发布时间 : 星期六 文章2008年湖南省益阳市中考数学试卷及答案更新完毕开始阅读
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