2008年湖南省益阳市中考数学试卷及答案

Ⅱa. 小聪想：要画出正方形DEFG，只要能计算出正方形的边长就能求出BD和CE的长，从而

确定D点和E点，再画正方形DEFG就容易了.

设△ABC的边长为2 ，请你帮小聪求出正方形的边长(结果用含根号的式子表示，不要求分母有理化) .

A G F B

D

图10(2)

E

C Ⅱb. 小明想：不求正方形的边长也能画出正方形. 具体作法是： ①在AB边上任取一点G’，如图作正方形G’D’E’F’；

②连结BF’并延长交AC于F； ③作FE∥F’E’交BC于E，FG∥F′G′交AB于G，GD∥G’D’交BC于D，则四边形DEFG即为所求.

你认为小明的作法正确吗？说明理由. A

G F

F′ G′ B C E′ D E D′

图10(3)

六、(本题10分)

23. 两个全等的直角三角形ABC和DEF重叠在一起，其中∠A=60°，AC=1. 固定△ABC不动，将△DEF进行如下操作：

(1) 如图11(1)，△DEF沿线段AB向右平移(即D点在线段AB内移动)，连结DC、CF、FB，四边形CDBF的形状在不断的变化，但它的面积不变化，请求出其面积.

C F

A B E D 图11(1)

(2)如图11(2)，当D点移到AB的中点时，请你猜想四边形CDBF的形状，并说明理由.

C F

A D B E 图11(2)

(3)如图11(3)，△DEF的D点固定在AB的中点，然后绕D点按顺时针方向旋转△DEF，使DF落在AB边上，此时F点恰好与B点重合，连结AE，请你求出sinα的值. (F) C

A D (F) B (E)

七、(本题12分)

24.我们把一个半圆与抛物线的一部分合成的封闭图形称为“蛋圆”，如果一条直线与“蛋圆”只有一个交点，那么这条直线叫做“蛋圆”的切线.

如图12，点A、B、C、D分别是“蛋圆”与坐标轴的交点，已知点D的坐标为(0，-3)，AB为半圆的直径，半圆圆心M的坐标为(1,0)，半圆半径为2.

(1) 请你求出“蛋圆”抛物线部分的解析式，并写出自变量的取值范围； (2)你能求出经过点C的“蛋圆”切线的解析式吗？试试看；

(3)开动脑筋想一想，相信你能求出经过点D的“蛋圆”切线的解析式.

y

C

A B x

M O

D

图12

2008年湖南省益阳市中考数学试卷

参考答案及评分意见

一、选择题（本题共10个小题，每小题3分，共30分）

题号 1 2 3 4 5 6

答案 D A D B C A

二、填空题(本题共6个小题，每个小题4分，满分24分)

题号

11

4

7 D 8 D 9 B 10 C 12 6

13 14 15 3 516

答案不惟一如：x?2 2xx?2x,x2?4答案 9.1×10

x2?2x108° （2,4）

xx2?4x?2x2?4x?4x?2x2?2xxx2?4x?4x?216题还有如下答案：2；2；2；2；2.,,,,,x?2x?2x?2x?2xx?4x?4x?4x?4x?4x?4x?2x（每空2 分）

三、解答题(本题共3个小题，每个小题6分，满分18分)

17.解：原式＝2＋1－9＋1 ·············································································· 4分

＝－5 ······················································································· 6分

18.解：（1）∵DE∥BC，

∴∠EDB＝∠DBC＝?ABC?40? ··················································· 3分

（2）∵AB＝BC， BD是∠ABC的平分线，∴D为AC的中点 ∵DE∥BC，∴E为AB的中点，

∴DE＝

19.解：（1）

1AB?6cm ········································································ 6分 2121··············································· 3分 (20?9?30?12?50?16?100?3)?41 ·

40(2) 41×1200=49200(元)

答：这40 名同学捐款的平均数为41元，这个中学的捐款总数大约是49200元 ······ 6分

四、解答题(本题共2个小题,每小题8分,共16分)

20.解：设原计划每天挖土石方x万立方米，增调人员和设备后每天挖y万立方米 ······· 1分

可列出方程组：??y?2x?1 ·························································· 5分

2x?(5?2)y?13.4? 解之得：??x?1.3

?y?3.6 答：原计划每天挖土石方1.3万立方米，增调人员和设备后每天挖3.6万立方米 ···· 8分 21.解：(1) 根据题意可知：y=4+1.5(x-2) ， ∴ y=1.5x+1(x≥2) ····················································· 4分 (2)依题意得：7.5≤1.5x+1＜8.5 ··························································· 6分 ∴

13≤x＜5 ································································· 8分 3

五、(本题10分)

22．Ⅰ.证明：∵DEFG为正方形，

∴GD=FE，∠GDB=∠FEC=90° ···················································· 2分

∵△ABC是等边三角形，∴∠B=∠C=60° ······································· 3分 ∴△BDG≌△CEF(AAS) ······························································ 5分 Ⅱa.解法一：设正方形的边长为x，作△ABC的高AH，

A 求得AH?3 ······································· 7分

x3?x G F 由△AGF∽△ABC得：? ············ 9分

23B

D

E

解图10(2)

H

C 解之得：x?232?3(或x?43?6) ······· 10分

解法二：设正方形的边长为x，则BD? 在Rt△BDG中，tan∠B=

∴

2?x ········································· 7分 2GD， BDx?3 ································································ 9分 2?x2232?3解之得：x?(或x?43?6) ··································· 10分

解法三：设正方形的边长为x，

则BD?2?x,GB?2?x ····················································· 7分 22?x2) ·································· 9分 2 由勾股定理得：(2?x)2?x2?( 解之得：x?43?6 ························································· 10分 Ⅱb.解： 正确 ······························································································· 6分 由已知可知，四边形GDEF为矩形 ······················································· 7分

∵FE∥F’E’ ，

A G G’ F’ F