2007年湖南省永州市中考数学真题及答案.doc

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2007 年湖南省永州市中考数学真题及答案 I 卷考生注意：1、本试卷共十道大题，其中正卷八大题，满分 100 分；另附加题 2 道，20 分，合计 120 分，时量 120 分钟． 2、本试卷分 I 卷和Ⅱ卷，I 卷为选择填空题 1—2 页；Ⅱ卷为解答题 3—8 页． 3、考生务必将 I 卷的答案写在Ⅱ卷卷首的答案栏内，交卷时只交Ⅱ卷．一、填空题(每小题 3 分，共 8 个小题，24 分．请将答案填在Ⅱ卷卷首的答案栏内．) 1． 3 0.001  ________． 2．因式分解：a3－a＝_______． 3．观察下列图形，根据变化规律推测第 100 个与第_______个图形位置相同． 4．如图，已知△ABC中，∠A＝40°，剪去∠A后成四边形，则∠1＋∠2＝_______． 5．图形：①线段，②等边三角形，③平行四边形，④矩形，⑤梯形，⑥圆．其中既是轴对称图形又是中心对称图形的序号是_______． 6．如图，添上条件：_______，则△ABC∽△ADE． 7．夏雪同学每次数学测试成绩都是优秀，则在这次中考中他的数学成绩_______(填“可能”，“不可能”，“必然”)是优秀． 8．如图，要把线段 AB平移，使得点 A到达点 (4 2) A ，，点 B到达点 B'，那么点 B'的坐标是_______．二、选择题(每小题 3 分，共 8 个小题，24 分．每小题只有一个正确选项，请将正确选项的代号填入Ⅱ卷卷首的答案栏内．) 1 x  1 2 9．函数 y  A． x  1 2 的自变量的取值范围是( ) B． x  1 2 C． x  1 2 D． x  的全体实数 1 2 10．2006 年 9 月在长沙市举行的“中国中部投资贸易博览会”中，永州市的外贸成交总额达 31264 万元人民币，用科学记数法(保留三个有效数字)表示这个数据(单位：万元)，正确的是( ) A．3.12×104 B．3.13×104 C．31.2×103 D．31.3×103 11．下列命题是假命题的是( ) A．四个角相等的四边形是矩形 B．对角线互相平分的四边形是平行四边形

C．四条边相等的四边形是菱形 D．对角线互相垂直且相等的四边形是正方形 12．下列运算中，正确的是( ) A．x2007＋x2008＝x4015 B．20070＝0 C．  22   3      4 9 D． (  a ) ( ·  a ) 2   a 3 13．如图所示， AB CD∥ ，∠E＝27°，∠C＝52°，则 EAB 的度数为( ) A．25° B．63° C．79° D．101° 14．用三个正方体，一个圆柱体，一个圆锥的积木摆成如图※所示的几何体，其正视图为( ) 15．在一周内体育老师对某运动员进行了 5 次百米短跑测试，若想了解该运动员的成绩是否稳定，老师需要知道他 5 次成绩的( ) A．平均数 B．方差 C．中位数 D．众数 16．永州市内货摩(运货的摩托)的运输价格为：2 千米内运费 5 元；路程超过 2 千米的，每超过 1 千米增加运费 1 元，那么运费 y元与运输路程 x千米的函数图象是( ) 三、解答题（本题 2 个小题，每小题 6 分，共 12 分） 17、计算： 1  2    1   1 2007 0     sin 30 2   °·   1 2     18 ．

18、解不等式组： 5    6   2( x  ( ≤ 1 6 19) 9  3 x  x  5) ，并在数轴上表示不等式的解集． x  5[ x  2( x  3)] 四、作图题：(本题 6 分，不写作法，保留作图痕迹) 19．近年来，国家实施“村村通”工程和农村医疗卫生改革，某县计划在张村、李村之间建一座定点医疗站 P ，张、李两村座落在两相交公路内(如图所示)．医疗站必须满足下列条件：①使其到两公路距离相等， ②到张、李两村的距离也相等，请你通过作图确定 P 点的位置．五．(本题共 8 分) 20．某校对初中三年级同学的视力进行了调查，如图是根据调查结果绘制的条形统计图．请根据统计图回答下列问题： (1)求视力在 1.2—1.5 的人数． (2)求视力在 0.9 以下的人数所占的比例． (3)根据统计图显示的信息，用一句话发表你的感想．六．(本题共 8 分) 21．已知一次函数与反比例函数的图象都经过 ( 2  ，和 ( 2)n，两点． 1) （1）求这两个函数的解析式．

（2）画出这两个函数的图象草图．七、应用题：(本题共 8 分) 22．为净化空气，美化环境，我市冷水滩区在许多街道和居民小区都种上了玉兰和樟树，冷水滩区新建的某住宅区内，计划投资 1.8 万元种玉兰树和樟树共 80 棵，已知某苗甫负责种活以上两种树苗的价格分别为：玉兰树 300 元／棵，樟树 200 元／棵，问可种玉兰树和樟树各多少棵? 八、综合题(本题共 10 分) 23． AB 是 O 的直径， D 是 O 上一动点，延长 AD 到C 使CD AD (1)证明：当 D 点与 A 点不重合时，总有 AB BC (2)设 O 的半径为 2， AD x ， BD y ，用含 x的式子表示 y． (3) BC 与 O 是否有可能相切?若不可能相切，则说明理由；若能相切，则指出 x为何值时相切．，连结 BC BD，．．附加题：(本题 2 个小题，每小题 10 分，共 20 分) 九．24．如图所示是永州八景之一的愚溪桥，桥身横跨愚溪，面临潇水，桥下冬暖夏凉，常有渔船停泊桥下避晒纳凉．已知主桥拱为抛物线型，在正常水位下测得主拱宽 24m，最高点离水面 8m，以水平线 AB 为 x 轴， AB 的中点为原点建立坐标系． ①求此桥拱线所在抛物线的解析式． ②桥边有一浮在水面部分高 4m，最宽处 12 2m 的河鱼餐船，试探索此船能否开到桥下?说明理由．

ABC 25、在梯形 ABCD 中， AB CD∥ ，（1）求 DC 的长；（2） E 为梯形内一点， F 为梯形外一点，若 BF DE 说明理由．（3）在（2）的条件下，若 BE EC BE EC  ， :  °， 90 AB  ， 5 BC  ， tan 10 ADC  ． 2 ， FBC    CDE ，试判断 ECF△ 的形状，并，求 DE 的长． 4 :3

湖南永州市 2007 初中毕业学业考试试卷数学参考答案及评分标准一、填空题（每小题 3 分） 1．0.1 2． ( a a  1)( a  1) 3．1（或 4） 4． 220 5．①④⑥ 6． BC DE∥ 或 ABC    ADE 或 AB AC AD AE  等 7．可能 8． (7 4)，二、填空题（每小题 3 分） 9．D 10．B 三、解答题： 11．D 12．D 13．C 14．A 15．B 16．B 17．解：原式  2 1 1      4 3 2 1 2 ······························································ 4 分  2 3 2 1 1 2       2 2 ·································································································· 6 分 x ≥ ···················································································2 分 x  ··················································································· 4 分 0 解不等式②得 4 18．解不等式①得 0 ···································································· 6 分 19．画出角平分线····························································································3 分原不等式组的解集为0 x ≤ 4 4 作出垂直平分线·······································3 分张 P 李 20．解：①25 分······························································································ 3 分  100% 47.5%  ············································································· 6 分 ② 95 200 ③只要与主题有关就给分············································································ 8 分 21．解①设反比例函数为 my  ， x 则 m      ·················································································· 2 分 2 ( 1) 2 反比例函数的解析式为 y  ··································································· 3 分 2 x 1n  ② ( 2)n ，在反比例函数上，设一次函数为 y  kx b  因为图象经过 ( 2  ，，，两点 1) (1 2) y 2 1 1 2 x 2 O1 1 2

········································· 5 分    2 k b      2 k b  1 k    b 1 1 一次函数为 y x  ·················································································· 6 分 1 ②如图：··································································································8 分 22．解：设种玉兰树 x 棵，樟树 y 棵，则····························································· 1 分 x   300     y x 80 200 y  18000 ··············································································· 5 分解之得： x    y 20 60 ······················································································7 分答：可种玉兰树 20 棵，樟树 60 棵．····························································· 8 分  ⊥ ··························································· 1 分 23．（1） AB 为 O 直径， BD AC  又 DC AD BD 是 AC 的垂直平分线   AB AC ····························································································· 3 分（2）在 Rt ABD△ 中， 2 BD  2 AB  2 AD ···················································· 5 分 2   y 2 4  2 x ·························································································· 6 分 y 2 x 16   ························································································ 7 分即（3） BC 与 O 有可能相切········································································ 8 分当 BC 与 O 相切时， BC 45 AB BC  ··········································································9 分 A  AB⊥ ，   x 2 2 AB  2 2 ················································································10 分 24．解：（1） ( 12 0)  ，，，，，．设抛物线为 (12 0) (0 8) C A B y  2 ax  bx  c C 点坐标代入得： 8c  ··············································································2 分 A B，点坐标代入得： 144   144  a a   12 b 12 b 8 0   8 0   解得 1 18    a     0 b ，所求抛物线为 y   21 x 18 ·····················································4 分  ··············································· 6 分 8

（2）当 4 y  时得 2 x  ， 18 4 x   6 2 ······················································ 8 分高出水面 4m 处，拱宽12 2 所以此船在正常水位时不可以开到桥下·························································10 分 25．解（1）过 A 点作 AG DC⊥ ，垂足为 G ······················································· 1 分（船宽） 12 2 m m   10 ································································· 2 分    10 CDE BC DC ······························································· 4 分  FBC  ·················································································· 5 分   ····································································6 分，  FCB ABC   90       AB CD BCD  ∥ ， 四边形 ABCG 为矩形   2 5 CG AB AG BC ， AG tan ADG   DG 5 DG DC DG CG     ，（2） DE BF   ， BFC DEC △ ≌△ ECD    ， EC CF  ECF   90     ， 90   ECD BCE ECF  △ （3）过 F 点作 FH BE⊥  ⊥ ， ⊥ ， 四边形 ECFH 是正方形， EC CF CE CE CF  BE  是等腰直角三角形········································································ 7 分 FH EC  ················································· 8 分 6  BE EC :  4 :3 ，  BEC   90 2 2 2  BE BC EC  8 6 EC ， ··················································································· 9 分 2 BH BE EH   BE       DE BF   2 FH  BH 2  2 10 ··························································10 分

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2007年湖南省永州市中考数学真题及答案.doc

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