发布时间 : 星期三 文章2012湖北省咸宁市数学中考试题 2更新完毕开始阅读3ef1983a10661ed9ad51f3e9
24．（本题满分12分）
如图，在平面直角坐标系中，点C的坐标为（0，4），动点A以每秒1个单位长的速度，从点O出发沿x轴的正方向运动，M是线段AC的中点．将线段AM以点A为中心，沿顺时针方向旋转90?，得到线段AB．过点B作x轴的垂线，垂足为E，过点C作y轴的垂线，交直线BE于点D．运动时间为t秒． （1）当点B与点D重合时，求t的值； y y
（2）设△BCD的面积为S，当t为何值
C D C M O A B E x O 备用图
x 25时，S??
4
（3）连接MB，当MB∥OA时，如果抛
物线y?ax2?10ax的顶点在△ABM
（第24题）
内部（不包括边），求a的取值范围．
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湖北省咸宁市2012年初中毕业生学业考试
数学试题参考答案及评分说明
说明：
1．如果考生的解答正确，思路与本参考答案不同，可参照本评分说明制定相应的评分细则评分．
2．每题都要评阅完毕，不要因为考生的解答中出现错误而中断对该题的评阅．当考生的解答在某一步出现
错误，影响了后继部分时，如果该步以后的解答未改变这道题的内容和难度，则可视影响的程度决定后面部分的给分，但不得超过后面部分应给分数的一半；如果这一步以后的解答有较严重的错误，就不给分．
3．为阅卷方便，解答题的解题步骤写得较为详细，但允许考生在解答过程中，合理地省略非关键性的步骤． 4．解答右端所注分数，表示考生正确做到这一步应得的累加分数． 5．每题评分时只给整数分数．
一．精心选一选（每小题3分，本大题满分24分）
题号 答案
1 B 2 D 3 C 4 D 5 B 6 C
7 A
8 A
二．细心填一填（每小题3分，本大题满分24分） 9．a(a?2) 10．x?3 11．360 12．210 13．1100 14．140 15．28 16．①④（多填、少填或错填均不给分） 三．专心解一解（本大题满分72分） 17．解：原式?3?22?4?32 ··············································································· 4分
?2?1． ·············································································································· 6分
（说明：第一步中写对3?22得1分，写对?4得2分，写对?32得1分，共4分）
x8?1?18．解：原方程即：． ························································· 1分 x?2(x?2)(x?2)方程两边同时乘以(x?2)(x?2)，得 x(x?2)?(x?2)(x?2)?8． ··············································································· 4分
化简，得 2x?4?8． 解得 x?2． ········································································································ 7分 检验：x?2时(x?2)(x?2)?0，x?2不是原分式方程的解，原分式方程无解．
·········································································· 8分
19．解：（1）∵点A（1，6），B（a，2）在y2?∴
m的图象上， xm······························································································· 1分 ?6，m?6． ·
1mm·························································································· 2分 ?2，a??3． ·
a2∵点A（1，6），B（3，2）在函数y1?kx?b的图象上，
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?k?b?6,∴? ······································································································· 4分 ?3k?b?2.?k??2,解这个方程组，得?
b?8.?∴一次函数的解析式为y1??2x?8，反比例函数的解析式为y2?6． ··········· 6分 x（2）1≤x≤3． ···································································································· 8分 20．解：不赞成小蒙同学的观点． ··············································································· 1分
记七、八年级两名同学为A，B，九年级两名同学为C，D． 画树形图分析如下：
D 第一名： C B A
第二名： B C D A C D A B D A B C 第三名： CD BD BC CD AD AC BD AD AB BC AC AB
·········································································· 5分
由上图可知所有的结果有12种，它们出现的可能性相等，满足前两名是九年级同学的结果有2种，所以前两名是九年级同学的概率为21·························································· 9分 ?． ·12621．（1）解：∵BF与⊙O相切， A ∴BF?AB． ·································································· 1分
而BF∥CD，∴CD?AB． 又∵AB是直径，∴CE?ED． ······································ 2分 O 连接CO，设OE?x，则BE?9?x． E C D F B （第21题） 由勾股定理可知：CO2?OE2?BC2?BE2?CE2，
即92?x2?62?(9?x)2，x?7． ································· 4分 因此CD?2CO2?OE2?292?72?82． ············· 5分
（2）∵四边形BDCF为平行四边形， ∴BF?CD． 而CE?ED?11··························································· 7分 CD， ∴CE?BF．·22∵BF∥CD， ∴△AEC∽△ABF． ···································································· 8分
AEEC1···················································· 9分 ??． ∴点E是AB的中点． ·
ABBF21.622．（1）解法一：由图2可知甲步行的速度为···························· 1分 ?2（km/h） ·
0.8∴
因此甲在每个景点逗留的时间为
1.8?0.8?2.6?1.6············································································· 3分 ?0.5（h） ·
2解法二：甲沿A→D步行时s与t的函数关系式为s?2t． ································ 1分 设甲沿D→C步行时s与t的函数关系式为s?2t?b． 则2?1.8?b?2.6． ∴b??1．
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∴s?2t?1． ········································································································· 2分 当s?1.6时，2t?1?1.6，t?1.3．
因此甲在每个景点逗留的时间为1.3?0.8?0.5（h）． ······································· 3分 补全图象如下： ······································································································ 5分
s/(km) 4 3 2.6 2 1.6 1 O 0.8 1.8 2.3 （第22题）
（2）解法一：甲步行的总时间为3?0.5?2?2（h）． ∴甲的总行程为2?2?4（km）． ·············· 7分 ∴C，E两点间的路程为4?1.6?1?0.8?0.6（km）．
·············································· 8分
解法二：设甲沿C→E→A步行时 s与t的函数关系式为s?2t?m．
3 t/(h) 则2?2.3?m?2.6．
∴m??2．
∴s?2t?2． ········································································································· 6分 当t?3时，s?2?3?2?4． ··············································································· 7分 ∴C，E两点间的路程为4?1.6?1?0.8?0.6（km）．······································· 8分 （3）他们的约定能实现． 乙游览的最短线路为：A→D→C→E→B→E→A（或A→E→B→E→C→D→A），总行程为1.6?1?0.6?0.4?2?0.8?4.8（km）． ······························································ 9分
∴乙游完三个景点后回到A处的总时间为4.8?0.5?3?3.1（h）． 3∴乙比甲晚6分钟到A处． ················································································ 10分
（说明：图象的第四段由第二段平移得到，第五段与第一、三段平行，且右端点的横坐标为3，如果学生补全的图象可看出这些，但未标出2.3也可得2分．第3问学生只说能实现约定，但未说理由不给分．） 23．（1）作图如下： ···································································································· 2分
A H B G D F F E 图2
C B
图3
E
C A H G D （2）解：在图2中，EF?FG?GH?HE?22?42?20?25，
∴四边形EFGH的周长为85． ··········································································· 3分 在图3中，EF?GH?22?12?5，FG?HE?32?62?45?35． ∴四边形EFGH的周长为2?5?2?35?85． ············································ 4分 猜想：矩形ABCD的反射四边形的周长为定值． ················································ 5分 （3）证法一：延长GH交CB的延长线于点N．
∵?1??2，?1??5， G D A ∴?2??5． 1 F 3 2 5 H 21世纪教育网4 -- 中国最大型、最专业的中小学教育资源门户网站。 版权所有@21世纪教育网 M K C B N E
图4