25．(本题满分14分) 如图，直角梯形ABCD中，AD∥BC，?BCD?90°，且CD?2AD，tan?ABC?2，
过点D作DE∥AB，交?BCD的平分线于点E，连接BE． （1）求证：BC?CD；
（2）将△BCE绕点C，顺时针旋转90°得到△DCG，连接EG.． 求证：CD垂直平分EG.
（3）延长BE交CD于点P． A D 求证：P是CD的中点．
E G
B C
（第25题图）
26．(本题满分14分)
，B两点，与y轴交于C点，且经过点 如图，抛物线y?ax?bx?3与x轴交于A(2，?3a)，对称轴是直线x?1，顶点是M．
（1） 求抛物线对应的函数表达式；
（2） 经过C,M两点作直线与x轴交于点N，在抛物线上是否存在这样的点P，使
以点P，A请求出点P的坐标；，C，N为顶点的四边形为平行四边形？若存在，
若不存在，请说明理由；
（3） 设直线y??x?3与y轴的交点是D，在线段BD上任取一点E（不与B，D重
合），经过A，B，E三点的圆交直线BC于点F，试判断△AEF的形状，并说明理由；
（4） 当E是直线y??x?3上任意一点时，（3）中的结论是否成立？（请直接写出
结论）．
y 2A O 1 ?3 C B x M （第26题图）
2009年烟台市初中学生学业考试 数学试题参考答案及评分意见
本试题答案及评分意见，供阅卷评分使用．考生若写出其它正确答案，可参照评分意见相应评分．
一、选择题（本题共12个小题，每小题4分，满分48分） 题号 答案 1 B 2 D 3 C 4 C 5 B 6 A 7 A 8 B 9 B 10 B 11 D 12 C 二、填空题（本题共6个小题，每小题4分，满分24分） 13．
1 14．?2 15．17 16．1 17．20 18．①，③，④ 4三、解答题（本题共8个小题，满分78分） 19．（本题满分6分） 解：18?93?6??(3?2)0?(1?2)2 233··························································· 2分 2?(1?2)?1?|1?2|． ·
23································································ 4分 ?32?2?1?2?1?2?1． ·
23··································································································· 6分 ?2?1 ·
2?32?20．（本题满分8分） 解：（1）（2）
1 ··································································································· 1分 21 ········································································································ 3分 3（3）根据题意，画树状图： ············································································· 6分
开始
第一次 1 2 3 4 第二次 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
（第20题图）
由树状图可知，共有16种等可能的结果：11，12，13，14，21，22，23，24，31，32，33，34，41，42，43，44．其中恰好是4的倍数的共有4种：12，24，32，44． 所以，P（4的倍数）?41?． ····································································· 8分 164
或根据题意，画表格： ···················································································· 6分
第一次 第二次 1 2 3 4 1 11 21 31 41 2 12 22 32 42 3 13 23 33 43 4 14 24 34 44 由表格可知，共有16种等可能的结果，其中是4的倍数的有4种，所以，
P（4的倍数）?41············································································· 8分 ?． ·
16421．（本题满分8分）
解：（1）a?1?(10%?15%?30%?15%?5%)?25%． ···································· 1分 初一学生总数：20?10%?200（人）． ····························································· 2分 （2）活动时间为5天的学生数：200?25%?50（人）． 活动时间为7天的学生数：200?5%?10（人）． ················································ 3分 频数分布直方图（如图） 人数
60
50 40 30 20 10
2天 3天 4天 5天 6天 7天 时间
···················· 4分 （第21题图）
（3）活动时间为4天的扇形所对的圆心角是360°?30%?108°． ··························· 5分 （4）众数是4天，中位数是4天． ···································································· 7分 （5）该市活动时间不少于4天的人数约是
． ················································· 8分 6000?(30%?25%?15%?5%)?4500（人）22．（本题满分8分）
解：过点C作CE⊥AB于E．
D
?D?90°?60??30°，?ACD?90°?30°?60°， ??CAD?90°．
1CD?10，?AC?CD?5． ························· 3分
2在Rt△ACE中，
5AE?ACsin?ACE?5sin30°?， ··············· 4分
25CE?ACcos?ACE?5cos30°?3， ·············5分
2在Rt△BCE中，
A B
C
B （第22题图）
?BCE?45°，?BE?CEtan45°??AB?AE?BE?5··················································· 6分 3， ·
2555． ?3?(3?1)≈6.8（米）
222所以，雕塑AB的高度约为6.8米． ··································································· 8分
23．（本题满分10分） 解：（1）根据题意，得y?(2400?2000?x)?8?4?即y????x??， 50?22········································································· 2分 x?24x?3200． ·
2522（2）由题意，得?x?24x?3200?4800．
25整理，得x?300x?20000?0．····································································· 4分 解这个方程，得x1?100，x2?200． ································································ 5分 要使百姓得到实惠，取x?200．所以，每台冰箱应降价200元． ···························· 6分 （3）对于y??当x??222x?24x?3200， 2524?150时， ·········································································· 8分 2??2?????25?150??y最大值?(2400?2000?150)?8?4???250?20?5000．
50??所以，每台冰箱的售价降价150元时，商场的利润最大，最大利润是5000元．········· 10分
24．（本题满分10分）
H （1）证明：连接OC，
························ 1分 HC?HG，??HCG??HGC． ·
·········· 2分 HC切⊙O于C点，??1??HCG?90°， ·D M ····································· 3分 OB?OC，??1??2， ·C G ······················ 4分 ?HGC??3，??2??3?90°．·B A O F ··························· 5分 ??BFG?90°，即DE⊥AB． ·
E （2）连接BE．由（1）知DE⊥AB． （第24题图） AB是⊙O的直径，
?BD?BE． ······························································································· 6分
···················································································· 7分 ??BED??BME． ·
四边形BMDE内接于⊙O，??HMD??BED． ··········································· 8分
??HMD??BME．
····························· 9分 ?BME是△HEM的外角，??BME??MHE??MEH． ·
··································································· 10分 ??HMD??MHE??MEH． ·
25．（本题满分14分） 证明：（1）延长DE交BC于F． AD∥BC，AB∥DF，
··························· 1分 ?AD?BF，?ABC??DFC． ·
在Rt△DCF中，tan?DFC?tan?ABC?2，
P E
CD?2，即CD?2CF． CFB C F ····················· 3分 CD?2AD?2BF，?BF?CF． ·
（第25题图）
11?BC?BF?CF?CD?CD?CD，
22即BC?CD． ······························································································· 4分 （2）CE平分?BCD，??BCE??DCE． 由（1）知BC?CD，CE?CE，?················ 6分 △BCE≌△DCE，?BE?DE． ·由图形旋转的性质知CE?CG，BE?DG，···································· 8分 ?DE?DG． ·
···································· 9分 ?C，D都在EG的垂直平分线上，?CD垂直平分EG． ·
（3）连接BD．由（2）知BE?DE，??1??2．
······················································ 11分 AB∥DE．??3??2．??1??3． ·
AD∥BC，??4??DBC．
由（1）知BC?CD．??DBC??BDC，??4??BDP． ···························· 12分 又BD?BD，?······································ 13分 △BAD≌△BPD，?DP?AD． ·
11······································· 14分 AD?CD，?DP?CD．?P是CD的中点． ·
22?28．（本题满分14分）
??3a?4a?2b?3，?解：（1）根据题意，得?b ··············2分
??1.??2ay D E N ?a?1，解得?
?b??2.?抛物线对应的函数表达式为y?x2?2x?3． ········3分
（2）存在．
在y?x?2x?3中，令x?0，得y??3．
2令y?0，得x?2x?3?0，?x1??1，x2?3．
2A O 1 N x F C P M （第26题图）
2009年烟台市中考数学试题(含答案)
