宁夏回族自治区2009年初中毕业暨高中阶段招生
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宁夏回族自治区2009年初中毕业暨高中阶段招生数 学 试 题注意事项：1．考试时间120分钟，全卷总分120分． 2．答题前将密封线内的项目填写清楚． 3．答卷一律使用黑、蓝钢笔或圆珠笔．4．凡使用答题卡的考生，答卷前务必将答题卡上的有关项目填写清楚．选择题的每小题选出答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案．不使用答题卡的考生，将选择题的答案答在试卷上．一、选择题（下列每小题所给的四个答案中只有一个是正确的，每小题3分，共24分） 1．下列运算正确的是（ ）A ．3412a a a =· B ．623(6)(2)3a a a -÷-= C ．22(2)4a a -=- D ．23a a a -=-2．某旅游景点三月份共接待游客25万人次，五月份共接待游客64万人次，设每月的平均增长率为x ，则可列方程为（ ）A ．225(1)64x +=B ．225(1)64x -=C ．264(1)25x +=D ．264(1)25x -= 3．把不等式组21123x x +>-⎧⎨+⎩≤的解集表示在数轴上，下列选项正确的是（ ）A ． B．C ．D ．4．某班抽取6名同学参加体能测试，成绩如下：85，95，85，80，80，85．下列表述错误．．的是（）A ．众数是85B ．平均数是85C ．中位数是80D ．极差是15 5．一次函数23y x =-的图象不经过（ ）A ．第一象限B ．第二象限C ．第三象限D ．第四象限6．如图，是一个几何体的三视图，根据图中标注的数据可求得这个几何体的体积为（ ） A ．24π B ．32π C ．36π D ．48π10 1- 11-10 1- 10 1- 主视图 左视图俯视图 （6题图）（7题图）7．在44⨯的正方形网格中，已将图中的四个小正方形涂上阴影（如图），若再从其余小正方形中任选一个也涂上阴影，使得整个阴影部分组成的图形成轴对称图形．那么符合条件的小正方形共有（ ）A ．1个B ．2个C ．3个D ．4个8．二次函数2(0)y ax bx c a =++≠的图象如图所示，对称轴是直线1x =，则下列四个结论错误．．的是（ ） A ．0c > B ．20a b += C ．240b ac -> D ．0a b c -+> 二、填空题（每小题3分，共24分） 9．分解因式：32m mn -= ．10．在Rt ABC △中，9032C AB BC ∠===°，，，则cos A 的值是 ． 11．已知：32a b +=，1ab =，化简(2)(2)a b --的结果是 ． 12．某商品的价格标签已丢失，售货员只知道“它的进价为80元，打七折售出后，仍可获利5%”．你认为售货员应标在标签上的价格为 元．13．用一个半径为6，圆心角为120°的扇形围成一个圆锥的侧面，则圆锥的高为 ． 14．如图，梯形ABCD 的两条对角线交于点E ，图中面积相等的三角形共有 对．15．如图，ABC △的周长为32，且AB AC AD BC =⊥，于D ，ACD △的周长为24，那么AD 的长为 ．16．如图，O ⊙是边长为2的等边三角形ABC 的内切圆，则图中阴影部分的面积为 ．三、解答题（共24分） 17．（6分）11(2009)12-⎛⎫-+ ⎪⎝⎭．（8题图）A DB E （14题图） （15题图） A B CD B （16题图）18．（6分） 解分式方程：1233xx x+=--． 19．（6分）已知正比例函数1y k x =1(0)k ≠与反比例函数22(0)k y k x=≠的图象交于A B 、两点，点A 的坐标为(21)，．（1）求正比例函数、反比例函数的表达式； （2）求点B 的坐标． 20．（6分）桌子上放有质地均匀，反面相同的4张卡片．正面分别标有数字1、2、3、4，将这些卡片反面朝上洗匀后放在桌面上，先从中任意抽出1张卡片，用卡片上所标的数字作为十位上的数字，将取出的卡片反面朝上放回洗匀；再从中任意抽取1张卡片，用卡片上所标的数字作为个位数字．试用列表或画树状图的方法分析，组成的两位数恰好能被3整除的概率是多少？四、解答题（48分） 21．（6分）在“首届中国西部（银川）房·车生活文化节”期间，某汽车经销商推出A B C D 、、、四种型号的小轿车共1000辆进行展销．C 型号轿车销售的成交率为50%，其它型号轿车的销售情况绘制在图1和图2两幅尚不完整的统计图中． （1）参加展销的D 型号轿车有多少辆？ （2）请你将图2的统计图补充完整；（3）通过计算说明，哪一种型号的轿车销售情况最好？ （4）若对已售出轿车进行抽奖，现将已售出A B C D 、、、四种型号轿车的发票（一车一票）放到一起，从中随机抽取一张，求抽到A 型号轿车发票的概率．22．（6分）如图：在Rt ABC △中，90ACB ∠=°，CD 是AB 边上的中线，将ADC △沿AC 边所在的直线折叠，使点D 落在点E 处，得四边形ABCE ．求证：EC AB ∥．型号DC 20% B20%A 35% 各型号参展轿车数的百分比（图2） （图1） EC B A D已知：如图，AB 为O ⊙的直径，AB AC BC =，交O ⊙于点D ，AC 交O ⊙于点45E BAC ∠=，°．（1）求EBC ∠的度数；（2）求证：BD CD =．24．（8分）如图，抛物线21222y x x =-++与x 轴交于A B 、两点，与y 轴交于C 点． （1）求A B C 、、三点的坐标；（2）证明ABC △为直角三角形；（3）在抛物线上除C 点外，是否还存在另外一个点P ，使ABP △是直角三角形，若存在，请求出点P 的坐标，若不存在，请说明理由．C如图1、图2，是一款家用的垃圾桶，踏板AB （与地面平行）或绕定点P （固定在垃圾桶底部的某一位置）上下转动（转动过程中始终保持AP A P BP B P ''==，）．通过向下踩踏点A 到A '（与地面接触点）使点B 上升到点B '，与此同时传动杆BH 运动到B H ''的位置，点H 绕固定点D 旋转（DH 为旋转半径）至点H '，从而使桶盖打开一个张角HDH '∠． 如图3，桶盖打开后，传动杆H B ''所在的直线分别与水平直线AB DH 、垂直，垂足为点M C 、，设H C '=B M '．测得6cm 12cm 8cm AP PB DH '===，，．要使桶盖张开的角度HDH '∠不小于60°，那么踏板AB 离地面的高度至少等于多少cm ？（结果保留两位有效数字））（图1）D（图2）D（图3）已知：等边三角形ABC 的边长为4厘米，长为1厘米的线段MN 在ABC △的边AB 上沿AB 方向以1厘米/秒的速度向B 点运动（运动开始时，点M 与点A 重合，点N 到达点B 时运动终止），过点M N 、分别作AB 边的垂线，与ABC △的其它边交于P Q 、两点，线段MN 运动的时间为t 秒．（1）线段MN 在运动的过程中，t 为何值时，四边形MNQP 恰为矩形？并求出该矩形的面积；（2）线段MN 在运动的过程中，四边形MNQP 的面积为S ，运动的时间为t ．求四边形MNQP 的面积S 随运动时间t 变化的函数关系式，并写出自变量t 的取值范围．C PQBA M N宁夏回族自治区2009年初中毕业暨高中阶段招生数学试卷参考答案二、填空题（每小题3分，共24分）三、解答题（共24分） 17．（6分）计算：解：原式=121+ ······························································································ 4分 =····································································································································· 6分 18．（6分）解分式方程：解：去分母得：12(3)x x -=- ···························································································· 3分 整理方程得：37x -=-73x =······································································································································ 5分 经检验73x =是原方程的解．∴原方程的解为73x =． ······································································································· 6分 19．（6分）解：（1）把点(21)A ，分别代入1y k x =与2k y x=得 112k =，22k =． ·················································································································· 2分 ∴正比例函数、反比例函数的表达式为：122y x y x==，． ············································· 3分 （2）由方程组122y xy x⎧=⎪⎪⎨⎪=⎪⎩得1121x y =-⎧⎨=-⎩，2221x y =⎧⎨=⎩．B ∴点坐标是(2,1)--． ········································································································ 6分20．（6分） 解：列表：树状图： (3)分∴能被3整除的两位数的概率是516． ·················································································· 6分 四、解答题（共48分） 21（6分） 解：（1）100025%250⨯=（辆） ······················································································ 1分 （2）如图，（100020%50%100⨯⨯=） ······ ········· 2分（3）四种型号轿车的成交率：16898A 100%48%B 100%49%350200⨯=⨯=：：C 50%： 130D 100%52%250⨯=： ∴D 种型号的轿车销售情况最好． ······························ 4分 （4）168168211689810013049662==+++．∴抽到A 型号轿车发票的概率为2162． ················································································· 6分 22．（6分）证明：CD 是AB 边上的中线，且90ACB ∠=°， CD AD ∴=．CAD ACD ∴∠=∠． ············································································································ 2分 又ACE △是由ADC △沿AC 边所在的直线折叠而成的， ECA ACD ∴∠=∠．············································································································· 4分 ECA CAD ∴∠=∠．············································································································· 5分 EC AB ∴∥． ······················································································································· 6分 23．（8分） （1）解：AB 是O ⊙的直径， 90AEB ∴∠=°．又45BAC ∠= °， 45ABE ∴∠=°． 又AB AC = ，11 2 3 4 14 13 12 11 1 2 3 4 24 23 22 21 21 2 3 4 34 33 32 31 31 3 44 43 42 41 4开始 C型号67.5ABC C ∴∠=∠=°． 22.5EBC ∴∠=°． ··············································································································· 4分 （2）证明：连结AD ． AB 是O ⊙的直径， 90ADB ∴∠=°． AD BC ∴⊥． 又AB AC = ， BD CD ∴=． ························································································································ 8分24．（8分）解：（1） 抛物线2122y x x =-+与x 轴交于A B 、两点，21202x x ∴-+=．即240x -=．解之得：12x x ==∴点A B 、的坐标为(A B ）． ····································································· 2分将0x =代入21222y x x =-++，得C 点的坐标为（0，2）········································ 3分（2）AC BC AB ===222AB AC BC ∴=+，则90ACB ∠=°，ABC ∴△是直角三角形． ······································································································ 6分（3）将2y =代入21222y x x =-++得212222x x -++=，120x x ∴=，P ∴点坐标为． ·········································································································· 8分 25．（10分）过点A '作A N AB '⊥垂足为N 点， 在Rt H CD '△中，若HDH '∠不小于60°，则sin 602H C H D '︒='≥即2H C H D ''=≥··········································· 5分B M HC ''= ≥················································· 6分 Rt Rt A NP B MP '' △∽△A N A PB M B P''∴=''3.5cm A P B M A N B P '''∴=='·································································· 9分 ∴踏板AB 离地面的高度至少等于3.5cm ． ········································································ 10分 26．（10分）（1）过点C 作CD AB ⊥，垂足为D ． 则2AD =，当MN 运动到被CD 垂直平分时，四边形MNQP 是矩形， 即32AM =时，四边形MNQP 是矩形， 32t ∴=秒时，四边形MNQP 是矩形．tan 60PM AM = °=MNQP S ∴=四边形·············································································································· 4分（2）1°当01t <<时，1()2MNQP S PM QN MN =+四边形·11)2t ⎤=++⎦2=+··································································· 6分 2°当12t ≤≤时1()2MNQP S PM QN MN =+四边形·1)12t ⎤=+-⎦·=·········································································· 8分 3°当23t <<时，C PQB A M D NC P Q BA M N CPQA M N1()2MNQP S PM QN MN =+四边形·1))2t t ⎤=--⎦=···························································· 10分CPQBAMN。
