBO，ABO  DBO ． ························· 分 ··········· ·········· ··· 9 ·········· ··········· ···

 AB  DB，  在 △ ABG 和 △DBG 中， ABG  DBG，  BG  BG， 

△ ABG ≌△DBG ， AGB  DGB  90 ． BO  AD ． ············ 分 ··········· 10 ·········· ·

29．解：（1）在 y  x  1 中，当 y  0 时， x  1  0 ，

 x  1 ，点 B 的坐标为 (1，． ····························· 1 分 ·········· ··········· ········ 0) ·····························

在y

3 3 x  3 中，当 y  0 时，  x  3  0， x  4 ，点 C 的坐标为（4，0） ···2 分  ． ··· ··· 4 4

8   y  x  1， x  7 ，   由题意，得  解得  3  y   4 x  3．  y  15 ．   7 

 8 15  ··········· ·········· ··········· · ·········· ··········· ··········· ·  点 A 的坐标为  ，  ． ·································3 分 7 7 

（2）当 △CBD 为等腰三角形时，有以下三种情况，如图（1）．设动点 D 的坐标为 ( x，y ) ． y D2 D3 A D 1 M4 M2 B O M1 C D4 图（1）图（2） x y D2A E2 E1 B O D1 C x

，， 0) 由（1），得 B(1 0) C (4，， BC  5 ．

①当 BD1  D1C 时，过点 D1 作 D1M1  x 轴，垂足为点 M 1 ，则 BM 1  M 1C 

1 BC ． 2

5 5 3 3  BM 1  ，OM 1   1  ，x  ． 2 2 2 2 3 3 15  3 15 

 y     3  ，点 D1 的坐标为  ，  ．···················· 4 分 ··········· ········· ·········· ·········· 4 2 8 2 8 

②当 BC  BD2 时，过点 D2 作 D2 M 2  x 轴，垂足为点 M 2 ，则 D2 M 2  M 2 B  D2 B ．

2 2 2

3  M 2 B   x 1 ， D2 M 2   x  3，D2 B  5 ， 4

 3   ( x  1)2    x  3   52 ．  4 

解，得 x1  

12 3  12  24 ，x2  4 （舍去）．此时， y        3  ． 5 4  5 5

 12 24  ··········· ·········· ·········· ·········· ··········· ·········  点 D2 的坐标为   ，  ． ·······························6 分  5 5 

③当 CD3  BC ，或 CD4  BC 时，同理可得 D3 (0，，D4 (8， 3) ． ·········· 9 分 ·········· ·········· 3)  由此可得点 D 的坐标分别为 D1  ， ，D2  

 3 15  2 8 

 12 24  ， ，D3 (0，，D4 (8， 3) ． 3)   5 5 

评分说明：符合条件的点有 4 个，正确求出 1 个点的坐标得 1 分，2 个点的坐标得 3 分，3 个点的坐标得 5 分，4 个点的坐标得满分；与所求点的顺序无关．（3）存在．以点 E，D，O，A 为顶点的四边形是平行四边形有以下三种情形，如图（2）． ①当四边形 AE1OD1 为平行四边形时，

BE1 3 2 ． ··········· ······· 分 ··········· ······· ················· 10  CD1 20 BE1 2 ． ·················· 11 分 ··········· ······· ·········· ········  CD2 10 BE2 27 2 ． ··········· ······ 分 ··········· ······ ················ 12  CD1 20

②当四边形 AD2 E1O 为平行四边形时，

③当四边形 AOD1E2 为平行四边形时，

评分说明：1．如你的正确解法与上述提供的参考答案不同时，可参照评分说明进行估分． 2．如解答题由多个问题组成，前一问解答有误或未答，对后面问题的解答没有影响．可依据参考答案及评分说明进行估分．