贺州市2018年初中毕业升学考试试卷及答案

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

3（2）在Rt△ODE中，由OD?，DE=2

2得OE?E 2 O D 1 B

A 5 ····················································· 5分 2C 3 又∵O、E分别是CB、CA的中点

5∴AB=2·OE?2??5

2第25题图

∴所求AB的长是5cm． ········································································· 7分 26．（本题满分7分）

解：（1）18?2?6?30（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 ……

?(18?30)x≥1800?400(18?30)x≤1800?350 ········································································· 4分

?x≥2916?5 解之得：x≤30?24

15所以不等式组的解集为：29≤x≤30 ·················································· 5分

624∵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 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，

1 ∴DN?DB ··············································2分

2 ∵∠EDF=90°，△BCD是等边三角形． ∴∠ADG=30°，而∠A=30°．

E

F C 45° G A 30° M

D

N

B

第27题图①

∴GA=GD．

∵GM⊥AB

1∴AM?AD·········································· 3分

2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 28．（本题满分10分）

1解：（1）抛物线y??x2?x?2与y轴的交于点B，

4令x=0得y=2．

∴B（0，2） ······································ 1分

11 ∵y??x2?x?2??(x?2)2?3

44∴A（—2，3） ····································· 3分

（2）当点P是 AB的延长线与x轴交点时，

· A A 30° M

D

E 45° G 1 C F H B

N 第27 题图②

y B H O P x PA?PB?AB． ······································ 5分

当点P在x轴上又异于AB的延长线与x轴的交点时， 在点P、A、B构成的三角形中，PA?PB?AB．

综合上述：PA?PB≤AB ····································································· 7分 （3）作直线AB交x轴于点P，由（2）可知：当PA—PB最大时，点P是所求的点 ··· 8分

作AH⊥OP于H． ∵BO⊥OP，

∴△BOP∽△AHP

AHHP ∴ ······················································································· 9分 ?BOOP由（1）可知：AH=3、OH=2、OB=2，

∴OP=4，故P（4，0） ······································································· 10分 注：求出AB所在直线解析式后再求其与x轴交点P（4，0）等各种方法只要正确也相应给分．

第28题图