2008年安徽省芜湖市中考数学试题及答案

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∴ y?1?2x?3x?3x2(x?0)． ··············································· 8分 222．（本小题满分9分） 解：（1）用列举法：（ AB，DEF） ， （ AB，DEG） ， （ AB，DFG） ， （ AB，EFG） ， （ AC，DEF） ， （ AC，DEG）， （ AC，DFG） （ AC，EFG）， （ BC，DEF） ， （ BC，DEG）， （ BC，DFG）， （ BC，EFG） 共12种可能的选择方式． ············ 6分 用树形图法：

·········································································································································· 6分 （2） 小宝恰好上午选中A．太空世界，同时下午选中G． 海螺湾这两个项目的概率为P?61?． ························································································································· 9分 12223．（本小题满分12分） （1） 证明：由已知DE⊥DB，⊙O是Rt△BDE的外接圆，∴BE是⊙O的直径，点O是BE的中点，连结OD， ················································································································ 1分 ∵?C?90，∴?DBC??BDC?90． 又∵BD为∠ABC的平分线，∴?ABD??DBC． ∵OB?OD，∴?ABD??ODB． ∴?ODB??BDC?90，即∴?ODC?90 ····································································· 4分 又∵OD是⊙O的半径， ∴AC是⊙O的切线． ··························································· 5分 （2） 解：设⊙O的半径为r，

22222在Rt△ABC中， AB?BC?CA?9?12?225，

????∴AB?15 ············································································· 7分 ∵?A??A，?ADO??C?90，∴△ADO∽△ACB．

?AOOD15?rr??． ．∴ABBC1594545∴r?．∴BE? ······················································ 10分

84∴

?又∵BE是⊙O的直径．∴?BFE?90．∴△BEF∽△BAC

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45EFBE3∴···························································································· 12分 ??4?． ·ACBA15424．（本小题满分15分） 解： （1）

过C点向x轴作垂线，垂足为D，由位似图形性质可知： △ABO∽△ACD， ∴AOBO4??． ADCD9由已知A(?4,0)，B(0,4)可知： AO?4,BO?4． ∴AD?CD?9．∴C点坐标为(5,9)． ····················· 2分 直线BC的解析是为：

y?4x?0? 9?45?0化简得： y?x?4 ····················································· 3分 （2）设抛物线解析式为y?ax2?bx?c(a?0)，由题4?c??意得：?9?25a?5b?c ， ?b2?4ac?0?1?a??225?a1?1?4??解得： ?b1??4?b2?5?c?4??1?c2?4 ??∴解得抛物线解析式为y1?x2?4x?4或y2?又∵y2?124x?x?4． 255124x?x?4的顶点在x轴负半轴上，不合题意，故舍去． 2552∴满足条件的抛物线解析式为y?x?4x?4 ····································································· 5分 （准确画出函数y?x?4x?4图象） ················································································ 7分 （3） 将直线BC绕B点旋转与抛物线相交与另一点P，设P到 直线AB的距离为h， 故P点应在与直线AB平行，且相距32的上下两条平行直线l1和l2上． ······················· 8分 由平行线的性质可得：两条平行直线与y轴的交点到直线BC的距离也为32． 如图，设l1与y轴交于E点，过E作EF⊥BC于F点，

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在Rt△BEF中EF?h?32，?EBF??ABO?45，

∴BE?6．∴可以求得直线l1与y轴交点坐标为(0,10)···················································· 10分 同理可求得直线l2与y轴交点坐标为(0,?2) ······································································ 11分 ∴两直线解析式l1:y?x?10；l2:y?x?2．

??y?x2?4x?4?y?x2?4x?4根据题意列出方程组： ⑴?；⑵?

?y?x?10?y?x?2?x1?6?x2??1?x3?2?x4?3

∴解得：?；?；?；?

y?0y?16y?9y?1?1?2?4?3∴满足条件的点P有四个，它们分别是P········ 15分 2(?1,9)，P4(3,1) ·1(6,16)，P3(2,0)，P [注：对于以上各大题的不同解法，解答正确可参照评分！]

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