27．（8分）如图，在⊿ABC中，∠ACB=90°，AC=BC，延长AB至点D，使DB=AB，连接CD，以CD为直
角边作等腰直角三角形CDE，其中∠DCE=90°，连接BE. （1）求证：⊿ACD≌⊿BCE；
（2）若AC=3cm，则BE= ▲ cm.
28．（10分）如图，已知一次函数y?x?1的图象与y轴交于点A，一次函数y?kx?b 的图象；经过点B（0，－1），并且与x轴以及y?x?1的图象分别交于点C、D． (1)若点D的横坐标为1，求四边形AOCD的面积（即图中阴影部分的面积）；
(2)在第(1)小题的条件下，在y轴上是否存在这样的点P，使得以点P、B、D为顶点的三角形是等腰
三角形．如果存在，求出点P坐标；如果不存在，说明理由．
(3)若一次函数y?kx?b的图象与函数y?x?1的图象的交点D始终在第一象限，则系数k的取值范
围是 ▲ ．（请直接写出结果，无需书写解答过程！）
（第28题图）
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参考答案及评分标准
说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．
一、选择题（每小题2分，共计16分）
题号 答案 1 B 2 D 3 A 4 C 5 B 6 D 7 C 8 A 二、填空题（本大题共10小题，每小题2分，共20分。） 9．±4 10．百 11．35 12． 49 13． 20 14．合格 15．10cm 16．?18．（0，0），（0，1），（0，
1 17．-6或4 23），（0，-3）(可以对2或3个1分，对4个2分) 4三、解答题（本大题共10题，共64分．） 19．原式=﹣1+2﹣1 ······································································ 2分 =． ····················································································· 4分 20．解：（x+1）=16， 2x+1=±4， ·················································································· 2分 x1=3，x2=﹣5． ········································································· 4分 21．全等. ······················································································· 1分 理由如下： ∵两三角形纸板完全相同， ∴BC=BF，AB=BD，∠A=∠D， ························································ 2分 ∴AB－BF=BD－BC，即AF=DC. ························································ 3分 在△AOF和△DOC中， ?AF?DC? ??A??D??AOF??DOC?∴△AOF≌△DOC（AAS）. ····························································· 6分 22．（本题6分）证明： ∵AD∥BC ∴∠1=∠B，∠2=∠C ···························································· 2分 6 ∵AD平分∠EAC ∴∠1=∠2 ··········································································· 4分 ∴∠B=∠C ··············································································· 5分 ∴AB=AC ·············································································· 6分 23．（本题6分）（对1个2分，对2个4分，对3个6分)
24．（本题4分） 能（1分),（1，0）（4分）
25（本题6分）（1）k=1 ································································· 1分
y=x-3 ········································································· 2分
(2) 在 ························································································ 3分
x=2,y=-1代入y=x-3 ································································ 4分 (3) y≤0; x-3≤0 ········································································· 5分
x≤3 ···················································································· 6分
26．（本题8分）作CD的垂直平分线和∠AOB的平分线， 两线的交点即为所作的点P；
CD的垂直平分线和∠AOB的平分线对1个3分，结论1分
27. （本题8分）
(1)证明：∵ △CDE为等腰直角三角形，∠DCE=90°,
∴ CD=CE. ·································································· 1分 又∵ ∠ACB=90°, ∴ ∠ACB=∠DCE. ······················································ 2分 ∴ ∠ACB+∠BCD=∠DCE+∠BCD.∴ ∠ACD=∠BCE. ··························································· 3分 又∵ AC=BC,∴ △ACD≌△BCE. ····································· 5分 (2)62. ·················································································· 8分
注：由于AC=BC=3 cm,在Rt△ACB中，根据勾股定理可以求出AB=32 cm，则AD=2AB= 62cm．因为△ACD≌△BCE,所以BE=AD=62 cm．
28．（本题10分） 解：（1）由题意知：D（1,2），
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∴直线BD的函数关系式为y?3x?1 ··············································· 1分 ∴A（0,1），C（,0). ······························································ 2分 ∴S四边形AOCD?S?AOD?S?COD?131115··················· 3分 ?1?1???2?. ·
2236（2）①当DP=DB时，P（0,5）；
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