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
8
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 2C , 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
2
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.