09年莆田中考试卷

200年莆田市初中毕业、升学考试试卷数 学（满分：150分，考试时间：120分钟）一、细心填一填（本大题共10小题，每小题4分，共40分．直接把答案填在题中的横线上．） 1．3-的相反数是 ．2．2009年莆田市参加初中毕业、升学考试的学生总人数约为43000人，将43000用科学记数法表示是___________．3．在组成单词“Probability ”（概率）的所有字母中任意取出一个字母，则取到字母“b ”的概率是 ．4．如图，A B 、两处被池塘隔开，为了测量A B 、两处的距离，在AB 外选一适当的点C ，连接AC BC 、，并分别取线段AC BC 、的中点E F 、，测得EF =20m ，则AB =__________m ．5．一罐饮料净重500克，罐上注有“蛋白质含量≥0.4%”，则这罐饮料中蛋白质的含量至少为__________克．6．如图，菱形ABCD 的对角线相交于点O ，请你添加一个条件： ，使得该菱形为正方形．7．甲、乙两位同学参加跳高训练，在相同条件下各跳10次，统计各自成绩的方差得22S S <乙甲，则成绩较稳定的同学是___________．（填“甲”或“乙”）8．已知1O ⊙和2O ⊙的半径分别是一元二次方程()()120x x --=的两根，且122OO =，则1O ⊙和2O ⊙的位置关系是 ．9．出售某种文具盒，若每个获利x 元，一天可售出()6x -个，则当x = 元时，一天出售该种文具盒的总利润y 最大．10．如图，在x 轴的正半轴上依次截取112233445OA A A A A A A A A ====，过点12345A A A A A 、、、、分别作x 轴的垂线与反比例函数()20y x x=≠的图象相交于点12345P P P P P 、、、、，得直角三角形111223OP A A P A A P A AP A A PA2、、、、，并设其面积分别为12345S S S S S 、、、、，则5S 的值为 ． 二、精心选一选（本大题共6小题，每小题4分，共24分，每小题给出的四个选项中有且只有一个是正确的，请把正确选项的代号写在题后的括号内，答对的得4分；答错、不答或答案超过一个的一律得0分）． 11x 的取值范围是（ ）A ．x ≥0B ．0x <C ．0x ≠D ．0x >(第4题图) ABCDD C BAO （第6题图）O（第10题图） 212．下列各式运算正确的是（ ）A ．22a a a ÷= B ．()2224aba b =C ．248a a a ·= D ．55ab b a -= 13．如图是一房子的示意图，则其左视图是（ ）A ．B ．C．Ｄ．14．某班5位同学参加“改革开放30周年”系列活动的次数依次为12333、、、、，则这组数据的众数和中位数分别是（ ）A ．22、B． 2.43、 C ． 32、 D ．33、 15．不等式组2410x x <⎧⎨+>，的解集在数轴上表示正确的是（ ）A BC D16．如图1，在矩形MNPQ 中，动点R 从点N 出发，沿N →P →Q →M 方向运动至点M 处停止．设点R 运动的路程为x ，MNR △的面积为y ，如果y 关于x 的函数图象如图2所示，则当9x =时，点R 应运动到（ ）A ．N 处B ．P 处C ．Q 处D ．M 处三、耐心做一做（本大题共9小题，共86分．解答应写出必要的文字说明、证明过程或演算步骤．）17．（8分）计算：0133⎛⎫⎪⎝⎭．18．（8分）先化简，再求值：2244242x x x x x x +++÷---，其中1x =．19．（8分）已知：如图在ABCD中，过对角线BD 的中点O 作直线EF 分别交DA 的延长线、AB DC BC 、、的延长线于点E M N F 、、、．（1）观察图形并找出一对全等三角形：△________≌△____________，请加以证明；（第16题图）（图1）（2）在（1）中你所找出的一对全等三角形，其中一个三角形可由另一个三角形经过怎样的变换得到？20．（8分）（1）根据下列步骤画图．．并标明相应的字母:（直接在图１中画图） ①以已知线段AB （图1）为直径画半圆O ；②在半圆O 上取不同于点A B 、的一点C ，连接AC BC 、； ③过点O 画OD BC ∥交半圆O 于点D ． （2）尺规作图．．：（保留作图痕迹，不要求写作法、证明） 已知：AOB ∠（图2）． 求作：AOB ∠的平分线．21．（8分）某校课题研究小组对本校九年级全体同学体育测试情况进行调查，他们随机抽查部分同学体育测试成绩（由高到低分A B C D 、、、四个等级），根据调查的数据绘制成如下的条形统计图和扇形统计图．请根据以上不完整的统计图提供的信息，解答下列问题：（1）该课题研究小组共抽查了__________名同学的体育测试成绩，扇形统计图中B 级所占的百分比b =___________；（2）补全条形统计图；（3）若该校九年级共有400名同学，请估计该校九年级同学体育测试达标（测试成绩C 级以上，含C 级）约有___________名．22．（10分）已知，如图，BC 是以线段AB 为直径的O ⊙的切线，AC 交O ⊙于点D ，过点D 作弦DE AB ⊥，垂足为点F ，连接BD BE 、．． （1）仔细观察图形并写出四个不同的正确结论：①________，②________ ，③________，④____________（不添加其它字母和辅助线，不必证明）；图2OBABA图1 （第20题图）E B M ODN C（第19题图）A等级（第21题图）（2）A ∠=30°，CD=3，求O ⊙的半径r ．23．（10分）面对全球金融危机的挑战，我国政府毅然启动内需，改善民生．国务院决定从2009年2月1日起，“家电下乡”在全国范围内实施，农民购买人选产品，政府按原价购买总额的．．．．．13%．．．给予补贴返还．某村委会组织部分农民到商场购买人选的同一型号的冰箱、电视机两种家电，已知购买冰箱的数量是电视机的2倍，且按原价购买冰箱总额为40000元、电视机总额为15000元．根据“家电下乡”优惠政策，每台冰箱补贴返还的金额比每台电视机补贴返还的金额多65元，求冰箱、电视机各购买多少台？ （1）设购买电视机台，依题意填充下列表格： （2）列出方程（组）并解答．24.（12分）已知：等边ABC △的边长为a ． 探究（1）：如图１，过等边ABC △的顶点A B C 、、依次作AB BC CA 、、的垂线围成MNG △，求证：MNG △是等边三角形且．MN =；探究（2）：在等边ABC △内取一点O ，过点O 分别作OD AB OE BC OF CA ⊥⊥⊥、、，垂足分别为点D E F 、、． ①如图2，若点O 是ABC △的重心，我们可利用三角形面积公式及等边三角形性质得到两个正确结论（不必证明）：结论1．OD OE OF ++=；结论2．32AD BE CF a ++=；②如图3，若点O 是等边ABC △内任意一点，则上述结论12、是否仍然成立？如果成立，请给予证明；如果不成立，请说明理由．（第22题图）25．（14分）已知，如图1，过点()01E -，作平行于x 轴的直线l ，抛物线214y x =上的两点A B 、的横坐标分别为-1和4，直线AB 交y 轴于点F ，过点A B 、分别作直线l 的垂线，垂足分别为点C 、D ，连接CF DF 、．（1）求点A B F 、、的坐标； （2）求证：CF DF ⊥；（3）点P 是抛物线214y x =对称轴右侧图象上的一动点，过点P 作PQ PO ⊥交x 轴于点Q ，是否存在点P 使得OPQ △与CDF △相似？若存在，请求出所有符合条件的点P 的坐标；若不存在，请说明理由．200年莆田市初中毕业、升学考试试卷 数学试卷参考答案及评分标准说明：（一）考生的解法与“参考答案”不同时，可参照“答案的评分标准”的精神进行评分（二）如解答的某一步计算出现错误，这一错误没有改变后续部分的考查目的，可酌情给分，但原则上不超过后面应得分数的二分之一；如属严重的概念性错误，就不给分． （三）以下解答各行右端所注分数表示正确做完该步骤应得的累计分数． （四）评分的最小单位是１分，得分或扣分都不能出现小数． 一、细心填一填（本大题共10小题，每小题4分，共40分．）1．3 2．44.310⨯（不必考虑有效数字） 3．2114．40 5．2 6．AB BC ⊥或AC BD =或AO BO =等 7．甲 8．相交 9．3 10．15二、精心选一选（本大题共6小题，每小题4分，共24分．） 11．A 12．B 13．C 14．D 15．A 16．C（图1）备用图（第25题图）N MAG C B AF E BDA F EB D（图1）（图2） （图3）（第24题图）O AF E B D（图4）O O三、耐心做一做（本题共9小题，共86分）17．（1）解：原式=341+ ······························································································· 6分=············································································································ 8分注：33=2分）4=（2分），13⎛⎫ ⎪⎝⎭=1（2分）18.解：原式=()()()222222x x xx x x +-⨯-+-+ ················································································· 6分=1x - ················································································································ 7分当1x =时原式=110-= ····························································································· 8分注：()()()22222442422?22x x x x x x x x x x +-++=+-=+-÷=⨯-+、、？（各2分） 19． （1）DOE BOF ①△≌△； ····························2分证明：∵四边形ABCD 是平行四边形∴AD BC ∥ ··············································3分 ∴EDO FBO E F ∠=∠∠=∠， ·································4分又∵OD OB =∴()DOE BOF AAS △≌△ ··················································································· 5分BOM DON ②△≌△ ····························································································· 2分证明：∵四边形ABCD 是平行四边形∴AB CD ∥ ·············································································································· 3分∴MBO NDO BMO DNO ∠=∠∠=∠， ······························································· 4分 又∵BO DO =∴()BOM DON AAS △≌△ ················································································· 5分ABD CDB ③△≌△；···························································································· 2分证明：∵四边形ABCD 是平行四边形∴AD CB AB CD ==， ··························································································· 3分又∵BD DB = ·········································································································· 4分∴()ABD CDB SSS △≌△ ····················································································· 5分 （2）绕点O 旋转180°后得到或以点O 为中心作对称变换得到． ································ 8分 20．（1）正确完成步骤①、②、③，各得1分，字母标注完整得1分，满分4分．（2）说明：①以点O 为圆心，以适当长为半径作弧交OA OB 、于两点C D 、 ·············· 5分②分别以点C D 、为圆心，以大于12CD 长为半径作弧，两弧相交于点E ·························································································· 7分EB M O DNC（第19题图）A③作射线OE ·································································································· 8分21．（1）80 ························································································2分40% ···································································································4分 （2）补全条形图（如右图） ···························································6分（3）380 ······························································································· 8分 22．（1）BC AB AD BD ⊥⊥，，DF FE BD BE ==，，BDF BEF △≌△，BDF △∽BAD △，BDF BEF ∠=∠，A E DE BC ∠=∠，∥等 （每写出一个正确结论得１分，满分４分．）（2）解:AB 是O ⊙的直径90ADB ∴∠=° ·································· 5分 又30E ∠= ° 30A ∴∠=° ······················································································6分12BD AB r ∴== ············································································7分 又BC 是O ⊙的切线 90CBA ∴∠=° ···················································································· 8分 60C ∴∠=︒在Rt BCD △中，CD =tan 60BD rDC ∴==° ············································································································· 9分 2r ∴= ·········································································································································· 10分23（2）解:依题意得2x -65x= ······························································· 7分解得10x = ··································································································································· 8分经检验10x =是原分式方程的解 ·································································································9分 220x ∴=．答:冰箱、电视机分别购买20台、10台 ······································· 10分24．证明：如图1，ABC △为等边三角形 60ABC ∴∠=°BC MN BA MG ⊥⊥ ，∴90CBM BAM ∠=∠=°（第22题图）NMAG CB （图1）A 图1（第20题图） 图2 O B A E D O CCD（第21题图）9030ABM ABC ∴∠=∠=︒°- ·····················································1分 9060M ABM ∴∠=︒∠=︒- ··························································2分 同理：60N G ∠=∠=︒ MNG ∴△为等边三角形． ··········································································································· 3分在Rt ABM △中，sin sin 60AB a BM a M ===︒在Rt BCN △中，tan tan 60BC a BN N ===︒ ···································································· 4分MN BM BN ∴=+= ····················································································································· 5分 （2）②：结论1成立.证明；方法一：如图2，连接AO BO CO 、、 由ABC AOB BOC AOC S S S S =++△△△△=()12a OD OE OF ++ ··············· 7分 作AH BC ⊥，垂足为H ，则sin sin 602AH AC ACB a a =∠=⨯︒=1122ABC S BC AH a ∴==△· ()1122a OD OE OF a ∴++=OD OE OF ∴++=··········································································································· 8分 方法二：如图3，过点O 作GH BC ∥，分别交AB AC 、于点G H 、，过点 H 作HM BC ⊥于点M ， 6060DGO B OHF C ∴∠=∠=∠=∠=°，° AGH ∴△是等边三角形 GH AH ∴=··········································································· 6分 OE BC ⊥ OE HM ∴∥∴四边形OEMH 是矩形 HM OE ∴= ············································································ 7分在Rt ODG △中，sin sin 60OD OGDGO OG =∠=︒=·· 在Rt OFH △中，sin sin 60OF OHOHF OH =∠=︒=·· A FCE BD（图2）OHAF CE BD （图3）OM HG在Rt HMC △中，sin sin 602HM HC C HC HC ==︒=··222OD OE OF OD HM OF HC ∴++=++=++)GH HC AC =+==····························· 8分 （2）②:结论2成立．证明：方法一：如图４，过顶点A B C 、、依次作边AB BC CA 、、的垂线围成MNG △，由（1）得M N G △为等边三角形且MN = ······················································· 9分 过点O 分别作OD MN '⊥于D ',OE NG '⊥于NG 于点E OF MG ''⊥，于点F '由结论1得：32OD OE OF MN a '+'+'=== ········································································ 10分 又OD AB AB MG OF MG ⊥⊥'⊥ ，，90ADO DAF OF A ∴∠=∠'=∠'=︒ ∴四边形ADOF '为矩形 OF ∴'=AD同理：OD BE '=，OE CF '= ··································································································· 11分32AD BE CF OD OE OF a ∴++='+'+'=····················································································· 12分 方法二：（同结论1方法二的辅助线） 在Rt OFH △中，tan 3OF FH OHF ==∠在Rt HMC △中，sin 3HM HC C == ··························· 9分CF HC FH ∴=+=同理：AD BE =+=，···················································· 10分 AD BE CF ∴++=333333OF OD +++++)OD OE OF ++ ················································································································ 11分AF CE BD（图3）OM HG A F CEBD（图4）O F 'D 'MGN E '。