20．（本题8分）
解：在Rt△BDC中，sinC＝
BD， ································································ 1分 BC
∴BD＝BC·sinC＝BC·sin25°＝120×0.42＝50.4 m． ······························ 3分 在Rt△AFB中，sin∠ABF＝
AF
， ·························································· 4分 AB
∴AF＝AB·sin∠ABF＝AB·sin50°＝70×0.77＝53.9 m． ·························· 6分 ∴AE＝AF＋FE＝AF＋BD＝50.4＋53.9＝104.3 m．
答：陵墓的垂直高度AE的长为104.3 m． ··············································· 8分
21．（本题7分）
1
解：（1） ． ························································································ 2分
3
（2）用树状图列出所有可能出现的结果：
第一次
第二次
第三次
所有可能出现的结果
B
A
C A
B
C A
C
B
C B C A B A
（A，B，C） （A，C，B） （B，A，C） （B，C，A） （C，A，B） （C，B，A）
开 始
一共有6种可能的结果，它们是等可能的，其中符合要求的有1种．
1
P(一次性对应打开a、b、c三把电子锁)＝．
61
答：一次性对应打开a、b、c三把电子锁的概率为 ． ························ 7分
6
22．（本题8分）
解：（1）不一定．
例如：芒果手机4月份销售200台，则5月份销售量为240台；四星手机4月份销售200台，则5月份销售量为100台；大米手机4月份销售50台，则5月份销售量为100台，从而可知大米手机5月份的销售量不是三个品牌手机中最高的． ····· 4分
（2）
200×20%－80×50%＋120×100%
＝30%．
200＋80＋120
答：该卖场5月份三个品牌手机销售量的平均增长率是30%． ·············· 8分
23．（本题9分）
解：（1）6000，200； ·············································································· 2分 （2）设AB所在直线的函数表达式为y＝kx＋b，
将点A(0，6 000)，B(30，0)代入y＝kx＋b得：
???b＝6000，?b＝6000，? 解得 ? ?30k＋b＝6，?k＝－200，??
∴AB所在直线的函数表达式为y＝－200x＋6 000． ····························· 5分 （3）设小君骑公共自行车时与奥体中心的距离为y1 m，
则y1＝－300(x－5)＋6 000， 当y1＝0时，x＝25. 30－25＝5.
∴小君先到达奥体中心，小君要等小敏5分钟. ······························· 9分
24．（本题8分）
解法一：设每件商品的售价上涨x元，
(210－10x)(50＋x－40)＝2200 ·························································· 4分 解得x1 ＝1，x2＝10， ··································································· 6分 ∴当x＝1时，50＋x＝51，当x＝10时，50＋x＝60； ·························· 7分
解法二：设每件商品的售价为x元，
[210－10(x－50)] (x－40)＝2200 ······················································· 4分 解得x1 ＝51，x2＝60， ································································· 7分 答：当每件商品的售价定为51或60元时，每个月的利润恰为2200元． · 8分
25．（本题10分）
解：（1）连接OA、OD，
在⊙O中，OA＝OD，
∵在△AOD中，点P是AD的中点， ∴OP⊥AD．
∵四边形ABCD是正方形， ∴AD∥BC， ∴OP⊥BC，
且OF是⊙O半径，
E A P O D B F C （第25题） ·∴BC与⊙O相切． ····································································· 3分
∵在△AOD中，点P是AD的中点，
∴AP＝DP＝1．
在Rt△AOP中，∠APO＝90°， ∵AP2＋OP2＝AO2，
5
∴12＋(2－r)2＝r2，求得r＝ ．······················································ 6分
4（2）120°或60°． ··········································································· 10分
26．（本题9分）
解：（1）当y＝0时，
4ac－b24(m2－1)－(2m＋1)2－4m－5
＝＝＝0， ···································· 3分 4a44
5
∴m＝－． ················································································ 5分
42m＋1－4m－5
（2）函数y＝x2＋(2m＋1)x＋m2－1的顶点坐标为（－，）
24
2m＋1－4m－5
设顶点在直线y1＝kx＋b上，则－k＋b＝ 243
求得k＝1，b＝ ，
4
3
不论m取何值，该函数图像的顶点都在直线y1＝x－ 上． ·················· 9分
4
27．（本题10分）
解：（1）B； ························································································· 1分 ENND（2），； ·················································································· 3分
EBAD（3）如图①，画图正确； ····································································· 6分 （4）如图②，过点Q作MN∥BC，交AB、AC分别于点M、N，
∵MN∥BC，
∴△AMQ∽△ABD，△AQN∽△ADC， MQAQAQQN∴＝ ，＝， BDADADDC
∴
MQQN
＝ ． ············································································· 7分 BDDC
∵点D是BC的中点， ∴BD＝CD， ∴MQ＝NQ． ∵MN∥BC，
∴△PMQ∽△PBC，△EQN∽△EBC， MQPQNQEQPQEQ∴＝ ，＝， ∴＝， BCPCCBEBPCEB∴
PQEQ＝， ··············································································· 8分 QCQB
又∵∠PQE＝∠CQB，
∴△PQE∽△CQB， ····································································· 9分 ∴∠EPQ＝∠BCQ， ∴PE∥BC，
即l∥BC．················································································· 10分
B A A P l Q B M P Q C D
l E N D
（图①）
C
（图②）
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