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2008年湖南省长沙市中考数学真题及答案.doc

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2008 年湖南省长沙市中考数学真题及答案 考生注意:本试卷共 26 道小题,时量 120 分钟,满分 120 分. 三 四 五 一 二 17-1 20-2 9 2 23 24 25 26 总分 合分人 复分人 题 次 得 分 得 分 评卷人 复评人 一、填空题(本题共 8 个小题,每小题 3 分,满分 24 分) 1、-8 的绝对值是 . 2、函数 y= 2x 中的自变量 x 的取值范围是 . 3、△ABC 中,∠A=55,∠B=25,则∠C= . 4、方程 2 x 11 的解为 x = . 5、如 图 , P 为 菱 形 ABCD 的 对 角 线 上 一 点 , PE⊥ AB 于 点 E, PF⊥ AD 于 点 F, PF=3cm, 则 P 点到 AB 的距离是 A cm. D P B F E C D A B C (第 5 题) (第 6 题) 6、如图,在 Rt△ABC 中,∠C=90,AB=10cm,D 为 AB 的中点,则 CD= 7、已知 a、b为两个连续整数,且 a< 7 <b,则 b 8、在一次捐款活动中,某班 50 名同学人人拿出自己的零花钱,有捐 5 元、10 元、20 元的,还有捐 50 元 cm. a  = . 和 100 元的。右边的统计图反映了不同捐款数的人数比例,那么该班同学平均每人捐款 元. 100 元 12% 50 元 16% 5 元 8% 10 元 20% 20 元 44% (第 8 题) 得 分 评卷人 复评人 二、选择题(本题共 8 个小题,每小题 3 分,满分 24 分) 请将你认为正确的选项的代号填在下面的表格里:
题 号 答 案 9 10 11 12 13 14 15 16 9、下面计算正确的是( A、 2 1  2 B、 ) 4  2 C、( 3nm  )2= 6nm  10、要反映长沙市一周内每天的最高气温的变化情况,宜采用( D、 6  mm ) 2  4 m A、条形统计图 B、扇形统计图 C、折线统计图 D、频数分布直方图 11、若点 P( a , a4 )是第二象限的点,则 a 必须满足( ) A、 a <4 B、 a >4 C、 a <0 D、0< a <4 12、如图是每个面上都有一个汉字的正方体的一种展开图,那么在正 “迎”相对的面上的汉字是( ) 方 体 的 表 面 , 与 讲 文 明 迎 奥 A、文 B、明 C、奥 D、运 运 13、在同一平面直角坐标系中,函数 y A、0 个 B、1 个 x C、2 个 D、3 个 1 与函数 x y  的图象交点 (第 12 题) 个数是( ) 14、在同一时刻,身高 1.6 米的小强在阳光下的影长为 0.8 米,一棵大树的影长为 4.8 米,则树的高度为 ( ) A、4.8 米 B、6.4 米 C、9.6 米 D、10 米 15、如图,P 为⊙O 外一点,PA 切⊙O 于点 A,且 OP=5,PA=4,则 sin∠APO 等于( ) 4 A、 5 P 3 B、 5 A · O 4 C、 3 3 D、 4 . . (第 16 题) 16、二次函数 A、 a <0 (第 15 题) ax  c 2   y bx B、 abc >0 的图象如图所示,则下列关系式不正确的是( ) C、 cba  >0 D、 b 2  >0 4 ac 得 分 评卷人 复评人 三、解答题(本题共 6 个小题,每小题 6 分,满分 36 分) 17、计算: 273  sin2 30  ( 1 0) 15  .
18、先化简,再求值: 2 1 a  4 2 2 a a 1a ,其中 2 . 19、在下面的格点图中,每个小正方形的边长均为 1 个单位,请按下列要求画出图形: (1)画出图①中阴影部分关于 O 点的中心对称图形; (2)画出图②中阴影部分向右平移 9 个单位后的图形; (3)画出图③中阴影部分关于直线 AB 的轴对称图形. (图①) (图②) (图③) 20、解不等式组: 1   01  2   34 14 x  x   x ,并将其解集在数轴上表示出来. -6 -5 -4 -3 -2 -1 0 1 2 3 21、当 m 为何值时,关于 x 的一元二次方程 x 42  mx   1 2 0 多少? 有两个相等的实数根?此时这两个实数根是
22、某商场开展购物抽奖活动,抽奖箱中有 4 个标号分别为 1、2、3、4 的质地、大小相同的小球,顾客任 意摸取一个小球,然后放回,再摸取一个小球,若两次摸出的数字之和为“8”是一等奖,数字之和为 “6”是二等奖,数字之和为其它数字则是三等奖,请分别求出顾客抽中一、二、三等奖的概率. 得 分 评卷人 复评人 四、解答题(本题共 2 个小题,每小题 8 分,满分 16 分) 23、(本题满分 8 分) “5·12”汶川大地震后,灾区急需大量帐篷。某服装厂原有 4 条成衣生产线和 5 条童装生产线,工厂 决定转产,计划用 3 天时间赶制 1000 顶帐篷支援灾区。若启用 1 条成衣生产线和 2 条童装生产线,一天可 以生产帐篷 105 顶;若启用 2 条成衣生产线和 3 条童装生产线,一天可以生产帐篷 178 顶. (1)每条成衣生产线和童装生产线平均每天生产帐篷各多少顶? (2)工厂满负荷全面转产,是否可以如期完成任务?如果你是厂长,你会怎样体现你的社会责任感? 得 分 评卷人 复评人 24、(本题满分 8 分) 如图,在□ABCD 中,BC=2AB=4,点 E、F 分别是 BC、AD 的中点. (1)求证:△ABE≌△CDF; (2)当四边形 AECF 为菱形时,求出该菱形的面积. A F D B E C 得 分 评卷人 复评人 五、解答题(本题共 2 个小题,每小题 10 分,满分 20 分) 25、(本题满分 10 分) 在平面直角坐标系中,一动点 P( x ,y)从 M(1,0)出发,沿由 A(-1,1),B(-1,-1),C(1,-1),
D(1,1)四点组成的正方形边线(如图①)按一定方向运动。图②是 P 点运动的路程 s(个单位)与运动 时间 t (秒)之间的函数图象,图③是 P 点的纵坐标 y与 P 点运动的路程 s之间的函数图象的一部分. ·P (图①) (图②) (1)s与 t 之间的函数关系式是: (2)与图③相对应的 P 点的运动路径是: 首次到达点 B; (图③) ; ;P 点出发 秒 (3)写出当 3≤s≤8 时,y与 s之间的函数关系式,并在图③中补全函数图象. 得 分 评卷人 复评人 26、(本题满分 10 分) 如图,六边形 ABCDEF 内接于半径为 r(常数)的⊙O,其中 AD 为直径,且 AB=CD=DE=FA. (1)当∠BAD=75时,求BC⌒的长; (2)求证:BC∥AD∥FE; (3)设 AB= x ,求六边形 ABCDEF 的周长 L 关于 x 的函数关系式,并指出 x 为何值时,L 取得最大值. B A O· 2008 年长沙市初中毕业学业考试试卷 数学参考答案及评分标准 F C E D 一、填空题 1、8 5、3 二、选择题 2、x≥2 6、5 3、100 7、5 4、3 8、31.2 题号 答案 9 D 10 C 11 C 12 A 13 A 14 C 15 B 16 C 三、解答题
17.原式=3+2× 2 1 -1·················································································· (3 分) =3+1-1·························································································(4 分) =3·······························································································(6 分) 2  =  a 2 a  2 a 1   2 a  2 a   2 a   ·····································································(2 分) 18.原式=  a   2 2 a a     2 2 a a   1 ······························································································ (4 分) 2 a 2 .·········································································(6 分) 1a 时,原式= 5 当 2 ········································································ (3 分) 2 = = 19.图略(“2008”字样),三部分图形各 2 分,共 6 分. 20.由  1  x  2    4 3  1 ≤ 0 x  14  x 得 x x  2   5 ,······························································ (4 分) 21.由题意,△=(-4)2-4(m- 2 即 16-4m+2=0,m= 2 不等式组的解集为-5<x≤2.································································ (5 分) 解集在数轴上表示略.········································································· (6 分) 1 )=0·································································· (2 分) 9 .················································································ (4 分) 9 时,方程有两个相等的实数根 x1=x2=2.········································· (6 分) 1 ,·····································································(2 分) 3 ,·····································································(4 分) 3 .······································································ (6 分) 22.抽中一等奖的概率为 16 抽中二等奖的概率为 16 抽中三等奖的概率为 4 当 m= 2 105  178  四、解答题 23.(1)设每条成衣生产线和童装生产线平均每天生产帐篷各 x、y 顶,则············ (1 分) 2 y x    2 3 y x   解得 x=41,y=32. 答:每条成衣生产线平均每天生产帐篷 41 顶,每条童装生产线平均每天生产帐篷 32 顶. ·············································································································· (5 分) ,······················································································· (3 分) (2)由 3(4×41+5×32)=972<1000 知,即使工厂满负荷全面转产,还不能如期完成任务. ········································································································(7 分) 可以从加班生产、改进技术等方面进一步挖掘生产潜力,或者动员其它厂家支援等,想法尽早完成生 产任务,为灾区人民多做贡献.··································································· (8 分) 24.(1)证明略;······················································································· (4 分) (2)当四边形 AECF 为菱形时,△ABE 为等边三角形,······································· (6 分) 四边形 ABCD 的高为 3 ,············································································ (7 分) ∴菱形 AECF 的面积为 2 3 .·······································································(8 分) 五、解答题 25.(1)S= t2 1 (t≥0)·················································································· (2 分)
(2)M→D→A→N,····················································································· (4 分) 10···········································································································(5 分) (3)当 3≤s<5,即 P 从 A 到 B 时,y=4-s;··················································· (6 分) 当 5≤s<7,即 P 从 B 到 C 时,y=-1;··························································· (7 分) 当 7≤s≤8,即 P 从 C 到 M 时,y=s-8.························································· (8 分) 补全图象略.·························································································· (10 分) 26.(1)连结 OB、OC,由∠BAD=75,OA=OB 知∠AOB=30,································· (1 分) ∵AB=CD,∴∠COD=∠AOB=30,∴∠BOC=120,···············································(2 分) r2 .····················································································· (3 分) 故BC⌒的长为 3 (2)连结 BD,∵AB=CD,∴∠ADB=∠CBD,∴BC∥AD,···································(5 分) 同理 EF∥AD,从而 BC∥AD∥FE.····························································(6 分) (3)过点 B 作 BM⊥AD 于 M,由(2)知四边形 ABCD 为等腰梯形,从而 BC=AD-2AM=2r-2AM. ∵AD 为直径,∴∠ABD=90,易得△BAM∽△DAB AB2 x2 ······································· (8 分) x x2 ,同理 EF=2r- r ,∴BC=2r- r = r ∴AM= AD 2 x2  2  4 r   )= ∴L=4x+2(2r- r ,其中 0<x< r2 ······· (9 分) 2 2 xr 6 r  2 ∴当 x=r 时,L 取得最大值 6r. x =  4 r 2  xr (10 分)  (7 分)
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