2013年南京市白下区中考一模数学试题及答案(4)

23．（本题8分）

（1）证明：∵四边形ABCD中，E、F、G、H分别是AD、BC、BD、AC的中点，

1111

∴FG＝CD，HE＝CD，FH＝AB，GE＝AB． ·············································· 2分

2222∵AB＝CD，

∴FG＝FH＝HE＝EG． ················································································ 3分 ∴四边形EGFH是菱形． ·············································································· 4分 （其他方法参照给分）

（2）解：∵四边形ABCD中，G、F、H分别是BD、BC、AC的中点，

∴GF∥DC，HF∥AB． ···················································································· 5分 ∴∠GFB＝∠DCB，∠HFC＝∠ABC． ∴∠HFC＋∠GFB＝∠ABC＋∠DCB＝90°．

∴∠GFH＝90°． ·························································································· 6分 ∴菱形EGFH是正方形． ················································································· 7分 11

∵AB＝1，∴EG＝AB＝．

22

11

∴正方形EGFH的面积＝()2＝． ····································································· 8分

24

24．（本题8分）

解：（1）直线BD与⊙O相切．理由如下：

连接OB．

∵CA是⊙O的直径，

∴∠ABC＝90°． ······································· 1分

∵OB＝OC，

∴∠OBC＝∠C．

又∵∠DBA＝∠C，

∴∠DBA＋∠OBA＝∠OBC＋∠OBA＝∠ABC＝90°． ············································· 2分

∴OB⊥BD．

又∵直线BD经过半径OB的外端点B， ······························································ 3分

∴直线BD与⊙O相切． ·················································································· 4分

（2）∵∠DBO＝90°，AD＝AO＝1，

∴AB＝OA＝OB＝1．

∴△AOB是等边三角形．

9

（第24题）

D A O C B ∴∠AOB＝60°． ···························································································· 5分

60π×12π

∴S扇形OBA＝＝． ················································································· 6分

3606 ∵在Rt△DBO中，BD＝DO2－BO2＝3，

113

∴S?DBO＝OB·BD＝×1×3＝． ································································ 7分

222

∴S阴影＝S? DBO－S扇形OBA＝

25．（本题8分）

解：（1）解法一：设第一个月单价降低x元，批发商销售完这批T恤获得的总利润为y

元． ······························································································································· 1分 根据题意，得y＝(70－50－x)(200＋10x)＋(40－50)×[500－(200＋10x)]

＝－10x2＋100x＋1000． ································································ 4分 批发商销售这批T恤可能亏本，理由如下：（答案不唯一，以下方法供参考）

方法一：当x＝17（或18或19）时，y＜0． ·························································· 5分 方法二：当y＝0时，x＝55＋5（负根舍去）．

又因为当55＋5＜x＜20时，y随x的增大而减小，

所以当x＝17或18或19时，y＜0． ························································ 5分 解法二：设第一个月单价降低x元，当月出售T恤获得的利润为y1元，清仓剩

余T恤获得的利润为y2元． ··············································································· 1分 根据题意，得y1＝(70－50－x)(200＋10x)＝－10x2＋4000， ······································· 3分 y2＝(40－50)×[500－(200＋10x)]＝100x－3000． ···················································· 4分 批发商销售这批T恤可能亏本，理由如下：（答案不唯一，以下方法供参考）

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