庆阳市2009年初中毕业学业监测与高中阶段学校招生考试数 学 试 卷友情提示：1．抛物线2y ax bx c =++的顶点坐标是2424b ac b a a ⎛⎫-- ⎪⎝⎭，． 2．扇形面积公式：2π360n R S =扇形；其中，n 为扇形圆心角度数，R 为圆的半径． 本试卷满分为150分，考试时间为120分钟．一、选择题：本大题共10小题，每小题3分，共30分．每小题给出的四个选项中，只有一项是符合题目要求的，将此选项的代号填入题后的括号内． 1．8的立方根是（ ）A ．2B ．2-C ．±2D ．2．方程240x -=的根是（ ） A ．2x =B ．2x =-C ．1222x x ==-，D ．4x =3．图1中不是中心对称图形的是（ ）A ．B ．C ．D ．图14．下列说法中，正确的是（ ） A ．“明天降雨的概率是80％”表示明天有80％的时间降雨 B ．“抛一枚硬币正面朝上的概率是0.5”表示每抛硬币2次就有1次出现正面朝上 C ．“彩票中奖的概率是1%”表示买100张彩票一定有1张会中奖 D ．在同一年出生的367名学生中，至少有两人的生日是同一天 5．将抛物线22y x =向下平移1个单位，得到的抛物线是（ ） A ．22(1)y x =+B ．22(1)y x =-C ．221y x =+D ．221y x =-6．如图2，晚上小亮在路灯下散步，在小亮由A 处径直走到B 处这一过程中，他在地上的影子（ ） A ．逐渐变短 B ．先变短后变长 C ．先变长后变短 D ．逐渐变长7．如图3，在宽为20米、长为30米的矩形地面上修建两条同样宽的道路，余下部分作为耕地．若耕地面积需要551米2，则修建的路宽应为（ ） A ．1米 B ．1.5米 C ．2米 D ．2.5米图2 图3 图4 图58．如图4，在平行四边形ABCD 中，E 是AB 的中点，CE 和BD 交于点O ，设△OCD 的面积为m ，△OEB） A ．5m =B．m =C．m =D ．10m =9．如图5，⊙O 的半径为5，弦AB =8，M 是弦AB 上的动点，则OM 不可能为（ ） A ．2 B ．3 C ．4 D ．510．图6（1）是一个横断面为抛物线形状的拱桥，当水面在l 时，拱顶（拱桥洞的最高点）离水面2m ，水面宽4m ．如图6（2）建立平面直角坐标系，则抛物线的关系式是（ ） A ．22y x =- B ．22y x = C ．212y x =-D ．212y x =二、填空题：本大题共10小题，每小题4分，共40分．把答案填在题中的横线上． 11在实数范围内有意义的x 应满足的条件是 ． 12．若关于x 的方程2210x x k ++-=的一个根是0，则k = ．13．如图7，将正六边形绕其对称中心O 旋转后，恰好能与原来的正六边形重合，那么旋转的角度至少是 度．14．若100个产品中有95个正品、5个次品，从中随机抽取一个，恰好是次品的概率是 ． 15．如图8，直线AB 与⊙O 相切于点B ，BC 是⊙O 的直径，AC 交⊙O 于点D ，连结BD ，则图中直角三角形有 个．16．从地面垂直向上抛出一小球，小球的高度h （米）与小球运动时间t （秒）的函数关系式是29.8 4.9h t t =-，那么小球运动中的最大高度为 米．17．如图9，菱形ABCD 的边长为10cm ，DE ⊥AB ，3sin 5A =，则这个菱形的面积图6（1） 图6（2）图7 图8= cm 2．18．如图10，两个等圆⊙O 与⊙O ′外切，过点O 作⊙O ′的两条切线OA 、OB ，A 、B 是切点，则∠AOB = ．图9 图10 图11 图1219．如图11，正方形OEFG 和正方形ABCD 是位似形，点F 的坐标为（1，1），点C 的坐标为（4，2），则这两个正方形位似中心的坐标是 ． 20．图12为二次函数2y ax bx c =++的图象，给出下列说法：①0ab <；②方程20ax bx c ++=的根为1213x x =-=，；③0a b c ++>；④当1x >时，y 随x 值的增大而增大；⑤当0y >时，13x -<<．其中，正确的说法有 ．（请写出所有正确说法的序号） 三、解答题（一）：本大题共5小题，共38分．解答时，应写出必要的文字说明、证明过程或演算步骤．21．（62sin 45°．22．（7分）一位美术老师在课堂上进行立体模型素描教学时，把由圆锥与圆柱组成的几何体（如图13所示，圆锥在圆柱上底面正中间放置）摆在讲桌上，请你在指定的方框内分别画出这个几何体的三视图（从正面、左面、上面看得到的视图）．图13主视图 左视图 俯视图23．（8分）如图14，在平面直角坐标系中，等腰Rt △OAB 斜边OB 在y 轴上，且OB =4． （1）画出△OAB 绕原点O 顺时针旋转90°后得到的三角形；（2）求线段OB 在上述旋转过程中所扫过部分图形的面积（即旋转前后OB 与点B 轨迹所围成的封闭图形的面积）．24．（8分）某企业2006年盈利1500万元，2008年克服全球金融危机的不利影响，仍实现盈利2160万元．从2006年到2008年，如果该企业每年盈利的年增长率相同，求： （1）该企业2007年盈利多少万元？（2）若该企业盈利的年增长率继续保持不变，预计2009年盈利多少万元？ 25．（9分）一只不透明的袋子中，装有2个白球（标有号码1、2）和1个红球，这些球除颜色外其他都相同．（1）搅匀后从中摸出一个球，摸到白球的概率是多少？（2）搅匀后从中一次摸出两个球，请用树状图（或列表法）求这两个球都是白球的概率．图14四、解答题（二）：本大题共4小题，共42分．解答时，应写出必要的文字说明、证明过程或演算步骤． （1）26．（10分）如图15（1），一扇窗户打开后用窗钩AB 可将其固定． （1）这里所运用的几何原理是（ ） （A ）三角形的稳定性 （B ）两点之间线段最短 （C ）两点确定一条直线 （D ）垂线段最短（2）图15（2）是图15（1）中窗子开到一定位置时的平面图，若∠AOB =45°， ∠OAB =30°，OA =60cm ，求点B 到OA 边的距离．1.7，结果精确到整数）27．（10分）如图16，网格中的每个小正方形的边长都是1，每个小正方形的顶点叫做格点． △ACB 和△DCE 的顶点都在格点上，ED 的延长线交AB 于点F ． （1）求证：△ACB ∽△DCE ；（2）求证：EF ⊥AB ．28．（10分）如图17，在边长为2的圆内接正方形ABCD 中，AC 是对角线，P 为边CD 的中点，延长AP 交圆于点E ． （1）∠E = 度；（2）写出图中现有的一对不全等的相似三角形，并说明理由； （3）求弦DE 的长．图15（1） 图15（2）图16 图1729．（12分）如图18，在平面直角坐标系中，将一块腰长为5的等腰直角三角板ABC 放在第二象限，且斜靠在两坐标轴上，直角顶点C 的坐标为（1-，0），点B 在抛物线22y ax ax =+-上．（1）点A 的坐标为 ，点B 的坐标为 ； （2）抛物线的关系式为 ；（3）设（2）中抛物线的顶点为D ，求△DBC 的面积；（4）将三角板ABC 绕顶点A 逆时针方向旋转90°，到达AB C ''△的位置．请判断点B '、C '是否在（2）中的抛物线上，并说明理由．附加题：如果你的全卷得分不足150分，则本题的得分记入总分，但记入总分后全卷得分不得超过150分，超过按150分算． 30．（10分）图19是二次函数2122y x =-+的图象在x 轴上方的一部分，若这段图象与x 轴所围成的阴影部分面积为S ，试求出S 取值的一个范围．图18 图19庆阳市2009年初中毕业学业监测与高中阶段学校招生考试数学试卷参考答案与评分标准11．x ＞1 12．1 13．60 14．20115．3 16．4.9 17．60 18．60° 19．（2-，0） 20． ①②④三、解答题（一）：本大题共5小题，共38分． 21．本小题满分6分解： 原式=22⨯························································································ 4分 =0． ························································································································ 6分22．本小题满分7分解：正确的三视图如图所示：主视图正确； ········································································ 2分 左视图正确； ········································································ 2分俯视图正确． ····································································· 3分 说明：俯视图中漏掉圆心的黑点扣1分．23．本小题满分8分解：（1）画图正确（如图）； ······················································· 4分 （2）所扫过部分图形是扇形，它的面积是：290π44π360⨯=． ································································· 8分24．本小题满分8分 解：（1）设每年盈利的年增长率为x ， ················································································ 1分根据题意，得21500(1)2160x +=． ·········································································· 3分 解得120.2 2.2x x ==-，（不合题意，舍去）．···························································· 5分1500(1)1500(10.2)1800x ∴+=+=．答：2007年该企业盈利1800万元． ············································································· 6分 （2） 2160(10.2)2592+=．答：预计2009年该企业盈利2592万元． ··································································· 8分 25． 本小题满分9分 解 （1）p （一个球是白球）=23······················································································ 3分 （2）树状图如下（列表略）：开始········································································································································· 6分P ∴（两个球都是白球）2163== ． ············································································· 9分 四、解答题（二）：本大题共4小题，共42分． 26．本小题满分10分 解：（1）A ． ··············································································· 3分 （2）如图，过点B 作BC ⊥OA 于点C ， ··························· 4分 ∵ ∠AOB =45°，∴∠CBO =45°，BC =OC ． ·························· 5分 设BC =OC =x ，∵∠OAB =30°， ∴ AC =BC ×tan60°=3x ． ··················································· 7分 ∵ OC +CA =OA ，∴x +3x =60， ······································ 8分 ∴ x =3160+≈22（cm ）．即点B 到OA 边的距离是22 cm ． ··············································································· 10分 27． 本小题满分10分 证明：（1）∵ 3,2AC DC = 63,42BC CE == ··············································· 2分∴ .AC BC DC CE = 又 ∠ACB =∠DCE =90°， ··················································· 3分 ∴ △ACB ∽△DCE ． ······················································ 5分 （2） ∵ △ACB ∽△DCE ，∴ ∠ABC ＝∠DEC ． ······································································ 6分 又 ∠ABC ＋∠A =90°，∴ ∠DEC ＋∠A =90°． ······························································· 8分 ∴ ∠EF A =90°． ∴ EF ⊥AB ． ······················································································ 10分 28．本小题满分10分 解：（1）45． ················································································ 2分 （2）△ACP ∽△DEP ． ···························································· 4分 理由：∵∠AED =∠ACD ，∠APC =∠DPE ，∴ △ACP ∽△DEP ． ································································ 6分（3）方法一：∵ △ACP ∽△DEP ， ∴ .AP AC DP DE =······························· 7分 又 AP =522=+DP AD ，AC =2222=+DC AD ， ·········································· 9分 ∴ DE =5102． ········································································································ 10分方法二：如图2，过点D 作DF AE ⊥于点F ．在Rt ADP △中， AP ··················· 7分 又1122ADP S AD DP AP DF == △， ·························· 8分 白2 红 白1 白1 红 白2白1 白2 红BCO图2图1∴ DF =552． ············································································································ 9分∴ 51022==DF DE ． ························································································ 10分29．本小题满分12分 解： （1）A （0，2）， B （3-，1）． ················································································· 2分 （2）211222y x x =+-． ······························································································ 3分 （3）如图1，可求得抛物线的顶点D （11728--，）． ················································· 4分 设直线BD 的关系式为y kx b =+， 将点B 、D 的坐标代入，求得54k =-，114b =-，∴ BD 的关系式为51144y x =--． ················································································· 5分设直线BD 和x 轴交点为E ，则点E （115-，0），CE =65．∴ △DBC 的面积为1617152588⨯⨯+=（1）． ······························································ 7分（4）如图2，过点B '作B M y '⊥轴于点M ，过点B 作BN y ⊥轴于点N ，过点C '作C P y '⊥轴于点P ． ······················································································································· 8分在Rt △AB ′M 与Rt △BAN 中，∵ AB =AB ′， ∠AB ′M =∠BAN =90°-∠B ′AM ， ∴ Rt △AB ′M ≌Rt △BAN ． ·································································································· 9分 ∴ B ′M =AN =1，AM =BN =3， ∴ B ′（1，1-）． ···························································· 10分图1图2同理△AC ′P ≌△CAO ，C ′P =OA =2，AP =OC =1，可得点C ′（2，1）； ·························· 11分 将点B ′、C ′的坐标代入211222y x x =+-，可知点B ′、C ′在抛物线上． ················· 12分 （事实上，点P 与点N 重合）附加题：如果你的全卷得分不足150分，则本题的得分记入总分，但记入总分后全卷得分不得超过150分，超过按150分算． 30．本小题满分10分 解：方法一：由题意，可知这段图象与x 轴的交点为A （-2，0）、B （2，0），与y 轴的交点为C （0，2）． ······················································· 2分显然，S 在ABC ∆面积与过A 、B 、C 三点的⊙O 半圆面积之间． ··································· 3分 ∵ ABC S △=4， ··································································· 4分 12O S =2π， ······························································· 5分 ∴ 4<S <2π． ······································································ 6分说明：关于半圆⊙O 的面积大于图示阴影部分面积的证明，如下（对学生不要求）： 设P （x ，y ）在图示抛物线上，则 OP 2=x 2+y 2=（4-2y ）+y 2=（y -1）2+3． ∵ 0≤y ≤2， ∴ 3≤OP 2≤4． ∴ 点P 在半圆x 2+y 2=3、x 2+y 2=4所夹的圆环内， 以及点P为内圆周点（1）与外圆周点A 、B 、C ． ∴ 半圆⊙O 的面积大于图示阴影部分的面积． 由于内半圆的面积为12O S -3π2， ∴3π2<S <2π． 如果学生能得出此结论，可在上面结论基础上，加4分．方法二：由题意，可知这段图象与x 轴的交点为A （-2，0）、B （2，0），与y 轴的交点为C （0，2）． ························································································· 2分2的两个半圆所夹的圆环内，以及过内半圆上点 P（1）与半外圆上点A 、B 、C ． ······················ 5分∴ S 在图示两个半圆面积之间． ····································· 7分即21π2⋅<S <2122π⋅． ········································· 9分 ∴ 3π2<S <2π． ·························································· 10分。