2003年辽宁省数学中考试题答案

数学试题答案及评分标准第1页（共7页）辽宁省2003年中等学校招生考试数学试题参考答案及评分标准一、（选两个或两个以上答案不给分）1．B2．C3．B4．C5．C6．C7．B8．D9．B10．A二、11．x≥1且x≠212．313．8，7.514．4215．七16．217．y2-8y-20=0（或写成y2-20=8y）18．3.619．（312）米20．15°或75°（注：13题错1个扣1分，顺序错不给分；17题写成分式方程不给分；19题不写单位扣1分；20题只写对一解扣1分）三、21．()()yxxyyxyxxyxyxyxy··································2分xyxy····························································································4分当x=2，y=3时，原式23523·························································6分（注：不化简，直接代数求值，按相应步骤给分）22．如图（O点找对）··················································3分（切线画对）··················································6分（注：不用尺规作图，不给分，没有保留作图痕迹不给分）ABMOC数学试题答案及评分标准第2页（共7页）23．（1）（频数）12，（频率）0.24······························································2分（2）补全频率分布直方图···································································4分（3）50····························································································6分（4）80.5～90.5·················································································8分（5）216人·····················································································10分（注：（1）中每空1分，（2）中直方图1个1分，（3）中样本容量写单位的扣1分）四、说明：本题给分点由两部分组成，一部分是图形设计（满分5分），按设计合理性和测量数据多少给分（5分、3分、1分、0分）；另一部分是依据图形计算（满分5分）。对不同设计方案（如1、2、3），同一图形可能字母标记不同，但只要计算正确即可得5分。根据不合理方案（如图d、e），计算正确给3分。24．解：方案1：（1）如图a（测三个数据）···················5分（2）解：设HG=x在Rt△CHG中CG=xcotβ在Rt△DHM中DM=（x－n）cotα∴xcotβ=（x－n）cotα····························8分∴x=cotcotcotn······································10分AHGBDCαβn方案1图aMHAGBDCαγnmM数学试题答案及评分标准第3页（共7页）方案2：（1）如图b（测四个数据）···················3分（2）解：设HG=x在Rt△AHM中AM=(x－n)cotγ在Rt△DHM中DM=（x－n）cotα∴(x－n)cotγ=（x－n）cotα+m·················6分∴x=cotcotcotcotnnm·························8分方案3：（1）如图c（测五个数据）···················1分（2）参照方案1（2）或方案2（2）给分······6分注：①如果在设计和计算中，考虑了测倾器高度，参照以上标准给分.②以下两种方案（图d、e）或其他与其相似的图形不给分，但如果计算正确给3分．AHGBDCαβnMHAGBDCαγnmβ方案3图cM图d图eAHGBDCγβnMm数学试题答案及评分标准第4页（共7页）五、25．解：（1）设s与t的函数关系式为s=at2+bt+c由题意得1.54222552.5abcabcabc（或1.54220abcabcc）解得1220abc∴s=2122tt······················································································4分（2）把s=30代入s=2122tt得30=2122tt解得t1=10，t2=-6（舍）答：截止到10月末公司累积利润可达到30万元·······································7分（3）把t=7代入，得s=212172710.522把t=8代入，得s=2182816216-10.5=5.5答：第8个月公司获利润5.5万元．······················································10分数学试题答案及评分标准第5页（共7页）六、26．解：设每周参观人数与票价之间的一次函数关系式为y=kx+b由题意得107000154500kbkb·····································································2分解得50012000kb∴y=-500x+12000·················································································4分根据题意，得xy=40000即x(-500x+12000)=40000···································································6分x2-24x+80=0解得x1=20x2=4·············································································8分把x1=20，x2=4分别代入y=-500x+12000中得y1=2000，y2=10000······································································10分因为控制参观人数，所以取x=20，y=2000··············································11分答：每周应限定参观人数是2000人，门票价格应是20元．·······················12分（注：其他方法按相应步骤给分）数学试题答案及评分标准第6页（共7页）七、27．（1）证明：①连结BD∵AB是⊙O的直径∴∠ADB＝90°∴∠AGC＝∠ADB＝90°又∵ACDB是⊙O内接四边形∴∠ACG＝∠B∴∠BAD＝∠CAG······································3分②连结CF∵∠BAD＝∠CAG∠EAG＝∠FAB∴∠DAE＝∠FAC又∵∠ADC＝∠F∴△ADE∽△AFC··············································································5分∴ADAE=AFAC∴AC·AD＝AE·AF··············································································6分（其他方法相应给分）（2）①图形正确················································8分②两个结论都成立，证明如下：①连结BC∵AB是直径∴∠ACB＝90°∴∠ACB＝∠AGC＝90°∵GC切⊙O于C图(a)BOAFDCGEl·BOA·EC(D)GF数学试题答案及评分标准第7页（共7页）∴∠GCA＝∠ABC∴∠BAC＝∠CAG（即∠BAD＝∠CAG）·······10分②连结CF∵∠CAG＝∠BAC，∠GCF＝∠GAC∴∠GCF＝∠CAE，∠ACF＝∠ACG－∠GFC，∠E＝∠ACG－∠CAE∴∠ACF＝∠E∴△ACF∽△AEC∴ACAF=AEAC∴AC2＝AE·AF（即AC·AD＝AE·AF）·········································12分（注：其他方法证明，按相应步骤给分）图(b)数学试题答案及评分标准第8页（共7页）八、28．解：（1）直线y=228x与x轴、y轴分别交于点C、P∴C（22，0），P(0,-8)∴cot∠OCD=22cot∠OPC=22∴∠OCD=∠OPC∵∠OPC+∠PCO=90°∴∠OCD+∠PCO=90°∴PC是⊙D的切线······································5分（2）设直线PC上存在一点E(x,y)，使S△EOP＝4S△CDO118412222x解得x=±2由228yx可知：当x=2时，y=-12，当x=-2时，y=-4··············································9分∴在直线PC上存在点E（2，-12）或（-2，-4）使S△EOP＝4S△CDO······10分（注：只求出一个点，扣2分）（3）解法一：作直线PF交劣弧AC于F，交⊙D于Q，连结DQ由切割线定理得：PC2＝PF·PQ·····················①在△CPD和△OPC中∵∠PCD＝∠POC＝90°∠CPD＝∠OPC∴△CPD∽△OPC∴PCPDPOPC＝yxOD(0,1)APCB··QF⌒数学试题答案及评分标准第9页（共7页）即PC2＝PO·PD·········································②由①、②得：PO·PD＝PF·PQ，又∵∠FPO＝∠DPQ∴△FPO∽△DPQPFPD=FODQ，即m9=n3∴m=3n···························································································13分（2n22）··················································································14分解法二：作直线PF交劣弧AC于F设F(x,y)，作FM⊥y轴，M为垂足，连结DF，∵m2-(8+y)2=x2n2-y2=x2∴m2-64-16y-y2=n2-y2即m2-64-16y=n2·········①又∵32-(1-y)2=x2∴32-(1-y)2=n2-y2解得y=282n··············②将②代入①，解得：m=3n，m=-3n(舍)∴m=3n··················································13分(2n22)··