2013年中考模拟题 数学试题参考答案及评分标准
一、选择题（每小题4分，共40分）
１．C ２．C ３．A； ４．C ５．D；６．A ７．A ８．B ９．B 1０．C
二、填空题（每小题4分，共20分）
１１．x（x－２）；１２．答案不唯一，小于或等于２的整数均可，如：2，1等；１３．＝； １４．40；１５．13π-26
三、解答题
１６． （１）（本题满分7分）
解：5x–12≤8x-6． ················································································· 3分
································································· 5分 ?3x≤6． ·
x≥-2 ． ····································································· 7分
（２）解：原式＝
x?yxy?xy(x?y)2
＝
x?y……………………………………………………4分
2代入，则
1 将x?原式＝ １７．
3，y?13?2?3?2……………………………………7分
（1）解：8?(3?1)??1?22?1?1?22
……………………8分
（２）解：设先安排整理的人员有x人，依题意得，
?60x2x(?15)?1 ……………………4分 600
9
解得， x＝１０．
答：先安排整理的人员有１０人．……………………8分
１８．证明: 过点C作CF⊥AB，垂足为F．……………… 1分
∵ 在梯形ABCD中，AB∥CD，∠A=90°， ∴ ∠D＝∠A＝∠CFA＝90°． ∴四边形AFCD是矩形．
AD=CF, BF=AB-AF=1．……………………………… 3分 在Rt△BCF中， CF2=BC2-BF2=8，
∴ CF=22．
∴ AD=CF=22．……………………………………………………………… 5分 ∵ E是AD中点， ∴ DE=AE=
12AD=2．…………………………………… 6分
在Rt△ABE和 Rt△DEC中， EB2=AE2+AB2=6， EC2= DE2+CD2=3， EB2+ EC2=9=BC2．
∴ ∠CEB＝90°．………………………………………………………9分 ∴ EB⊥EC．………………………… 10分
（其他不同证法，参照以上标准评分）
１９．（每小题各3分,共12分）
（１）５０ （２）３ （３）普遍增加了 （４）15
２０．（每小题3分,共12分）
（１）如图 （２）5
55255（３）∠CAD，
12(或∠ADC，)
（４）
２１．解：（1）点 M ········································································································ 1分 （2）经过t秒时，NB?t，OM?2t 则CN?3?t，AM?4?2t
10
∵?BCA=?MAQ=45
∴QN? CN ?3?t ∴PQ ?1? t ······································································· 2分 ∴S△AMQ?2?12AM?PQ?12(4?2t)(1?t)
······················································································································ 3分 ??t?t?2 ·
1?9?∴S??t2?t?2???t??? ···················································································· 5分
24??2∵0≤t≤2∴当t?12时，S的值最大． ······································································ 6分
（3）存在． ·················································································································· 7分 设经过t秒时，NB=t，OM=2t 则CN?3?t，AM?4?2t
∴?BCA=?MAQ=45 ······················································································ 8分 ①若?AQM?90，则PQ是等腰Rt△MQA底边MA上的高 ∴PQ是底边MA的中线 ∴PQ?AP?∴1?t?∴t?1212(4?2t)
12MA
??
∴点M的坐标为（1，0） ·························································································· 10分 ②若?QMA?90，此时QM与QP重合 ∴QM?QP?MA
∴1?t?4?2t
∴t?1
∴点M的坐标为（2，0） ·························································································· 12分
２２．（1）解：由
23x?83?0，得x??4． ?A点坐标为??4，0?．?由?2x?16?0，得x?8． 0?．?B点坐标为?8，∴AB?8???4??12． ·························································································· 2分
11
28?y?x?，?x?5，?由?解得∴C点的坐标为?5， ······································· 3分 6?．33??y?6．?y??2x?16．?∴S△ABC?12AB·yC?12 ······························································· 4分 ?12?6?36．23?8?83 （2）解：∵点D在l1上且xD?xB?8，?yD? ?8． ∴D点坐标为?8， ····························································································· 5分 8?．??2xE?16?8．?xE?4．又∵点E在l2上且yE?yD?8，
∴E点坐标为?4， ····························································································· 6分 8?．∴OE?8?4?4，EF?8． ················································································· 8分
（3）解法一：①当0≤t?3时，如图1，矩形DEFG与△ABC重叠部分为五边形
．过C作CM?AB于M，CHFGR（t?0时，为四边形CHFG）则Rt△RGB∽Rt△CMB．
yl2 yE C D R l1l2 yl1E D C R E l2 D C R l1A O F M G B x （图1）
∴
BGBM?RGCMA F O G M （图2）
t3RG6B x F A G O M B x （图3）
即，?∴RG?2t． ， ?Rt△AFH∽Rt△AMC，∴S?S△ABC?S△BRG?S△AFH?36?即S??
43t?212?t?2t?12 ?8?t???8?t?．32163t?443． ········································································· 14分
12
