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上海历年中考数学压轴题复习(试题附答案)

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所以xC?xD=﹣yH.

点评:本题主要考查了二次函数的应用、一次函数解析式的确定、图形面积的求法、函数图象的交点等知识点.

2005年上海市初中毕业生统一学业考试数学试卷

1、 (本题满分12分,每小题满分各为4分)

在△ABC中,∠ABC=90°,AB=4,BC=3,O是边AC上的一个动点,以点O为圆心作半圆,与边AB相切于点D,交线段OC于点E,作EP⊥ED,交射线AB于点P,交射线CB于点F。

(1) 如图8,求证:△ADE∽△AEP; (2) 设OA=x,AP=y,求y关于x的函数解析式,并写出它的定义域; (3) 当BF=1时,求线段AP的长. J 2006 年上海市初中毕业生统一学业考试数学试卷 25(本题满分14分,第(1)小题满分4分,第(2)小题满分7分,第(3)小题满分3分) 已知点P在线段AB上,点O在线段AB的延长线上。以点O为圆心,OP为半径作圆,点C是圆O上的一点。 (1) 如图9,如果AP=2PB,PB=BO。求证:△CAO∽△BCO; (2) 如果AP=m(m是常数,且m〉1),BP=1,OP是OA、OB的比例中项。当点C在圆O上运动时,求AC:BC的值(结果用含m的式子表示); (3) 在(2)的条件下,讨论以BC为半径的圆B和以CA为半径的圆C的位置关系,并写出相应m的取值范围。 C A P B O 图9

25.(1)证明:

AP?2PB?PB?BO?PO,?AO?2PO.

AOPO······························································································· (2分) ???2. ·

POBO······································································································· (1分) PO?CO, ·

AOCO.∠COA?∠BOC,?····················· (1分) △CAO∽△BCO. ·??COBO(2)解:设OP?x,则OB?x?1,OA?x?m,OP是OA,OB的比例中项,

······················································································· (1分) ?x2??x?1??x?m?, ·

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mm,即OP?. ··········································································· (1分) m?1m?11. ····································································································· (1分) ?OB?m?1OAOP, OP是OA,OB的比例中项,即?OPOBOAOC. ············································································· (1分) OP?OC,??OCOB得x?设圆O与线段AB的延长线相交于点Q,当点C与点P,点Q不重合时,

················································· (1分) ∠AOC?∠COB,?△CAO∽△BCO. ·ACOC. ······································································································· (1分) ??BCOBACOCOPAC????m;当点C与点P或点Q重合时,可得?m, BCOBOBBC?当点C在圆O上运动时,AC:BC?m; ····················································· (1分) (3)解:由(2)得,AC?BC,且AC?BC??m?1?BC?m?1?, AC?BC??m?1?BC,圆B和圆C的圆心距d?BC, 显然BC??m?1?BC,?圆B和圆C的位置关系只可能相交、内切或内含. 当圆B与圆C相交时,?m?1?BC?BC??m?1?BC,得0?m?2, ······················································································ (1分) m?1,?1?m?2; ·当圆B与圆C内切时,?m?1?BC?BC,得m?2; ··································· (1分) 当圆B与圆C内含时,BC??m?1?BC,得m?2. (1分) 2007年上海市初中毕业生统一学业考试 25.(本题满分14分,第(1)小题满分4分,第(2),(3)小题满分各5分) 已知:∠MAN?60,点B在射线AM上,AB?4(如图10).P为直线AN上一动点,以BP为边作等边三角形BPQ(点B,P,Q按顺时针排列),O是△BPQ的外心. (1)当点P在射线AN上运动时,求证:点O在∠MAN的平分线上; (2)当点P在射线AN上运动(点P与点A不重合)时,AO与BP交于点C,设AP?x,

ACAO?y,求y关于x的函数解析式,并写出函数的定义域;

(3)若点D在射线AN上,AD?2,圆I为△ABD的内切圆.当△BPQ的边BP或BQ与圆I相切时,请直接写出点A与点O的距离.

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25.(1)证明:如图4,连结OB,OP,

·································································· 1分 O是等边三角形BPQ的外心,?OB?OP, ·

圆心角?BOP?360?120. 3当OB不垂直于AM时,作OH?AM,OT?AN,垂足分别为H,T. 由?HOT??A??AHO??ATO?360,且?A?60,

?AHO??ATO?90,??HOT?120.

··············································································································· 1分 ??BOH??POT. ···································································································· 1分 ?Rt△BOH≌Rt△POT. ····································································· 1分 ?OH?OT.?点O在?MAN的平分线上. ·当OB?AM时,?APO?360??A??BOP??OBA?90. 即OP?AN,?点O在?MAN的平分线上. 综上所述,当点P在射线AN上运动时,点O在?MAN的平分线上. 图4 (2)解:如图5, 图5 AO平分?MAN,且?MAN?60, ······································································································ 1分 ??BAO??PAO?30. ·由(1)知,OB?OP,?BOP?120, ??CBO?30,??CBO??PAC. ·········································································· 1分 ?BCO??PCA,??AOB??APC. ·

?△ABO∽△ACP. ABAO.?ACAO?ABAP.?y?4x. ····························································· 1分 ??ACAP定义域为:x?0. ····················································································································· 1分 (3)解:①如图6,当BP与圆I相切时,AO?23; ···················································· 2分 ②如图7,当BP与圆I相切时,AO?4··································································· 1分 3; ·

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③如图8,当BQ与圆I相切时,AO?0. ··········································································· 2分

图6 图7

图8

2008年上海市中考数学试卷 25.(本题满分14分,第(1)小题满分5分,第(2)小题满分4分,第(3)小题满分5分) 已知AB?2,AD?4,?DAB?90,AD∥BC(如图13).E是射线BC上的动点(点,M是线段DE的中点. E与点B不重合)(1)设BE?x,△ABM的面积为y,求y关于x的函数解析式,并写出函数的定义域; (2)如果以线段AB为直径的圆与以线段DE为直径的圆外切,求线段BE的长; (3)联结BD,交线段AM于点N,如果以A,N,D为顶点的三角形与△BME相似,求线段BE的长.

D A A (1)取ABD 25.解:中点H,联结MH, 1M M为DE的中点,?MH∥BE,MH?(BE?AD). ········ (1分)

2又AB?BE,?MH?AB. ····················· (C 1分)

B B E C 备用图 图13 11··········· (2分)(1分) ?S△ABM?ABMH,得y?x?2(x?0); 22(2)由已知得DE?(x?4)2?22. ·················· (1分) 以线段AB为直径的圆与以线段DE为直径的圆外切, ?MH?1111······· (2分) AB?DE,即(x?4)??2?(4?x)2?22?. ?2222?44解得x?,即线段BE的长为; ··················· (1分)

33(3)由已知,以A,N,D为顶点的三角形与△BME相似, 又易证得?DAM??EBM. ······················ (1分) 由此可知,另一对对应角相等有两种情况:①?ADN??BEM;②?ADB??BME. ①当?ADN??BEM时,AD∥BE,??ADN??DBE.??DBE??BEM.

··············· (2分) ?DB?DE,易得BE?2AD.得BE?8;

②当?ADB??BME时,AD∥BE,??ADB??DBE. ??DBE??BME.又?BED??MEB,?△BED∽△MEB.

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?DEBE122?(x?4)2?22?(x?4)2. ,即BE2?EMDE,得x2??2BEEM解得x1?2,x2??10(舍去).即线段BE的长为2. ··········· (2分) 综上所述,所求线段BE的长为8或2.

2009年上海市初中毕业统一学业考试

25.(本题满分14分,第(1)小题满分4分,第(2)小题满分5分,第(3)小题满分5

分)

已知?ABC?90°为线段BD上的动点,点Q在射线,AB?2,BC?3,AD∥BC,PAB上,且满足PQAD(如图8所示). ?PCAB(1)当AD?2,且点Q与点B重合时(如图9所示),求线段PC的长; (2)在图8中,联结AP.当AD?为x,3,且点Q在线段AB上时,设点B、Q之间的距离2S△APQS△PBC?y,其中S△APQ表示△APQ的面积,S△PBC表示△PBC的面积,求y关于x的函数解析式,并写出函数定义域; (3)当AD?AB,且点Q在线段AB的延长线上时(如图10所示),求?QPC的大小. D (2009解:(1)AD=2,且Q点与D B点重合,根据题意,∠PBC=∠PDA,A A D A 年上海25题解析)。P 因为∠A=90PQ/PC=AD/AB=1,所以:△PQC为等腰直角三角形,BC=3,所以:PC=3/2, P (2)如图:添加辅助线,根据题意,两个三角形的面积可以分别表示成S1,S2,高分别是P H,h, Q (2-x)H/2=(2*3/2)/2-(x*H/2)-(3/2)*(2-h)/2 则:S1=B 2C C S2=3*h/2得: B 因为两S1/S2=y,消去C H,h,B (Q) 图8 图9 Y=-(1/4)*x+(1/2), 图10 Q 定义域:当点P运动到与D点重合时,X的取值就是最大值,当PC垂直BD时,这时X=0,连接DC,作QD垂直DC,由已知条件得:B、Q、D、C四点共圆,则由圆周角定理可以推知:三角形QDC相似于三角形ABD QD/DC=AD/AB=3/4,令QD=3t,DC=4t,则:QC=5t,由勾股定理得: 直角三角形AQD中:(3/2)^2+(2-x)^2=(3t)^2 直角三角形QBC中:3^2+x^2=(5t)^2

整理得:64x^2-400x+301=0(8x-7)(8x-43)=0 得x1=7/8x2=(43/8)>2(舍去)所以函数: Y=-(1/4)*x+1/2的定义域为[0,7/8] (3)因为:PQ/PC=AD/AB,假设PQ不垂直PC,则可以作一条直线PQ′垂直于PC,与AB交于Q′点,

则:B,Q′,P,C四点共圆,由圆周角定理,以及相似三角形的性质得:

PQ′/PC=AD/AB,

又由于PQ/PC=AD/AB所以,点Q′与点Q重合,所以角∠QPC=90

上海历年中考数学压轴题复习(试题附答案)

海量资源,欢迎共阅所以xC?xD=﹣yH.点评:本题主要考查了二次函数的应用、一次函数解析式的确定、图形面积的求法、函数图象的交点等知识点.2005年上海市初中毕业生统一学业考试数学试卷1、(本题满分12分,每小题满分各为4分)在△ABC中,∠ABC=90°,AB=4,BC=3,O是边AC上的一个动点,以点O为
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