18.解:(1)设每支钢笔x元,每本笔记本y元 ·························································· 1分 依题意得:??x?3y?18 ································································· 3分
?2x?5y?31 解得:??x?3 ··············································································· ４分 y?5? 答:每支钢笔3元,每本笔记本5元 ···················································· ５分
(2)设买a支钢笔,则买笔记本(48－a)本
依题意得:??3a?5(48?a)?200 ··················································· 7分
?48?a?a 解得:20?a?24 ······································································ 8分 所以,一共有５种方案． ·································································· 9分
即购买钢笔、笔记本的数量分别为:
20,28； 21,27； 22,26； 23,25； 24,24． ········································ 10分
五、解答题:本题满分12分．
19．(1)证明:由题意可得:△ABD≌△ABE,△ACD≌△ACF ····································· 1分
∴∠DAB＝∠EAB ,∠DAC＝∠FAC ,又∠BAC＝45°, ∴∠EAF＝90° ········································································· 3分 又∵AD⊥BC
∴∠E＝∠ADB＝90°∠F＝∠ADC＝90° ······································· 4分 又∵AE＝AD,AF＝AD ∴AE＝AF ················································································ 5分 ∴四边形AEGF是正方形 ···························································· 6分
(2)解:设AD＝x,则AE＝EG＝GF＝x ···························································· ７分
∵BD＝2,DC＝3 ∴BE＝2 ,CF＝3
∴BG＝x－2,CG＝x－3 ·································································· ９分 在Rt△BGC中,BG2＋CG2＝BC2 ∴( x－2)2＋(x－3)2＝52 ································································ 11分 化简得,x2－5x－6＝0 解得x1＝6,x2＝－1(舍) 所以AD＝x＝6 ··········································································· 12分
六、解答题:本题满分14分．
20．解:(1)设抛物线的解析式为:y1?a(x?1)?4 ··············································· 1分
把A(3,0)代入解析式求得a??1
所以y1??(x?1)?4??x?2x?3 ············································· 3分
设直线AB的解析式为:y2?kx?b
222
由y1??x?2x?3求得B点的坐标为(0,3) ··································· 4分 把A(3,0),B(0,3)代入y2?kx?b中 解得:k??1,b?3
所以y2??x?3·········································································· 6分 (2)因为C点坐标为(１,4)
所以当x＝１时,y1＝4,y2＝2 所以CD＝4-2＝2 ········································································· 8分
2S?CAB?1·················································· 10分 ?3?2?3(平方单位) ·
222(3)假设存在符合条件的点P,设P点的横坐标为x,△PAB的铅垂高为h,
则h?y1?y2?(?x?2x?3)?(?x?3)??x?3x ······················ 12分 由S△PAB=得:
9S△CAB 819?3?(?x2?3x)??3 282化简得:4x?12x?9?0 解得,x?将x?3 232代入y1??x?2x?3中, 2315解得P点坐标为(,) ······························································ 14分
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2009年益阳市普通初中毕业学业考试试卷
