25．(10分)
为极大地满足人民生活的需求，丰富市场供应，我区农村温棚设施农业迅速发展，温棚种植面积在不断扩大．在耕地上培成一行一行的矩形土埂，按顺序间隔种植不同农作物的方法叫分垄间隔套种．科学研究表明：在塑料温棚中分垄间隔套种高、矮不同的蔬菜和水果（同一种紧挨在一起种植不超过两垄），可增加它们的光合作用，提高单位面积的产量和经济效益．
现有一个种植总面积为540m的矩形塑料温棚，分垄间隔套种草莓和西红柿共24垄，种植的草莓或西红柿单种农作物的总垄数不低于10垄，又不超过14垄(垄数为正整数)，它们的占地面积、产量、利润分别如下： 西红柿 草莓 占地面积（m/垄） 30 15 22产量（千克/垄） 160 50 利润（元/千克） 1.1 1.6 （1）若设草莓共种植了x垄，通过计算说明共有几种种植方案？分别是哪几种？ （2）在这几种种植方案中，哪种方案获得的利润最大？最大利润是多少？
26． (10分)
如图，在边长为4的正方形ABCD中，点P在AB上从A向B运动，连接DP交AC于点Q．
（1）试证明：无论点P运动到AB上何处时，都有△ADQ≌△ABQ； （2）当点P在AB上运动到什么位置时，△ADQ的面积是正方形ABCD面积的
1； 6（3）若点P从点A运动到点B，再继续在BC上运动到点C，在整个运动过程中，当点P
运动到什么位置时，△ADQ恰为等腰三角形．
宁夏回族自治区2008年初中毕业暨高中阶段招生
数学试题参考答案及评分标准
说明：1． 除本参考答案外，其它正确解法可根据评分标准相应给分．
2． 涉及计算的题，允许合理省略非关键步骤．
3． 以下解答中右端所注的分数，表示考生正确做到这步应得的累计分．
一、选择题（3分×8=24分）
题号答案
二、填空题（3分×8=24分） 9．32 ；10． 25 ；11． 三、解答题(共24分) 17． 解：(B C D C A B B D 1 2 3 4 5 6 7 8 15002 ；12． 360 ；13．；14 ．210；15．10 ； 16．①②③ ．
2x?35321?)?(a2?1) a?1a?1=
2(a?1)?(a?1)?(a?1)(a?1)?a?3 ························································· 4分
(a?1)(a?1)当a?3?3时，
原式=3?3?3=3 ··························································································· 6分
B18．解：在Rt△ABC中, ∠C=90°, AB=15
BC4=, AB5∴ BC?12 ···················································· 3分 sinA=
AC?AB2?BC2?152?122?9
AC∴△ABC的周长为36 ·························································································· 5分
tanA=
BC4? ···································································································· 6分 AC319．解：（1） 被污染处的人数为11人 ············································································· 1分
设被污染处的捐款数为x元，则 11x+1460=50×38 解得 x=40
答：（1）被污染处的人数为11人，被污染处的捐款数为40元． ······························· 4分
（2）捐款金额的中位数是40元，捐款金额的众数是50元． ······························ -6分
20．解：（1）P(阴影）?P(白色）?31? ∴张红的设计方案是公平的． ························ 2分 62（2）能正确列出表格或画出树状图 ··································· 4分 ∵P(奇数）?4 P9?(偶数）554 ＞ ∴王伟的设计方案不公平- ······················· 6分 999四、解答题(共48分) 21．（1）方案三 ·················································································································· 2分
（2）正确填写下表 ······································································································· 4分
规律：商品标价接近600元的按促销方式②购买，商品标价接近800元的按促销方式①购买．或商品标价大于600元且小于720元按促销方式②购买，商品标价大于720元且小于800元按促销方式①购买 ········································································································· 6分 （其它表述正确，或能将两种购物方式抽象概括成一次函数并能正确解答的均可给分） 22．解：（1）如图，△OA1B1就是△OAB放大后的图象 ··················································· 2分 （2）由题意得： A1（4，0），B1（2，-4） 设线段A1B1所在直线的函数关系式为y?kx?b(k?0) 则??4x?b?0，?k?2， 解得? ?b??8?2k?b??4∴函数关系式为 y?2x?8 ······························································································· 6分 223．解：（1）x?2x?1?0 解得 x1?1?2， x2?1?2
∴图象与x轴的交点坐标为（1?2，0）和（1?2，0） ······································ 4分
b?24ac?b2?4?1?(?2)2???1 ???2 （2）?2a2?14a4?1∴顶点坐标为（1，?2）
将二次函数y?x2图象向右平移1个单位，再向下平移2个单位，
就可得到二次函数y?x2?2x?1的图象 ··································································· 8分
24．解：（1）图中共有三对全等三角形： DB≌AC②CE ③CB ·①△A△D△ABE≌△D△ABC≌△D······································ 3分
DB≌选择①△A△DAC证明
在⊙O中，∠ABD=∠DCA,∠BCA=∠BDA
CA=∠CAD ∴AD=∠BDA ∵BC∥AD ∴∠B∠C又∵AD?AD
DB≌AC ·∴△A△D························································· 5分 （2）图中与△ABE相似的三角形有： △DCE，△DBA， △ACD． ····································· 8分 25．解：（1）根据题意西红柿种了（24-x）垄
15x+30(24-x)≤540 解得 x≥12 ······················ 2分 ∵x≤14，且x是正整数 ∴x=12，13，14 ············· 4分 共有三种种植方案，分别是：
方案一：草莓种植12垄，西红柿种植12垄 方案二：草莓种植13垄，西红柿种植11垄 方案三：草莓种植14垄，西红柿种植10垄 ····························································· 6分 （2）解法一：方案一获得的利润：12×50×1.6+12×160×1.1=3072（元）
方案二获得的利润：13×50×1.6+11×160×1.1=2976（元） 方案三获得的利润：14×50×1.6+10×160×1.1=2880（元）
由计算知，种植西红柿和草莓各12垄，获得的利润最大，
最大利润是3072元 ················································································ 10分
解法二：若草莓种了x垄，设种植草莓和西红柿共可获得利润y元，则
y?1.6?50x?1.1?160(24?x)??96x?4224
∵k?-96＜0 ∴y随x的增大而减小 又∵12≤x≤14，且x是正整数
∴当x=12时，y最大=3072（元） ························································ 10分
26．（1）证明：在正方形ABCD中，
无论点P运动到AB上何处时，都有
AD=AB ∠DAQ=∠BAQ AQ=AQ
∴△ADQ≌△ABQ ···················································· 2分
(2)解法一：△ADQ的面积恰好是正方形ABCD面积的
1时， 6过点Q作QE⊥AD于E,QF⊥AB于F，则QE = QF
118AD?QE=S正方形ABCD= 2634 ∴QE= ···················································································································· 4分
3QEDEAP得 ?由△DEQ ∽△D 解得AP?2
APDA1∴AP?2时，△ADQ的面积是正方形ABCD面积的 ···································· 6分
6解法二：以A为原点建立如图所示的直角坐标系，过点Q作QE⊥y轴
于点E，QF⊥x轴于点F．
1184AD?QE=S正方形ABCD= ∴QE= 26334433 ∵点Q在正方形对角线AC上 ∴Q点的坐标为(，)
∴ 过点D（0，4），Q(
44,)两点的函数关系式为：y??2x?4 33 当y?0时，x?2 ∴P点的坐标为（2，0） ∴AP?2时，△ADQ的面积是正方形ABCD面积的
1． ···································· 6分 6（3）若△ADQ是等腰三角形，则有 QD=QA或DA=DQ或AQ=AD ①当点P运动到与点B重合时，由四边形ABCD是正方形知 QD=QA 此时△ADQ是等腰三角形
②当点P与点C重合时，点Q与点C也重合，
此时DA=DQ, △ADQ是等腰三角形 ····································· 8分 ③解法一：如图，设点P在BC边上运动到CP?x时，有AD=AQ
Q=∠CPQ ∵ AD∥BC ∴∠ADD=∠CQP ∠ADQ=∠AQD 又∵∠AQ
∴∠CQP=∠CPQ ∴ CQ=CP=x
∵AC=42 AQ = AD =4 ∴x?CQ?AC?AQ?42?4
即当CP?42?4时，△ADQ是等腰三角形 ········································· 10分 解法二：以A为原点建立如图所示的直角坐标系，设点P在BC上运动到BP?y时，有AD=AQ．
过点Q作QE⊥y轴于点E，QF⊥x轴于点F,则QE?QF 在Rt△AQF中，AQ?4，∠QAF=45° ∴QF=AQ?sin45°=22 ∴Q点的坐标为（22，22）
∴过D、Q两点的函数关系式：y?(1?2)x+4
当x=4时，y?8?42 ∴P点的坐标为（4，8-42）．
∴当点P在BC上运动到BP?8?42时，△ADQ是等腰三角形． ···························· 10分