（2007-2010年）济南中考数学试题及答案(8)

⑵如图2所示，点M、N分别在边AB、AC上，且

AM1AN1?，?，点P1、P2是边AB3AC3BC的三等分点，你认为∠MP1N+∠MP2N=∠A是否正确？请说明你的理由． AM1AN1⑶如图3所示，点M、N分别在边AB、AC上，且，，点P1、??AB2010AC2010P2、??、P2009是边BC的2010等分点，则∠MP1N+∠MP2N+??+∠MP2009N=____________．

（请直接将该小问的答案写在横线上．） A M M B N C

B C

B P1

??

P2 ?? P2009 C

第23题图3

A N

M A N

P

第23题图1

P1 P2

第23题图2

得 分 评卷人

24．(本小题满分9分)

如图所示，抛物线y??x2?2x?3与x轴交于A、B两点，直线BD的函数表达式为

y??3x?33，抛物线的对称轴l与直线BD交于点C、与x轴交于点E．

⑴求A、B、C三个点的坐标． ⑵点P为线段AB上的一个动点（与点A、点B不重合），以点A为圆心、以AP为半径的圆弧与线段AC交于点M，以点B为圆心、以BP为半径的圆弧与线段BC交于点N，分别连接AN、BM、MN．

①求证：AN=BM． ②在点P运动的过程中，四边形AMNB的面积有最大值还是有最小值？并求出该最大值或最小值.

y D l C M N A O E P 第24题图 B x

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济南市2010年初三年级学业水平考试

数学试题参考答案及评分标准

一、选择题 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 C D C B D A B B B C A C 二、填空题

13. (x?1)2 14. 70 15. x??9 16. 4 17. 13 三、解答题

18.(1)解：??x?2＞?x①

??2x≤4

②

解不等式①，得x＞?1， ··································································· 1分 解不等式②，得x≥－2， ···································································· 2分 ∴不等式组的解集为x＞?1. ···································································· 3分 (2) 证明：∵BC∥AD，AB=DC，

∴∠BAM=∠CDM， ····································································· 1分 ∵点M是AD的中点，

∴AM=DM， ················································································ 2分

∴△ABM≌△DCM， ···································································· 3分 ∴BM=CM. ················································································· 4分 19.(1)解：原式=5?2(5?2)(5?2)?(?3)0 ·

·························································· 1分 =5－2+1 ················································································ 2分 =5－1 ··················································································· 3分

(2)解：∵△ABC中，∠C=90o，∠B=30o，

∴∠BAC=60o，

∵AD是△ABC的角平分线，

∴∠CAD=30o， ··············································································· 1分 ∴在Rt△ADC中，AD?ACcos30? ··················································· 2分

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=3×23 ················································ 3分

=2 . ························································ 4分

20.解：a与b的乘积的所有可能出现的结果如下表所示： b a 1 2 －3 －4

·················································································································· 6分 总共有16种结果，每种结果出现的可能性相同，其中ab=2的结果有2种， ··························································································································· 7分

1∴a与 b的乘积等于2的概率是. ··························································· 8分

821.解：设BC边的长为x米，根据题意得 ··················································· 1分

32?x x?················································································· 4分 ?120，

2 解得：x1?12，x2?20， ··········································································· 6分

∵20＞16，

∴x2?20不合题意，舍去， ·································································· 7分

答：该矩形草坪BC边的长为12米. ··············································· 8分 22. 解：⑴∵点A的坐标为(－2，0)，∠BAD=60°，∠AOD=90°，

∴OD=OA·tan60°=23，

∴点D的坐标为（0，23）， ··························································· 1分 设直线AD的函数表达式为y?kx?b，

1 1 2 －3 －4 2 2 4 －6 －8 －3 －3 －6 9 12 －4 －4 －8 12 16 ????2k?b?0?k?3，解得， ?????b?23?b?23∴直线AD的函数表达式为y?3x?23. ········································ 3分 ⑵∵四边形ABCD是菱形， ∴∠DCB=∠BAD=60°， ∴∠1=∠2=∠3=∠4=30°， AD=DC=CB=BA=4， ········································································· 5分 如图所示：

y ①点P在AD上与AC相切时， P2 D AP1=2r=2， 2 3 ∴t1=2. ·································································································· 6分

②点P在DC上与AC相切时，

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C P1 P3 1 4 A O P4 第22题图

B x CP2=2r=2， ∴AD+DP2=6， ∴t2=6. ········································ 7分 ③点P在BC上与AC相切时， CP3=2r=2， ∴AD+DC+CP3=10， ∴t3=10. ········································ 8分 ④点P在AB上与AC相切时， AP4=2r=2， ∴AD+DC+CB+BP4=14， ∴t4=14， ∴当t=2、6、10、14时，以点P为圆心、以1为半径的圆与对角线AC相切. ··························································· 9分

23. ⑴证明：∵点M、P、N分别是AB、BC、CA的中点， ∴线段MP、PN是△ABC的中位线，

A

∴MP∥AN，PN∥AM， ······················ 1分

∴四边形AMPN是平行四边形， ······ 2分

M N ∴∠MPN=∠A. ······························· 3分

⑵∠MP1N+∠MP2N=∠A正确. ······················ 4分 1 2 如图所示，连接MN， ···························· 5分 AMAN1∵A=∠A， ??，∠ABAC3C B PP

1

2

∴△AMN∽△ABC，

MN1∴∠AMN=∠B，?，

BC3第23题图

1∴MN∥BC，MN=BC， ························ 6分

3∵点P1、P2是边BC的三等分点， ∴MN与BP1平行且相等，MN与P1P2平行且相等，MN与P2C平行且相等， ∴四边形MBP1N、MP1P2N、MP2CN都是平行四边形， ∴MB∥NP1，MP1∥NP2，MP2∥AC， ······································································ 7分 ∴∠MP1N=∠1，∠MP2N=∠2，∠BMP2=∠A， ∴∠MP1N+∠MP2N=∠1+∠2=∠BMP2=∠A. ···································································· 8分 ⑶∠A. ······················································ 9分

24.解：⑴令?x2?2x?3?0，

解得：x1??1,x2?3， ∴A(－1，0)，B(3，0) ································ 2分

N

y

D

l

M

C

F

A O E P B x

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第24题图

∵y??x2?2x?3=?(x?1)2?4， ∴抛物线的对称轴为直线x=1，

将x=1代入y??3x?33，得y=23， ∴C（1，23）. ···································· 3分 ⑵①在Rt△ACE中，tan∠CAE=

CE?3， AE∴∠CAE=60o，

由抛物线的对称性可知l是线段AB的垂直平分线， ∴AC=BC， ∴△ABC为等边三角形， ···································································· 4分 ∴AB= BC =AC = 4，∠ABC=∠ACB= 60o， 又∵AM=AP，BN=BP， ∴BN = CM， ∴△ABN≌△BCM， ∴AN=BM. ···························································································· 5分 ②四边形AMNB的面积有最小值． ···················································· 6分 设AP=m，四边形AMNB的面积为S，

3由①可知AB= BC= 4，BN = CM=BP，S△ABC=×42=43，

4∴CM=BN= BP=4－m，CN=m， 过M作MF⊥BC,垂足为F,

3(4?m)， 则MF=MC?sin60o=233211(4?m)=?m?3m， ·∴S△CMN=CN?MF=m?························· 7分 2422∴S=S△ABC－S△CMN

32m?3m） =43－（?4=3(m?2)2?33 ············································································· 8分 4∴m=2时，S取得最小值33. ··························································· 9分

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