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． 　；

28．(本题满分10分)

　如图，抛物线的顶点为A，与y 轴交于点B．

(1)求点A、点B的坐标．

(2)若点P是x轴上任意一点，求证：．

(3)当最大时，求点P的坐标．　　　　　　

贺州市2009年初中毕业升学考试数学评分标准

27．(本题满分8分)

图中是一副三角板，45°的三角板Rt△DEF的直角顶点D恰好在30°的三角板Rt△ABC斜边AB的中点处，∠A=30^o，∠E= 45^o，∠EDF=∠ACB=90 ^o ，DE交AC于点G，GM⊥AB于M．

(1)如图①，当DF经过点C　　 时，作CN⊥AB于N，求证：AM=DN．

(2)如图②，当DF∥AC时，DF交BC于H，作HN⊥AB于N，(1)的结论仍然成立，请你说明理由．