2008年湖北省襄阳市初中毕业、升学统一考试数学试题
一、选择题：本大题共10小题，每小题3分，共30分．在每小题给出的四个选项中，只有一项是符合题目要求的，请将其序号在答题卡上涂黑作答． 1．2的相反数是（ ） A．2
B．?2
C．
12 D．?12
2．下列运算正确的是（ ） A．x?x?x C．2a?3a??a
3412B．(?6x)?(?2x)?3x D．(x?2)?x?4
226233．如图1，已知AD与BC相交于点O，AB∥CD，如果
?B?40，?D?30，则?AOC的大小为（ ）
??A．60
?B．70
?C．80
?D．120
?4．下列说法正确的是（ ） A．4的平方根是2 C．8是无理数
3
2) ?3)向右平移5个单位长度到点(?2，B．将点(?2，3) D．点(?2，?3)关于x轴的对称点是(?2，5．在正方形网格中，△ABC的位置如图2所示，则cos?B的值为（ ） A．
12 B．
22 C．32 D．33
6．某种商品零售价经过两次降价后的价格为降价前的81％，
则平均每次降价（ ） A．10％ B．19％ C．9.5％ D．20％ 7．顺次连接等腰梯形四边中点所得四边形是（ ） A．菱形 B．正方形 C．矩形 D．等腰梯形
8．在一个可以改变体积的密闭容器内装有一定质量的二氧化碳，当改变容器的体积时，气体的密度也会随之改变，密度?（单位：kg/m3）是体积V（单位：m3）的反比例函数，它的图象如图3所示，当V?10m3时，气体的密度是（ ） A．5kg/m3 B．2kg/m3 C．100kg/m3 D，1kg/m3
9．如图4，是一个由若干个相同的小正方体组成的几何体的三视图，则组成这个几何体的小正方体的个数是（ ）
A．7个
B．8个
C．9个
D．10个
?10．如图5，扇形纸扇完全打开后，外侧两竹条AB，AC夹角为120，AB的长为30cm，贴纸部分BD的长为20cm，则贴纸部分的面积为（ ） A．100?cm C．800?cm
22
B．D．
40038003?cm ?cm
22二、填空题：本大题共6小题，每小题3分，共18分．把答案填在答题卡的相应位置上． 11．一方有难，八方支援．截至6月3日12时，中国因汶川大地震共接受国内外捐赠款物423.64亿元，用科学记数法表示为 元．
?12．如图6，?O中OA?BC，?CDA?25，则?AOB的度数
为 ．
13．当m? 时，关于x的分式方程
??1无解． x?314．如图7，一名男生推铅球，铅球行进高度y（单位：m）与水平距离x（单位：m）之间的关系是y??112x?22x?m23x?53．则他将铅球推出的距离是 m．
?15．如图8，张华同学在学校某建筑物的C点处测得旗杆顶部A点的仰角为30，旗杆底部B点的俯角为45．若旗杆底部B点到建筑物的水平距离BE?9米，旗杆台阶高1米，则旗杆顶点A离地面的高度为 米（结果保留根号）．
16．如图9，在锐角?AOB内部，画1条射线，可得3个锐角；画2条不同射线，可得6个锐角；画3条不同射线，可得10个锐角；??照此规律，画10条不同射线，可得锐角 个．
?
三、解答题：本大题共9小题，共72分．解答应写出文字说明，证明过程或演算步骤，并且写在答题卡上每题对应的答题区域内． 17．（本小题满分7分）
?x?16x?1??化简求值：?2，其中x??2x?8x?16x?4x?16??22?1．
18．（本小题满分6分）
为了了解学生课业负担情况，某初中在本校随机抽取50名学生进行问卷调查，发现被抽查的学生中，每天完成课外作业时间，最长不足120分钟，没有低于40分钟的．并将抽查结果绘制成了一个不完整的频数分布直方图，如图10所示． （1）请补全频数分布直方图；
（2）被调查50名学生每天完成课外作业时间的中位数在 组（填时间范围）；
（3）若该校共有1200名学生，请估计该校大约有 名学生每天完成课外作业时间在80分钟以上（包括80分钟）．
19．（本小题满分6分）
如图11-1，方格纸中有一透明等腰三角形纸片，按图中裁剪线将这个纸片裁剪成三部分．请你将这三部
分小纸片重新分别拼接成；（1）一个非矩形的平行四边形；（2）一个等腰梯形；（3）一个正方形．请在图11-2中画出拼接后的三个图形，要求每张三角形纸片的顶点与小方格顶点重合．
20．（本小题满分7分）
如图12，B，C，E是同一直线上的三个点，四边形ABCD与四边形CEFG都是正方形．连接BG，DE．
（1）观察猜想BG与DE之间的大小关系，并证明你的结论； （2）图中是否存在通过旋转能够互相重合的两个三角形？若存在，请指出，并说出旋转过程；若不存在，请说明理由．
21．（本小题满分7分）
在一个不透明的布袋中有4个完全相同的乒乓球，把它们分别标号为1，2，3，4，随机地摸出一个乒乓球然后放回，再随机地摸出一个乒乓球．求下列事件的概率： （1）两次摸出的乒乓球的标号相同；
（2）两次摸出的乒乓球的标号的和等于5．
22．（本小题满分7分）
“六一”儿童节前夕，某消防队官兵了解到汶川地震灾区一帐篷小学的小朋友喜欢奥运福娃，就特意购买了一些送给这个小学的小朋友作为节日礼物．如果每班分10套，那么余5套；如果前面的班级每个班分13套，那么最后一个班级虽然分有福娃，但不足4套．问：该小学有多少个班级？奥运福娃共有多少套？
23．（本小题满分10分）
我国是世界上严重缺水的国家之一．为了增强居民节水意识，某市自来水公司对居民用水采用以户为单
位分段计费办法收费．即一月用水10吨以内（包括10吨）的用户，每吨收水费a元；一月用水超过10吨的用户，10吨水仍按每吨a元收费，超过10吨的部分，按每吨b元（b?a）收费．设一户居民月用水x吨，应收水费y元，y与x之间的函数关系如图13所示． （1）求a的值；某户居民上月用水8吨，应收水费多少元？ （2）求b的值，并写出当x?10时，y与x之间的函数关系式；
（3）已知居民甲上月比居民乙多用水4吨，两家共收水费46元，求他们上月分别用水多少吨？
24．（本小题满分10分）
如图14，直线AB经过?O上的点C，并且OA?OB，CA?CB，?O交直线OB于E，D，连接
EC，CD．
（1）求证：直线AB是?O的切线；
（2）试猜想BC，BD，BE三者之间的等量关系，并加以证明； （3）若tan?CED?12，?O的半径为3，求OA的长．
25．（本小题满分12分）
如图15，四边形OABC是矩形，OA?4，OC?8，将矩形OABC沿直线AC折叠，使点B落在D处，AD交OC于E． （1）求OE的长；
（2）求过O，D，C三点抛物线的解析式；
（3）若F为过O，D，C三点抛物线的顶点，一动点P从点A出发，沿射线AB以每秒1个单位长度的速度匀速运动，当运动时间t（秒）为何值时，直线PF把△FAC分成面积之比为1:3的两部分？
2008年湖北省襄樊市初中毕业、升学统一考试
数学试题参考答案及评分标准
一、选择题（每小题3分，共30分） 题号 答案 1 B 2 C 3 B 4 D 5 B 6 A 7 A 8 D 9 C 10 D 二、填空题（每小题3分，共18分） 11．4.2364?10
10 12．50
?13．?6 14．10 15．10?33 16．66
三、解答题（共72分） 17．解：原式??2?x?4?x?4??·························································· （2分） ??(x?4)(x?4) ·
x?4?x?(x?4)?x(x?4) ······································································································ （3分） ?2x?4x?16． ·········································································································· （4分）
2当x?2?1时，原式?2(2?1)?4(2?1)?16 ················································· （5分）
2?18． ···························································································································· （7分）
18．（1）如图1．·········································································································· （2分）
（2）80－100． ·············································································································· （4分） （3）840 ·························································································································· （6分） 19．解：如图2所示．
说明：正确画出拼接图形每个2分，共6分．
20．解：（1）BG?DE． ·························································································· （1分） ?四边形ABCD和四边形CEFG都是正方形，
?GC?CE，BC?CD，?BCG??DCE?90． ··············································· （2分） ?△BCG≌△DCE． ·································································································· （3分） ?BG?DE． ················································································································ （4分）
?（2）存在．△BCG和△DCE． ··············································································· （5分）
△BCG绕点C顺时针方向旋转90后与△DCE重合． ············································ （7分）
?21．解：将两次摸乒乓球可能出现的结果列表如下：
···································································· （2分） 以上共有16种等可能结果． ························································································· （3分） （1）两次摸出的乒乓球标号相同的结果有4种，
故P(标号相同)?416?144． ································································································· （5分）
（2）两次摸出的乒乓球的标号的和等于5的结果有4种， 故P(标号的和等于5)?16?14． ··························································································· （7分）
22．解：设该小学有x个班，则奥运福娃共有(10x?5)套．
?10x?5?13(x?1)?4，?10x?5?13(x?1).由题意，得?143 ············································································· （3分）
解之，得?x?6． ··································································································· （5分）
?x只能取整数，?x?5，此时10x?5?55． ························································· （6分）
答：该小学有5个班级，共有奥运福娃55套． ··························································· （7分） 23．解：（1）当x≤10时，有y?ax．将x?10，y?15代入，得a?1.5． ···· （1分） 用8吨水应收水费8?1.5?12（元）． ······································································· （2分） （2）当x?10时，有y?b(x?10)?15． ·································································· （3分） 将x?20，y?35代入，得35?10b?15．b?2． ················································ （4分） 故当x?10时，y?2x?5． ······················································································· （5分） （3）因1.5?10?1.5?10?2?4?46，
所以甲、乙两家上月用水均超过10吨． ······································································ （6分） 设甲、乙两家上月用水分别为x吨，y吨，
?y?x?4，则? ······························································································· （8分）
2y?5?2x?5?46.?解之，得??x?16，?y?12. ·········································································································· （9分）
故居民甲上月用水16吨，居民乙上月用水12吨． ··················································· （10分） 24．解：（1）证明：如图3，连接OC． ··································································· （1分） ?OA?OB，CA?CB，?OC?AB． ·································································· （2分） ?AB是?O的切线． ··································································································· （3分） （2）BC?BD?BE． ············································ （4分）
?ED是直径，??ECD?90． ??E??EDC?90．
??2又??BCD??OCD?90，?OCD??ODC，
??BCD??E． ·········································································································· （5分）
?又??CBD??EBC，?△BCD∽△BEC． ·························································· （6分）
?BCBE?BDBC．?BC?BD?BE． ············································································· （7分）
122（3）?tan?CED?，?CDEC??12．
?12?△BCD∽△BEC，?BDBCCDEC． ································································· （8分）
设BD?x，则BC?2x．
又BC?BD?BE，?(2x)?x?(x?6)． ································································· （9分） 解之，得x1?0，x2?2．?BD?x?0，?BD?2．
?OA?OB?BD?OD?3?2?5． ······································································· （10分）
2225．解：（1）?四边形OABC是矩形，
??CDE??AOE?90，OA?BC?CD． ··························································· （1分）
?又??CED??OEA，?△CDE≌△AOE． ························································· （2分）
?OE?DE．
?OE?OA?(AD?DE)，
222即OE?4?(8?OE)，
解之，得OE?3． ··················································· （3分） （2）EC?8?3?5．如图4，过D作DG?EC于G， ?△DGE≌△CDE． ············································· （4分）
?DEEC?DGCD222，
DEEC?EGDE．?DG?125，EG?95．
?2412??D?，?． ······················································· （5分）
5??5因O点为坐标原点，故可设过O，C，D三点抛物线的解析式为y?ax?bx．
5??64a?8b?0，a??，???32???24?2 2412解之，得?5a?b?.????b?.55??5???4y??532x?2254x． ······································································································· （7分）
??5?2?（3）?抛物线的对称轴为x?4，?其顶点坐标为?4，?．
1?k?，?8k?b?0，?设直线AC的解析式为y?kx?b，则?解之，得?2
?b??4.?b??4.?
?y?12x?4． ·············································································································· （9分）
??1??设直线FP交直线AC于H?m，m?4?，过H作HM?OA于M．
2?△AMH∽△AOC．?HM:OC?AH:AC．
?S△FAH:S△FHC?1:3或3:1，
?AH:HC?1:3或3:1，?HM:OC?AH:AC?1:4或3:4．
?HM?2或6，即m?2或6．
?H1(2，?3)，H2(6，?1)． ······················································································· （10分）
直线FH1的解析式为y?114x?172．当y??4时，x?1811． ．
直线FH2的解析式为y???当t?74x?192．当y??4时，x?5471811秒或
547秒时，直线FP把△FAC分成面积之比为1:3的两部分． ····· （12分）
说明：只求对一个值的给11分．