25．（本题满分10分）已知□ABCD，对角线AC与BD相交于点O，点P在边AD上，过
点P分别作PE⊥AC、PF⊥BD，垂足分别为E、F，PE＝PF． （1）如图10，若PE＝3，EO＝1，求∠EPF的度数； （2）若点P是AD的中点，点F是DO的中点，
BF ＝BC＋32－4，求BC的长．
k2
26．（本题满分12分）已知点A(1，c)和点B (3，d )是直线y＝k1x＋b与双曲线y＝（k2＞
x
0）的交点．
（1）过点A作AM⊥x轴，垂足为M，连结BM．若AM＝BM，求点B的坐标； k
（2）设点P在线段AB上，过点P作PE⊥x轴，垂足为E，并交双曲线y＝2（k2＞0）
x
于点N．当
PN1
取最大值时，若PN＝ ，求此时双曲线的解析式． NE2
AEBOPFCD图10- 6 -
参考答案及评分标准
说明：
1．解答只列出试题的一种或几种解法．如果考生的解法与所列解法不同，可参照解答中评分标准相应评分； 2．评阅试卷，要坚持每题评阅到底，不能因考生解答中出现错误而中断对本题的评阅．如果考生的解答在某一步出现错误，影响后续部分而未改变本题的内容和难度，视影响的程度决定后继部分的给分，但原则上不超过后续部分应得分数的一半； 3．解答题评分时，给分或扣分均以1分为基本单位．
一、选择题（本大题共7小题，每小题3分，共21分）
题号 选项 1 A 2 C 3 A 4 D 5 B 6 C 7 B 二、填空题（本大题共10小题，每题4分，共40分） 8. a. 9. 50°.
10. m.
1
11. . 12. 3. 13. x＋y＞1.
2
14. 60.
15. 540°. 16. 5； 6. 17. ；2πr.
三、解答题（本大题共9小题，共89分） 18．（本题满分18分）
（1）解：4÷(－2) ＋(－1)×4
＝－2＋1×1·············································································· 4分 ＝－2＋1 ··················································································· 5分 ＝－1. ····················································································· 6分 （2）解：正确画出坐标系 ········································································ 8分 正确写出两点坐标 ····································································10分
画出直线 ··················································································12分
（3）证明：∵ AC∥DF，
??13分
∴ ∠ACB＝∠DFE. ??15分 又∵ ∠A＝∠D， ??16分 AC＝DF， ∴ △ABC≌△EDF.
19．（本题满分7分）
??17分 ??18分
2
0
ABCFDE?3x＋y＝4， ①解1：?
?2x－y＝1. ②
①＋②，得 ················································································· 1分 5x＝5， ····················································································· 2分 x＝1. ······················································································· 4分 将x＝1代入 ①，得 3＋y＝4， ················································································· 5分
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y＝1．······················································································· 6分
?x＝1，∴? ················································································· 7分 ?y＝1.
解2：由①得 y＝4－3x． ③·············································· 1分 将③代入②，得 2x－(4－3x) ＝1．····································································· 2分 得x＝1. ·················································································· 4分 将x＝1代入③ ，得
y＝4－3×1················································································ 5分 ＝1． ······················································································ 6分
?x＝1，∴? ················································································· 7分 ?y＝1.
20．（本题满分7分）
（1）解：∵ DE∥BC ，∴ △ADE∽△ABC. ??1分
∴ ∴
ADDE
＝ . ABBC
AD1
＝. AB3
AD1
＝，BD＝10， AB3
??2分 ??3分
AED（2）解1：∵
∴
CGBAD1
＝ ········································································· 4分
AD＋103∴ AD＝5 ················································································ 5分 经检验，符合题意. ∴ AB＝15. 在Rt△ABC中， ······································································· 6分 sin∠A＝
解2： ∵
∴
BC3
＝. ······································································ 7分 AB5
AD1
＝，BD＝10， AB3
AD1
＝ ········································································· 4分
AD＋103∴ AD＝5 ················································································ 5分 经检验，符合题意. ∵ DE∥BC，∠C＝90° ∴ ∠AED＝90°
在Rt△AED中， ······································································· 6分 sin∠A＝
ED3
＝. ······································································ 7分 AD5
解3：过点D作DG⊥BC，垂足为G. ∴ DG∥AC.
∴∠A＝∠BDG. ······································································ 4分 又∵ DE∥BC，∴四边形ECGD是平行四边形. ∴ DE＝CG. ············································································· 5分
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∴ BG＝6.
在Rt△DGB中， ······································································ 6分 ∴ sin∠BDG＝BD3
＝. ···························································· 7分 GB5
3
∴ sin∠A＝.
5
21．（本题满分7分） （1）解：A组数据的平均数是
0＋1－2－1＋0－1＋3
······························ 1分
7
＝0. ························································· 3分
（2）解1：选取的B组数据：0，－2，0，－1，3. ································ 4分
∵ B组数据的平均数是0. ······················································ 5分 ∴ B组数据的平均数与A组数据的平均数相同. ∴ SB2＝∴
1416
，SA2＝ . ·························································· 6分 57
1416 ＞.·············································································· 7分 57
∴ B组数据：0，－2，0，－1，3.
解2：B组数据：1，－2，－1，－1，3. ········································ 4分
∵ B组数据的平均数是0. ······················································ 5分
∴ B组数据的平均数与A组数据的平均数相同.
1616
∵SA2＝, SB2＝ . ·························································· 6分
75∴
1616＞ ················································································· 7分 57
∴ B组数据：1，－2，－1，－1，3.
22．（本题满分9分） （1）解:由题意得,
2
x＝(2x－2) ············································································· 1分 3∴ x＝4. ················································································· 2分 ∴ x2－1＝16－1＝15(小时). ···················································· 3分
答：乙车床单独加工完成这种零件所需的时间是15小时. ········ 4分
（2）解1：不相同. ················································································ 5分
若乙车床的工作效率与丙车床的工作效率相同，由题意得, ······ 6分 11
＝ . ········································································ 7分 x－12x－2
2∴
11＝. x＋12
∴ x＝1. ················································································ 8分 经检验，x＝1不是原方程的解. ∴ 原方程无解. ····················· 9分
答：乙车床的工作效率与丙车床的工作效率不相同.
解2：不相同. ················································································ 5分
若乙车床的工作效率与丙车床的工作效率相同，由题意得, ······ 6分
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x－1＝2x－2. ········································································· 7分
解得，x＝1. ············································································ 8分 此时乙车床的工作时间为0小时，不合题意. ··························· 9分 答：乙车床的工作效率与丙车床的工作效率不相同. 23．（本题满分9分） （1）证明1：∵∠BCD＝∠BAC，
︵︵
∴ BC＝BD . ∵ AB为⊙O的直径， ∴ AB⊥CD， CE＝DE. ∴ AC＝AD .
??2分 ??3分 ??4分 ??1分
2
CGFBDEOA 证明2：∵∠BCD＝∠BAC，
︵︵
∴ BC＝BD . ······································································ 1分 ︵︵
∵ AB为⊙O的直径， ∴ BCA＝BDA . ······························· 2分 ︵︵
∴ CA＝DA . ·········································································· 3分
∴ AC＝AD . ·········································································· 4分
证明3：∵ AB为⊙O的直径，∴ ∠BCA＝90°. ··························· 1分
∴ ∠BCD+∠DCA＝90°, ∠BAC+∠CBA＝90°
∵∠BCD＝∠BAC，∴∠DCA＝∠CBA ···································· 2分 ︵︵
∴ CA＝DA . ·········································································· 3分
∴ AC＝AD . ·········································································· 4分
（2）解1：不正确. ················································································ 5分
连结OC.
当 ∠CAB＝20°时， ································································ 6分 ∵ OC＝OA，有 ∠OCA＝20°.
∵ ∠ACB＝90°， ∴ ∠OCB＝70°. ······························· 7分 又∵∠BCF＝30°，
∴∠FCO＝100°， ··································································· 8分 ∴ CO与FC不垂直. ······························································ 9分
∴ 此时CF不是⊙O的切线.
解2：不正确. ················································································ 5分
连结OC.
当 ∠CAB＝20°时， ································································ 6分
∵ OC＝OA，有 ∠OCA＝20°.
∵ ∠ACB＝90°， ∴ ∠OCB＝70°. ······························· 7分 又∵∠BCF＝30°，
∴∠FCO＝100°， ··································································· 8分 在线段FC的延长线上取一点G，如图所示，使得∠COG＝20°. 在△OCG中， ∵∠GCO＝80°， ∴∠CGO＝80°.
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