2005年四川省绵阳市中考数学试题含答案

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24 (本小题满分14分)

还有最后一关，

如图9，在平行四边形ABCD中，AD=4 cm，∠A=60°，BD⊥AD. 一动点P从A出发，以每秒1 cm的速度沿A→B→C的路线匀速运动，过点P作直线PM，使PM⊥AD .

(1) 当点P运动2秒时，设直线PM与AD相交于点E，求△APE的面积；

(2) 当点P运动2秒时，另一动点Q也从A出发沿A→B→C的路线运动，且在AB上以每秒1 cm的速度匀速运动，在BC上以每秒2 cm的速度匀速运动. 过Q作直线QN，使QN∥PM. 设点Q运动的时间为t秒(0≤t≤10)，直线PM与QN截平行四边形ABCD所得图形的面积为S cm2 .

① 求S关于t的函数关系式； ② (附加题) 求S的最大值.

注：附加题满分4分，但全卷的得分不超过150分.

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绵阳市2005年高中阶段学校招生统一考试

数学试题参考答案及评分意见

说 明：

1. 解答题中各步骤所标记分数为考生解答到这一步所得的累计分数； 2. 给分和扣分都以1分为基本单位；

3. 参考答案都只给出一种解法，若考生的解答与参考答案不同，请根据解答情况参考评分意见给分 .

一、选择题：每小题3分，共10个小题，满分30分. 1-5. ABDCC；7-10. DABAD .

二、填空题：每小题4分，共6个小题，满分24分.

32011. ，6；12. 5；13. 2；14. ；15. 12；16. 如右图：

23三、解答题：共8个小题，满分96分 . 17. 画出图形(基本正确即可).·········································································· 6分 AC=26 mm，OC=50 mm. ················································································· 10分

(若量得AC=25 mm或27 mm，OC=49 mm或51 mm，同样给4分；若量得AC=24 mm或28 mm，OC=48 mm或52 mm，给2分；其余答案不给分)

18. 由题意有??2A?7B?8, ············································································· 4分

?3A?8B?10.(正确建立关于A、B的一个方程，给2分.)

6?A?,??5解得：? ····························································································· 10分

4?B??.?5?64即A、B的值分别为、? .

5519.(1) 全市共有300名学生参加本次竞赛决赛，最低分在20－39之间，最高分在120

－140之间 ··············································································································· 4分

(答出参赛人数1分，最低分和最高分同时答对1分) (2) 本次决赛共有195人获奖，获奖率为65％ . ··········································· 6分

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(3) 决赛成绩的中位数落在60—79分数段内. ················································ 8分 (4) 如“120分以上有12人；60至79分数段的人数最多；??”等. ······ 12分 (写出一条正确信息给1分)

1a?3(a?1)21a?3a2?2a?1??20. = ······················ 6分 ??a?1a2?1a2?4a?3a?1(a?1)(a?1)(a?3)(a?1)(每正确分解一个因式给1分) =

2 ··········································································································· 8分 (a?1)2由a2＋2a－8=0知，(a＋1)2=9， ····································································· 10分

2221a?3a2?2a?1∴ =，即的值为. ··································· 12分 ?2?22(a?1)a?1a?1a?4a?39921. (1) ∵方程有两个不相等的实数根，∴Δ=[-2(k＋1)]2－4k(k-1)>0，且k≠0，解得k>-1，且k≠0 .即k的取值范围是k>-1，且k≠0 . ············································· 6分

(2) 假设存在实数k，使得方程的两个实数根x1 , x2的倒数和为0. ··············· 8分

2(k?1)11k?1k?0，即则x1 ，x2不为0，且??0，解得k=-1 . ?0，且

k?1x1x2kk ·························································································································· 10分 而k=-1 与方程有两个不相等实根的条件k>-1，且k≠0矛盾， 故使方程的两个实数根的倒数和为0的实数k不存在 . ································ 12分

22. (1) 连AF，因A为BF的中点，∴∠ABE=∠AFB， 又∠AFB=∠ACB，∴ ∠ABE=∠ACB . ∵ BC为直径，∴∠BAC=90°，AH⊥BC，∴∠BAE=∠ACB， ∴∠ABE=∠BAE， ∴ AE=BE . ·········································································· 5分 (2) 设DE=x(x>0)，由AD=6，BEEF=32，AEEH=BEEF， ···················· 8分 有(6-x)(6+x)=32，由此解得x=2， 即DE的长为2 . ······································ 9分 (3) 由(1)、(2)有：BE=AE=6-2=4，

在RtΔBDE中，BD=42?22=23. ······························································ 12分

22

23. 设直角三角形ABC的三边BC、CA、AB的长分别为a、b、c，则c=a+b2 . (1) S1=S2+S3 . ······································································································ 2分 (2) S1=S2+S3 . 证明如下：

323232显然，S1=c，S2=a， S3=b，

4443232∴S2+S3=··································································· 5分 (a?b2)?c=S1 . ·

44(也可用三角形相似证明)

(3) 当所作的三个三角形相似时，S1=S2+S3 . 证明如下：

S2a2S3b2?,?. ∵ 所作三个三角形相似， ∴

S1c2S1c2S2?S3a2?b2???1,?S1?S2?S3. ····························································· 10分

S1c2(4) 分别以直角三角形ABC三边为一边向外作相似图形，其面积分别用S1、S2、S3表

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示，则S1＝S2＋S3 . ··································································································· 14分

(若仅考虑到特殊的多边形，给2分；若考虑到任意的相似多边形，给3分) 24. (1) 当点P运动2秒时，AP=2 cm，由∠A=60°，知AE=1，PE=3. ·························································································································· 2分

3. ································································································· 4分 2(2) ① 当0≤t≤6时，点P与点Q都在AB上运动，设PM与AD交于点G，QN与

∴ SΔAPE=

AD交于点F，则AQ=t，AF=

tt33，QF=t. t，AP=t+2，AG=1+，PG=3?222233∴ 此时两平行线截平行四边形ABCD的面积为S=. ···················· 8分 t?22当6≤t≤8时，点P在BC上运动，点Q仍在AB上运动. 设PM与DC交于点G，QNtt3，DF=4-，QF=BP=t-6，CP=10-t，PG=(10?t)3， t，222532t?103t?343. 而BD=43，故此时两平行线截平行四边形ABCD的面积为S=?8 ··························································································································· 10分

当8≤t≤10时，点P和点Q都在BC上运动. 设PM与DC交于点G，QN与DC交

与AD交于点F，则AQ=t，AF=

于点F，则CQ=20-2t，QF=(20-2t)3，CP=10-t，PG=(10?t)3.

332t?303t?1503. 2 ··························································································································· 14分

?33,(0?t?6)?t?22???532故S关于t的函数关系式为S???t?103t?343,(6?t?8)

8??332t?303t?1503.(8?t?10)?2??∴ 此时两平行线截平行四边形ABCD的面积为S=

73； ············································· 1分 2当6≤t≤8时，S的最大值为63； ································································ 2分

②(附加题)当0≤t≤6时，S的最大值为

当8≤t≤10时，S的最大值为63； ······························································ 3分 所以当t=8时，S有最大值为63 . ······························································ 4分 (如正确作出函数图象并根据图象得出最大值，同样给4分)