27．(本题满分8分)

证明：(1)∵∠A=30°，∠ACB=90°，D是AB的中点．

　　 ∴BC=BD， ∠B=60°

　　 ∴△BCD是等边三角形．····································· 1分

　　 又∵CN⊥DB，

　　 ∴ ····················································· 2分

　　 ∵∠EDF=90°，△BCD是等边三角形．

　　 ∴∠ADG=30°，而∠A=30°．

　　 ∴GA=GD．

∵GM⊥AB

∴················································· 3分

又∵AD=DB

∴AM=DN　 ··················································· 4分

(2)∵DF∥AC

∴∠1=∠A=30°，∠AGD=∠GDH=90°，

∴∠ADG=60°．

∵∠B=60°，AD=DB，

∴△ADG≌△DBH

∴AG=DH，···················································· 6分

又∵∠1=∠A，GM⊥AB，HN⊥AB，

∴△AMG≌△DNH．

∴AM=DN ．　 ·············································· 8分

26．(本题满分7分)

解：(1)(元)　　 ········································································· 1分

所以一个书包的价格是30元． ··············································································· 2分

(注：用其它方法解出正确答案也给予相应的分值)

(2)设还能为x 名学生每人购买一个书包和一件文化衫，根据题意得：····················· 3分

　　　　　　　　　　　　　　　　　　　　　　　　　 ……　 4分

　 解之得：

所以不等式组的解集为：　 ·························································· 5分

∵x为正整数，

∴x=30　 ················································································································ 6分

答：剩余经费还能为30名学生每人购买一个书包和一件文化衫．···························· 7分

25、(本题满分7分)

证明：(1)连结OD． ·························································································· 1分

由O、E分别是BC、AC中点得OE∥AB．

∴∠1=∠2，∠B=∠3，又OB=OD．

∴∠2=∠3．

而OD=OC，OE=OE

∴△OCE≌△ODE．

∴∠OCE=∠ODE．

又∠C=90°，故∠ODE =90°．　 ······························· 2分

∴DE是⊙O的切线．　 ·········································· 3分

(2)在Rt△ODE中，由，DE=2

得　 ····························································· 5分

又∵O、E分别是CB、CA的中点

∴AB=2·

　　 ∴所求AB的长是5cm．　 ····················································································· 7分

24．(本题满分6分)

(1)方法一：作 BC′= BC，DC′=DC．

　　 方法二：作∠C′BD=∠CBD，取BC′=BC，连结DC′．

　　 方法三：作∠C′DB=∠CDB，取DC′=DC，连结BC′．

　　 方法四：作C′与C关于BD对称，连结 BC′、DC′．

……

以上各种方法所得到的△BDC′都是所求作的三角形．

　　 只要考生尺规作图正确，痕迹清晰都给3分．

(2)解：∵△C′BD与△CBD关于BD对称，

∴∠EBD=∠CBD．

又∵矩形ABCD的AD∥BC

∴∠EDB=∠CBD．

∴∠EBD=∠EDB，BE = DE．

在Rt△ABE中，AB²+AE²=BE²，而AB=5，BC=12．

∴5²+(12-BE)²=BE² ·············································································· 5分　

∴所求线段BE的长是．··········································································· 6分

23、(本题满分6分)

解：(1)根据题意列表如下：

	1	2	3	4
1		(1，2)	(1，3)	(1，4)
2	(2，1)		(2，3)	(2，4)
3	(3，1)	(3，2)		(3，4)
4	(4，1)	(4，2)	(4，3)

由以上表格可知：有12种可能结果　 ·········································································· 3分

(注：用其它方法得出正确的结果，也给予相应的分值)

(2)在(1)中的12种可能结果中，两个数字之积为奇数的只有2种，

所以，P(两个数字之积是奇数)．······························································ 6分

22．(本题满分6分)

解：延长AC交 ON于点E，········································· 1分

∵AC⊥ON，

∠OEC=90°，································································ 2分

∵四边形ABCD是矩形，

∴∠ABC=90°，AD=BC，

又∵∠OCE=∠ACB，

∴∠BAC=∠O=25°， ··················································· 3分

在Rt△ABC中，AC=3，

∴BC=AC·sin25°≈1.27　 ················································· 5分

∴AD≈1.27　 ································································ 6分

(注：只要考生用其它方法解出正确的结果，给予相应的分值)

21．(本题共2小题；第(1)题5分，第(2)题5分，共10分)

　(1) 解：原式 ································································· 4分

　······························································································ 5分

(2) 解：方程两边同乘，得 ································································· 1分

　································································· 3分

　　　 解这个方程，得　 x=2　 ············································································· 4分

检验：当x=2时，=0，所以x=2是增根，原方程无解．················ 5分

9．　 45　 ；　　　　 10．2　 ；　 　　　　11．；　　 　　　12．

5．　 甲　　 ；　　　 6．；　 　　7．；　　　　 8．　 13　　 ；

1．　 2009　 ； 　　 2．；　 3． 2.124×10⁴　 ；　　　 4． 　；