二00九年南宁市中等学校招生考试数 学本试卷分第Ⅰ卷和第Ⅱ卷，满分120分，考试时间120分钟.注意：答案一律填写在答题卷上，在试题卷上作答无效．．．．．．．．．．考试结束，将本试卷和答题卷一并交回．第Ⅰ卷（选择题 共36分）一、选择题：（本大题共12小题，每小题3分，共36分）每小题都给出代号为A 、B 、C 、D 的四个结论，其中只有一个是正确的．使用机改卷的考生．．．．．．．．，请用2B 铅笔在答题卷上将选定的答案标号涂黑；使用非机改卷的六县考生．．．．．．．．．．．，请用黑（蓝黑）墨水笔将每小题选定的答案的序号填写在答题卷相应的表格内．1．13的相反数是（ ） A ．3B ．13C ．3-D ．13-2．图1是一个五边形木架，它的内角和是（ ）A ．720°B ．540°C ．360°D ．180°3．今年6月，南宁市举行了第五届泛珠三角区域经贸合作洽谈会.据估算，本届大会合同投资总额达2260亿元.将2260用科学记数法表示为（结果保留2个有效数字）（ ） A ．32.310⨯B ．32.210⨯C ．32.2610⨯D ．40.2310⨯4．与左边三视图所对应的直观图是（ ）5．不等式组11223x x ⎧⎪⎨⎪-<⎩≤的解集在数轴上表示为（ ）6．要使式子1x x+有意义，x 的取值范围是（ ） A ．1x ≠ B ．0x ≠ C ．10x x >-≠且 D ．10x x ≠≥-且7．如图2，将一个长为10cm ，宽为8cm 的矩形纸片对折两次后，沿所得矩形两邻边中点的连线（虚线）剪下，再打开，得到的菱形的面积为（ ）-1 01 2 A ．-1 01 2 B ．-1 01 2 C ．-1 01 2 D ．A ．B ．C ．D ．A ．210cm B ．220cmC ．240cmD ．280cm8．把多项式2288x x -+分解因式，结果正确的是（ ） A ．()224x -B ．()224x -C ．()222x -D ．()222x +9．在反比例函数1ky x-=的图象的每一条曲线上，y x 都随的增大而增大，则k 的值可以是（ ） A ．1- B ．0C ．1D ．210．如图3，AB O 是⊙的直径，弦30CD AB E CDB O ⊥∠=于点，°，⊙，则弦CD 的长为（ ） A ．3cm 2B ．3cm C． D ．9cm11．已知二次函数2y ax bx c =++（0a ≠）的图象如图4所示，有下列四个结论：20040b c b ac <>->①②③④0a b c -+<，其中正确的个数有（ ）A ．1个B ．2个C ．3个D ．4个12．从2、3、4、5这四个数中，任取两个数()p q p q ≠和，构成函数2y px y x q =-=+和，并使这两个函数图象的交点在直线2x =的右侧，则这样的有序数对()p q ，共有（ ） A ．12对B ．6对C ．5对D ．3对A BCD图2图3CABOE D图4第Ⅱ卷（非选择题，共84分）二、填空题：（本大题共6小题，每小题2分，共12分）13．如图5，直线a 、b 被c 所截，且11202a b ∠=∠=∥，°，则 °． 14．计算：()22a ba ÷ ．15．三角尺在灯泡O 的照射下在墙上形成影子（如图6所示）.现测得20cm 50cm OA OA '==，，这个三角尺的周长与它在墙上形成的影子的周长的比是 ．16．有五张分别印有圆、等腰三角形、矩形、菱形、正方形图案的卡片（卡片中除图案不同外，其余均相同），现将有图案的一面朝下任意摆放，从中任意抽取一张，抽到有中心对称图案的卡片的概率是 ．17．如图7，一艘海轮位于灯塔P 的东北方向，距离灯塔A 处，它沿正南方向航行一段时间后，到达位于灯塔P 的南偏东30°方向上的B 处，则海轮行驶的路程AB 为 _____________海里（结果保留根号）．18．正整数按图8的规律排列．请写出第20行，第21列的数字 ．考生注意：第三至第八大题为解答题，要求在答题卷．．．上写出解答过程． 三、（本大题共2小题，每小题满分6分，共12分） 19．计算：()1200911sin 602-⎛⎫-+-- ⎪⎝⎭°20．先化简，再求值：cab 12 图5 图6 A A O 灯 三角尺 投影 图7 BA C P 东北45° 30° 第一行第二行 第三行 第四行 第五行 第一列 第二列 第三列 第四列 第五列1 2 5 10 17 … 4 3 6 11 18 … 9 8 7 12 19 … 16 15 14 1320 … 25 24 23 22 21 … ……图8()2111211x x x ⎛⎫+÷-- ⎪--⎝⎭，其中x = 四、（本大题共2小题，每小题满分10分，共20分）21．为迎接国庆60周年，某校举行以“祖国成长我成长”为主题的图片制作比赛，赛后整请根据以上图表提供的信息，解答下列问题：（1）表中m n 和所表示的数分别为：__________m n ==，__________； （2）请在图9中，补全频数分布直方图； （3）比赛成绩的中位数落在哪个分数段？（4）如果比赛成绩80分以上（含80分）可以获得奖励，那么获奖率是多少？22．已知ABC △在平面直角坐标系中的位置如图10所示． （1）分别写出图中点A C 和点的坐标；（2）画出ABC △绕点C 按顺时针方向旋转90A B C '''°后的△；（3）求点A 旋转到点A '所经过的路线长（结果保留π）．图9 频数分数（分）图1023．如图11，PA 、PB 是半径为1的O ⊙的两条切线，点A 、B 分别为切点，60APB OP AB C O D ∠=°，与弦交于点，与⊙交于点．（1）在不添加任何辅助线的情况下，写出图中所有的全等三角形；（2）求阴影部分的面积（结果保留π）．六、（本大题满分10分）24．南宁市狮山公园计划在健身区铺设广场砖．现有甲、乙两个工程队参加竞标，甲工程队铺设广场砖的造价y 甲（元）与铺设面积()2m x 的函数关系如图12所示；乙工程队铺设广场砖的造价y 乙（元）与铺设面积()2m x 满足函数关系式：y kx =乙．（1）根据图12写出甲工程队铺设广场砖的造价y 甲（元）与铺设面积()2m x 的函数关系式； （2）如果狮山公园铺设广场砖的面积为21600m ，那么公园应选择哪个工程队施工更合算？ 七、（本大题满分10分）25．如图13-1，在边长为5的正方形ABCD 中，点E 、F 分别是BC 、DC 边上的点，且AE EF ⊥，2BE =. （1）求EC ∶CF 的值；（2）延长EF 交正方形外角平分线CP P 于点（如图13-2），试判断AE EP 与的大小关系，并说明理由；（3）在图13-2的AB 边上是否存在一点M ，使得四边形DMEP 是平行四边形？若存在，请给予证明；若不存在，请说明理由．图11图12)2图13-1 ADCB E 图13-2BCE DAF PF26．如图14，要设计一个等腰梯形的花坛，花坛上底长120米，下底长180米，上下底相距80米，在两腰中点连线（虚线）处有一条横向甬道，上下底之间有两条纵向甬道，各甬道的宽度相等．设甬道的宽为x米．（1）用含x的式子表示横向甬道的面积；（2）当三条甬道的面积是梯形面积的八分之一时，求甬道的宽；（3）根据设计的要求，甬道的宽不能超过6米.如果修建甬道的总费用（万元）与甬道的宽度成正比例关系，比例系数是5.7，花坛其余部分的绿化费用为每平方米0.02万元，那么当甬道的宽度为多少米时，所建花坛的总费用最少？最少费用是多少万元？图142009年南宁市中等学校招生考试 数学试题参考答案与评分标准二、填空题（本大题共6小题，每小题2分，共12分）13．60 14．32a b 15．25 16．4517．()40 18．420 三、（本大题共2小题，每小题满分6分，共 12分）19．解：()1200911sin 602-⎛⎫-+-- ⎪⎝⎭°=()12-+······················································································ 4分 =12-- ········································································································ 5分 3=- ··········································································································· 6分 20．解：()2111211x x x ⎛⎫+÷-- ⎪--⎝⎭ =()()11211x x x x x +--+-· ············································································· 3分 22x =+ ······································································································· 4分当x =22=+ ········································································ 5分4= ·················································································· 6分 四、（本大题共2小题，每小题满分10分，共20分）21．解：（1）900.3m n ==，； ······································································· 4分（2）图略． ·································································································· 6分 （3）比赛成绩的中位数落在：70分~80分． ························································ 8分 （4）获奖率为：6020100200+⨯%=40%（或0.3+0.1=0.4） ····································· 10分 22．解：（1）()04A ，、()31C ，； ······································································ 2分 （2）图略． ·································································································· 6分（3）AC =···························································································· 7分¼AA '= ························································································ 9分π2=···································································································· 10分 五、（本大题满分10分） 23．解：（1）ACO BCO APC BPC PAO PBO △≌△，△≌△，△≌△ ···················· 3分 （写出一个全等式子得1分） （2）PA Q 、PB 为O ⊙的切线PO ∴平分90APB PA PB PAO ∠=∠=，，° ················· 5分 PO AB ∴⊥ ····························································· 6分 ∴由圆的对称性可知：AOD S S =阴影扇形 ························ 7分Q 在Rt PAO △中，11603022APO APB ∠=∠=⨯=︒° 90903060AOP APO ∴∠=-∠=-︒=︒°° ························································· 8分 260π1360AODS S ⨯⨯∴==阴影扇形 ·········································································· 9分π6=·················································································· 10分 六、（本大题满分10分）24．解：（1）当0500x ≤≤时，设1y k x =甲，把()50028000，代入上式得：11280002800050056500k k =∴==， 56y x ∴=甲 ··································································································· 2分当500x ≥时，设2y k x b =+甲，把()50028000，、()100048000，代入上式得：2250028000100048000k b k b +=⎧⎨+=⎩ ····················································································· 3分 解得：2408000k b =⎧⎨=⎩··························································································· 4分408000y x ∴=+甲()()560500408000500x x y x x <⎧⎪∴=⎨+⎪⎩甲≤≥ ········································································ 5分 （2）当1600x =时，401600800072000y =⨯+=甲 ·········································· 6分1600y k =乙···································································· 7分①当y y <乙甲时，即：720001600k <得：45k > ··································································································· 8分②当y y >乙甲时，即：720001600k >得：045k << ······························································································ 9分③当y y =乙甲时，即720001600k =，45k ∴=答：当45k >时，选择甲工程队更合算，当045k <<时，选择乙工程队更合算，当45k =时，选择两个工程队的花费一样． ··································································· 10分 七、（本大题满分10分） 25．解：（1）AE EF ⊥Q2390∴∠+∠=°Q 四边形ABCD 为正方形90B C ∴∠=∠=° 1390∴∠+∠=°12∠=∠ ···························································1分90DAM ABE DA AB ∠=∠==Q °，DAM ABE ∴△≌△DM AE ∴=································································································· 9分 AE EP =Q DM PE ∴=∴四边形DMEP 是平行四边形． ···································································· 10分 （备注：作平行四边形DMEP ，并计算出AM 或BM 的长度，但没有证明点M 在AB 边上的扣1分）解法②：在AB 边上存在一点M ，使四边形DMEP 是平行四边形 ························· 8分 证明：在AB 边上取一点M ，使AM BE =，连接ME 、MD 、DP ． 90AD BA DAM ABE =∠=∠=，°Rt Rt DAM ABE ∴△≌△14DM AE ∴=∠=∠， ································ 9分 1590∠+∠=Q ° 4590∴∠+∠=°AE DM ∴⊥AE EP ⊥QDM EP ∴⊥∴四边形DMEP 为平行四边形 ······································································· 10分（备注：此小题若有其他的证明方法，只要证出判定平行四边形的一个条件，即可得1分） 八、（本大题满分10分）26．解：（1）横向甬道的面积为：()2120180150m 2x x += ··································· 2分 （2）依题意：2112018028015028082x x x +⨯+-=⨯⨯ ······································· 4分整理得：21557500x x -+=F A D CBE1 3 2B C E D A FP5 41M125150x x ==，（不符合题意，舍去） ····························································· 6分 ∴甬道的宽为5米．（3）设建设花坛的总费用为y 万元．()21201800.028******** 5.72y x x x x +⎡⎤=⨯⨯-+-+⎢⎥⎣⎦······································ 7分20.040.5240x x =-+当0.56.25220.04b x a =-==⨯时，y 的值最小． ··················································· 8分 因为根据设计的要求，甬道的宽不能超过6米，6x ∴=当米时，总费用最少． ·········································································· 9分最少费用为：20.0460.56240238.44⨯-⨯+=万元 ··········································· 10分。