资阳市2017-2018学年八年级下期末学业质量检测数学试题含答案新人教版

资阳市2017-2018学年度学业质量检测

八年级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应得的累计分数。

2. 参考答案一般只给出该题的一种解法，如果考生的解法和参考答案所给解法不同，请参照本答案及评分意见给分。

3. 考生的解答可以根据具体问题合理省略非关键步骤。

4. 评卷时要坚持每题评阅到底，当考生的解答在某一步出现错误、影响了后继部分时，如果该步以后的解答未改变问题的内容和难度，可视影响程度决定后面部分的给分，但不得超过后继部分应给分数的一半；如果这一步后面的解答有较严重的错误，就不给分；若是几个相对独立的得分点，其中一处错误不影响其他得分点的得分。

5. 给分和扣分都以1分为基本单位。

6. 正式阅卷前应进行试评，在试评中须认真研究参考答案和评分意见，不能随意拔高或降低给分标准，统一标准后须对全部试评的试卷予以复查，以免阅卷前后期评分标准宽严不同。

一、选择题（每小题3分，共10个小题，满分30分） 1-5.DBCAC；6-10.DBDDA

二、填空题：（本大题共6个小题，每小题3分，共18分） 11. -1；12. 6；13. 83； 14.x1?x2；15. 4；16. (2016，0)；

三、解答题：（本大题共8个小题，共72分）解答应写出必要的文字说明、证明过程或演算步骤.

17.原式=?1···················································································· 4分 ， ·

a?11··············································································· 8分 当a=3时，原式=?. ·418.（1）众数为8万车次，中位数为8万车次，平均数为8.5万车次； ················· 3分 8.5=255(万车次). 答估计4月份共租车255万车次； ···························· 5分 （2）30×

0.75÷9600=25%. 答全年租车费收入占总投入的25%. ························· 8分 （3）3200×

19.（1）在平行四边形ABCD中，∠BAD的角平分线AE交CD于点F，

························································································· 2分 易证AB=BE， ·

············································································· 4分 又∵AB=CD，∴BE=CD.

·················· 5分 （2）由BE=CD=AB，∠BEA=60°得△ABE为等边三角形，则AE=4， ·································· 6分 又∵BF⊥AE，∴AF=EF=2，根据勾股定理得BF=23， ·

················································································· 7分 易证△ADF?△ECF， ·

······················· 8分 ∴平行四边形ABCD的面积等于△ABE的面积，则面积等于43. ·20.（1）设A种运动鞋的进价为元，依题意得，

32002560， ?xx?20···························································································· 3分 解得=100， ·

··········································· 4分 经检验，=100是原分式方程的解，所以，=100； ·

则A运动鞋的售进价价为100元/双，B运动鞋的进价是80元/双， （2）设总利润为W，

································· 7分 则W=（250-100）m+（180-80）（200﹣m）=50m+20000， ······························································· 8分 ∵50＞0，W随m的增大而增大， ·

又∵90≤m≤105，所以，当m=105时，W有最大值，即此时应购进甲种运动鞋105双，···················································································· 9分 购进乙种运动鞋95双. ·

21.（1）依题意可知，点A的横坐标为-1，代入y2??3， x······································· 3分 求出A的坐标为（-1，3），则y1的解析式为y=-+2；·························································· 4分 （2） ∵y=2+b与轴交于点B（3，0）， ···································································· 6分 则直线BC的解析式为y=2-6， ·

28································································ 7分 求出点C的坐标为（，?）， ·331811··········································································· 9分 ∴SAOC=?(?1)?2=. ·

233······································································ 2分 22.（1）易证BM=MD=DN； ·

········································································· 4分 ∴四边形BMB?N为菱形； ·

（2）设BM=，在Rt△AMB′中，利用勾股定理求出=则DM=

13································· 5分 ， ·313=DN， ······················································································ 6分 313138?(6?)=，······································ 7分 333413. ··············································· 9分 3过点M作MQ⊥CD于点Q，则NQ=

在Rt△MNQ中，利用勾股定理可得MN=························································ 2分 23.（1）易求点A的坐标为（-4，-5）， ·则解析式为y?20. ··················································································· 3分 x······················ 4分 （2）如图，求出点E的坐标为（-2，-10），点F的坐标（4，5） ·分别过点E、F作EN⊥轴于点N，FM⊥GM于点M，FM也垂直于轴，证明△ENO?△FMG，………………………………………… 5分

设点G的坐标为（m，n），则5-n=10，m-4=-2，

则点G的坐标为（2，-5）；……………………………………6分 （3）由于OE为定值，则只需求出OF的最小值即可，设点F的坐

2020220222标为（a，.）根据勾股定理得，OF?a?2?(a?)?40.…7分

aaa显然当a?20.时，OF2最小，即a=25时，OF最小，OF=210， ··················· 8分 a················ 9分 因此，当点F的坐标为（25，25）时，四边形OEGF周长最小，·

············································································ 10分 最小值为426?410. ·24.（1）如图，易证△EBM1?△EFN1，则∠EFN1=90°，则四边形BEFG为矩形，即FN1⊥AB；……………………………………… 3分

（2） 如图，同理，△EBM2?△EFN2，则∠EFN2=90°，………5分 由于∠EFN1+∠EFN2=180°，所以点N2在直线FN1上； ……6分 （3） 易证四边形BEFG为正方形，易求BE=4；…………7分

11=2x?x2，当点M1在线段AB的延长线上时，S1=?x?(4?x）22

此时>0；………………………………………………………9分

当点M2在线段BA的延长线上时，

112①当3<<4时，S2=?x?(4?x）=2x?x……………………………………10分

22112·················································· 11分 ②当>4时，S3=?x?(x?4）=x?2x， ·

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