2008年辽宁省十二市初中毕业生学业考试题试题及答案

2008年辽宁省十二市初中毕业生学业考试数学试卷（六三制）*考试时间120分钟 试卷满分150分一、选择题（下列各题的备选答案中，只有一个是正确的，请将正确答案的序号填入下面表格内，每小题3分，共1．截止2008年6月7日12时，全国各地支援四川地震灾区的临时安置房已经安装了40600套．这个数用科学记数法表示为（ ） A ．50.40610⨯套 B ．44.0610⨯套C ．340.610⨯套D ．240610⨯套2．如图1，直线12l l ∥，l 分别与12l l ，相交，如果2120∠=， 那么1∠的度数是（ ） A ．30B ．45C ．60D ．753．下列事件中是必然事件的是（ ） A ．阴天一定下雨B ．随机掷一枚质地均匀的硬币，正面朝上C ．男生的身高一定比女生高D ．将油滴在水中，油会浮在水面上4．图2是由几个相同的小正方体搭成的一个几何体，它的俯视图是（ ）5．下列命题中正确的是（ ）A ．两条对角线互相平分的四边形是平行四边形B ．两条对角线相等的四边形是矩形C ．两条对角线互相垂直的四边形是菱形D ．两条对角线互相垂直且平分的四边形是正方形 6．若反比例函数(0)kyk x=≠的图象经过点(21)-，，则这个函数的图象一定经过点（ ） A ．122⎛⎫- ⎪⎝⎭，B ．(12)，C ．112⎛⎫- ⎪⎝⎭，D ．(12)-， 7．不等式组2133x x +⎧⎨>-⎩≤的解集在数轴上表示正确的是（ ）A ．B ．C ．D ．图2A ．B ．C ．D ．l l 1 l 212图18．图3是对称中心为点O 的正八边形．如果用一个含45角的直角三角板的角，借助点O （使角的顶点落在点O 处）把这个正八边形的面积n 等分． 那么n 的所有可能的值有（ ） A ．2个 B ．3个 C ．4个 D ．5个 二、填空题（每小题3分，共24分） 9．分解因式：34x y xy -= ．10．体育老师对甲、乙两名同学分别进行了8次跳高测试，经计算这两名同学成绩的平均数相同，甲同学的方差是2 6.4S =甲，乙同学的方差是28.2S =乙，那么这两名同学跳高成绩比较稳定的是 同学． 11．一元二次方程2210x x -+=的解是 ．12．如图4，D E ，分别是ABC △的边A B A C ，上的点，DE BC ∥，2ADDB=，则:A D E A B C S S =△△．13．如图5，假设可以在图中每个小正方形内任意取点（每个小正方形除颜色外完全相同），那么这个点取在阴影部分的概率是 ．14．一个圆锥底面周长为4πcm ，母线长为5cm ，则这个圆锥的侧面积是 ．15．如图6，观察下列图案，它们都是由边长为1cm 的小正方形按一定规律拼接而成的，依此规律，则第16个图案中的小正方形有 个．图616.如图7，直线y x =+x 轴、y 轴分别相交于A B ， 两点，圆心P 的坐标为(10)，，P 与y 轴相切于点O ．若将P沿x 轴向左移动，当P 与该直线相交时，横坐标为整数的点P有 个．三、（每小题8分，共16分）17．先化简，再求值：23111a a a a a a-⎛⎫- ⎪-+⎝⎭，其中2a =．AE CD B图4图案1图案2图案3图案4……图5图318．如图8所示，在网格中建立了平面直角坐标系，每个小正方形的边长均为1个单位长度，将四边形ABCD 绕坐标原点O 按顺时针方向旋转180后得到四边形1111A B C D ．（1）直接写出1D 点的坐标；（2）将四边形1111A B C D 平移，得到四边形2222A B C D ，若2(45)D ，，画出平移后的图形．（友情提示：画图时请不要涂错阴影的位置哦！）四、（每小题10分，共20分）19．如图9，有四张背面相同的纸牌A B C D ，，，，其正面分别画有四个不同的图形，小明将这四张纸牌背面朝上洗匀后随机摸出一张，放回后洗匀再随机摸出一张．（1）用树状图（或列表法）表示两次摸牌所有可能出现的结果（纸牌用A B C D ，，，表示）； （2）求两次摸牌的牌面图形既是中心对称图形又是轴对称图形的概率．20.如图10，AB 为O 的直径，D 为弦BE 的中点，连接OD 并延长交O 于点F ，与过B 点的切线相交于点C ．若点E 为AF 的中点，连接AE ． 求证：ABE OCB △≌△．图8图9图10ODBCF EA五、（每小题10分，共20分）21．某中学开展以“我最喜欢的职业”为主题的调查活动．通过对学生的随机抽样调查得到一组数据，下面两图（如图11、图12）是根据这组数据绘制的两幅不完整的统计图．请你根据图中所提供的信息解答下列问题：（1）求在这次活动中一共调查了多少名学生？（2）在扇形统计图中，求“教师”所在扇形的圆心角的度数． （3）补全两幅统计图．22．在“汶川地震”捐款活动中，某同学对甲、乙两班捐款情况进行了统计：甲班捐款人数比乙班捐款人数多3人，甲班共捐款2400元，乙班共捐款1800元，乙班平均每人捐款的钱数是甲班平均每人捐款钱数的45倍．求甲、乙两班各有多少人捐款？ 六、（每小题10分，共20分）23．如图13，某数学兴趣小组在活动课上测量学校旗杆高度．已知小明的眼睛与地面的距离()AB 是1.7m ，看旗杆顶部M 的仰角为45；小红的眼睛与地面的距离()CD 是1.5m ，看旗杆顶部M 的仰角为30．两人相距28米且位于旗杆两侧（点B N D ，，在同一条直线上）．请求出旗杆MN 的高度．1.41.7，结果保留整数）其它 教师 医生公务员 军人10% 20%15% 图11 图12 MN BA DC30° 45°图1324．2008年6月1日起，我国实施“限塑令”，开始有偿使用环保购物袋．为了满足市场需求，某厂家生产A B ，两种款式的布质环保购物袋，每天共生产4500个，两种购物袋的成本和售价如下表，设每天生产A 种购物袋x 个，每天共获利y 元．（1）求出y 与x （2）如果该厂每天最多投入成本10000元，那么每天最多获利多少元？ 七、（本题12分）25.如图14，在Rt ABC △中，90A ∠=，AB AC =，BC =另有一等腰梯形DEFG （GF DE ∥）的底边DE 与BC 重合，两腰分别落在AB AC ，上，且G F ，分别是AB AC ，的中点． （1）求等腰梯形DEFG 的面积；（2）操作：固定ABC △，将等腰梯形DEFG 以每秒1个单位的速度沿BC 方向向右运动，直到点D 与点C 重合时停止．设运动时间为x 秒，运动后的等腰梯形为DEF G''（如图15）．探究1：在运动过程中，四边形BDG G '能否是菱形？若能，请求出此时x 的值；若不能，请说明理由．探究2：设在运动过程中ABC △与等腰梯形DEFG 重叠部分的面积为y ，求y 与x 的函数关系式．八、（本题14分）26．如图16，在平面直角坐标系中，直线y =与x 轴交于点A ，与y 轴交于点C ，抛物线A F G (D )BC (E ) 图14F G A F ' G ' B D CE 图152(0)3y ax x c a =-+≠经过A B C ，，三点． （1）求过A B C ，，三点抛物线的解析式并求出顶点F 的坐标；（2）在抛物线上是否存在点P ，使ABP △为直角三角形，若存在，直接写出P 点坐标；若不存在，请说明理由；（3）试探究在直线AC 上是否存在一点M ，使得MBF △的周长最小，若存在，求出M 点的坐标；若不存在，请说明理由．2008年辽宁省十二市初中毕业生学业考试数学试卷（六三制）答案一、选择题（每小题3分，共24分）二、填空题（每小题3分，共24分） 9．(2)(2)xy x x +-10．甲11．121x x ==12．4:913．72514．210cm π（丢单位扣1分） 15．13616．3三、（每小题8分，共16分）17．解法一：原式223(1)(1)11a a a a a a a +---=⨯- ································································ 2分24a =+ ·································································································································· 6分当2a =时，原式2248=⨯+= ··························································································· 8分解法二：原式3(1)(1)(1)(1)11a a a a a a a a a a+-+-=⨯-⨯-+ ··············································· 2分 24a =+ ·································································································································· 6分 当2a =时，原式2248=⨯+= ··························································································· 8分18．解：（1）1(31)D -， ······················································································································ 2分 （2）2A ，222B C D ，，描对一个点给1分． ······································································ 6分 画出正确图形（见图1） ········································································································ 8分x四、（每小题10分，共20分） 19．（1························································ 6分（2）从表中可以得到，两次摸牌所有可能出现的结果共有16种，其中既是中心对称图形又是轴对称图形的有9种． ······························································································································ 8分 故所求概率是916． ··············································································································· 10分 19．（1）解法二：所以可能出现的结果：（A ，A ），（A ，B ），（A ，C ），（A ，D ），（B ，A ），（B ，B ），（B ，C ），（B ，D ），（C ，A ），（C ，B ），（C ，C ），（C ，D ），（D ，A ），（D ，B ），（D ，C ），（D ，D ）． ·························································································· 6分 （2）以下同解法1． 20.解：（1）证明：如图2． AB 是O 的直径．90E ∴∠= ··············································································· 1分又BC 是O 的切线，90OBC ∴∠=A B C DA ABC DB A BC DC A B C DD 开始第一次牌面的字母第二次牌面的字母 图1图2OD BC F EAE OBC ∴∠=∠ ········································································ 3分 OD 过圆心，BD DE =，EF FB ∴=BOC A ∴∠=∠． ·················································································································· 6分 E 为AF 中点，EF BF AE ∴==30ABE ∴∠= ······················································································································· 8分 90E ∠=12AE AB OB ∴== ··············································································································· 9分 ABE OCB ∴△≌△． ········································································································· 10分 五、（每小题10分，共20分） 21．（1）被调查的学生数为4020020=％（人） ········································································· 2分 （2）“教师”所在扇形的圆心角的度数为70115201010036072200⎛⎫----⨯⨯= ⎪⎝⎭％％％％ ··························································· 5分 （3）如图3，补全图 ·············································································································· 8分如图4，补全图 ····················································································································· 10分22．解法一：设乙班有x 人捐款，则甲班有(3)x +人捐款． ············································· 1分 根据题意得：24004180035x x⨯=+ ··················································································································· 5分 解这个方程得45x =． ·········································································································· 8分 经检验45x =是所列方程的根． ··························································································· 9分 348x ∴+=（人）答：甲班有48人捐款，乙班有45人捐款． ······································································· 10分 解法二：设甲班有x 人捐款，则乙班有(3)x -人捐款． ····················································· 1分 根据题意得：其它 教师医生 公务员军人10% 20%15%图3图435%20%24004180053x x ⨯=- ··················································································································· 5分 解这个方程得48x =． ·········································································································· 8分 经检验48x =是所列方程的根． ··························································································· 9分 345x ∴-=（人）答：甲班有48人捐款，乙班有45人捐款． ······································································· 10分 六、（每小题10分，共20分） 23．解法一：解：过点A 作AE MN ⊥于E ，过点C 作CF MN ⊥于F ， ··········································· 1分 则 1.7 1.50.2EF AB CD =-=-= ······················································································ 2分 在Rt AEM △中，90AEM ∠=，45MAE ∠=AE ME ∴= ........................................................................................................................... 3分 设AE ME x ==（不设参数也可） 0.2MF x ∴=+，28FC x =- . (5)分在Rt MFC △中，90MFC ∠=，30MCF ∠=tan MF CF MCF ∴=∠0.2(28)3x x ∴+=- ····················································· 7分 10.0x ∴≈ 12MN ∴≈ ···························································································································· 9分答：旗杆高约为12米． ······································································································· 10分 解法二：解：过点A 作AE MN ⊥于E ，过点C 作CF MN ⊥于F ， ···························· 1分 则 1.7 1.50.2EF AB CD =-=-= ······················································································ 2分 在Rt AEM △中，90AEM ∠=，45MAE ∠=AE ME ∴=设AE x =，则0.2MF x =+ ································································································ 3分在Rt MFC △中，90MFC ∠=，30MCF ∠=tan603(0.2)CF MF x ==+ ························································································· 5分 BN ND BD +=0.2)28x x ∴+= ·········································································································· 7分 解得10.2x ≈ 12MN ∴≈ ···························································································································· 9分 答：旗杆高约为12米． ······································································································· 10分 （注：其他方法参照给分） 24．解：（1）根据题意得：(2.32)(3.53)(4500)0.22250y x x x =-+--=-+ ························· 2分 （2）根据题意得：23(4500)10000x x +-≤ ··································································· 5分M N B A DC30° 45°图5EF。