河南省鹤壁市淇滨中学7年级34班数学拔高测试A4版.docx
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河南省鹤壁市淇滨中学7年级34班数学拔高测试A4版
2014年河南省鹤壁市淇滨中学7年级3、4班数学拔高测试
期末测评试卷
班级:
姓名:
注意事项:
1、本试卷分第1卷和第2卷,共6页,13个小题,满分100分。
请用0.5毫米黑色或蓝色圆珠笔直接答在试卷上。
2、答卷前请将考生信息填写清楚。
第1卷基础夯实(满分20分)
本部分试题在做题时,可直接选择/填空,不需写出步骤。
1、(4分)下面横排有12个方格,每个方格都有一个数字,若任何三个数字的和都是20,则x=。
5
A
B
C
D
E
F
x
G
H
P
10
(上海市竞赛题)
2、(4分)当b=1时,关于x的方程a(3x-2)+b(2x-3)=8x-7有无数解,则a=。
3、(4分)为确保信息安全,信息需加密传输,发送方由明文→密文(加密),接收方由密文→明文(解密),已知加密规则为:
明文x,y,z分别对应密文2x+3y,3x+4y,3z。
例如,明文1、2、3对应密文8、11、9,当接收方收到密文12、17、27时,则解密得到的明文为
、、。
(山东省泰安市中考题)
4、(4分)已知三角形三边长分别为2、x、13,若x为正整数,则这样的三角形有个。
(2011年河北省中考题)
5、
如图,∠CAD和∠CBD的平分线相交于点P,设∠CAD、∠CBD、∠C、∠D的度数依次为a、b、c、d,用仅含有两个字母的代数式表示∠P的度数:
。
(第21届江苏省中考题)
第2卷能力扩展(满分80分)
本部分试题在做题时,需给出详细、准确的步骤,步骤将纳入总分数。
6、(10分)已知三个非负数a、b、c满足3a+2b+c=5和2a+b-3c=1,若m=3a+b-7c,求m的最大值和最小值。
(江苏省竞赛题)
7、(10分)如图,已知P为平行四边形ABCD内一点,且S△PBC=5,S△PAB=2,那么S△PBD=。
(武汉市竞赛题)
8、
(10分)如图,已知四边形ABCD中,E、F是DC边的三等分点,G、H是AB边的三等分点。
求证:
S四边形GHFC=
S四边形ABCD。
(“华杯赛”试题)
S2
S1
13
49
35
9、(10分)右图中,在长方形内画了一些直线,已知边上3块面积分别为13、35、49,求阴影部分的面积。
10、(10分)如图,在等腰三角形ABC中,BC=AB=4,将△ABC沿BC方向平移到△A/B/C/,若平移距离为3,求两三角形重合的面积。
11、(10分)如图是一个六角星,已知∠AOE=60○,求∠A+∠B+∠C+∠D+∠E+∠F的度数。
(注:
作图有误差,AOD、BOE三点均在同一条直线上。
)
(“希望杯”邀请赛试题)
12、(10分)如图,在四边形ABCD中,E、F分别平分∠BEC、∠BFC,若∠ADC=60°,
∠ABC=80°,则∠EGF=度。
(湖北省黄冈市竞赛题)
13、(10分)8个人乘速度相同的两辆小汽车同时赶往火车站,每辆车乘4人(不包括司机),其中一辆小汽车在距火车站15km的地方出现故障,此时,距停止检票的时间还有42分钟,这时,唯一可以利用的交通工具是另一辆小汽车,已知包括司机在内这辆车限乘5人,且这辆车的平均速度是60km/时,人步行的平均速度是5km/时,试设计两种方案,通过计算说明这8个人能够在停止检票前赶到火车站.
(全国初中数学竞赛题)
2014年河南省鹤壁市淇滨中学7年级3、4班数学拔高测试
期末测评试卷参考答案及评分标准
第1卷基础夯实(满分20分)
题号
1
2
3
4
5
答案
5
2
3;2;9
3
第2卷能力扩展(满分80分)
6、
解:
由题意得
3a+2b=5-c
2a+b=1+3c············································3分
解得
a=7c-3
b=7-11c··············································5分
∴m=3c-2···············································6分
a≥0
∵b≥0
c≥0··············································7分
7c-3≥0
∴7-11c≥0
c≥0··············································8分
∴
≤c≤
············································9分
∴
≤m≤
··········································10分
7、∵S△PBC+S△PAD=
S四边形ABCD=S△ABD····························2分
∴S△PAC=S△ABD-S△PAB-S△PAB···································4分
∴S△PAC=S△PBC-S△PAB·········································6分
∵S△PBC=5,S△PAB=2········································8分
∴S△PAC=5-2=3···········································10分
8、连FG、DG、FB、DB,···································2分
∵S△EGF=S△EGD,S△HFG=S△HFB····································3分
∴S四边形DGBF=2S四边形GHFE①···································4分
∵DE=EF=FC,AG=GH=HB····································5分
∴S△DBC=3S△FBC,S△DBA=3S△DGA···································6分
∴S四边形ABCD=3(S△FBC+S△DGA)··································7分
得S△FBC+S△DGA=
S四边形ABCD②································8分
①+②,得S四边形ABCD=2S四边形GHFE+
S四边形ABCD·····················9分
故S四边形GHFE=
S四边形ABCD.···································10分
9、由题意可得
49+S1+35+13+S2=S3+S阴+S4①
49+S3+13+35+S4=S1+S阴+S2②·····························3分
由①、②得
49+35+13=S阴+S3+S4-S1-S2③·····························6分
49+35+13=S阴+S1+S2-S3-S4④·····························7分
③+④得
2S阴=2×(49+35+13)·····································8分
S阴=49+35+13············································9分
S阴=97·················································10分
10、∵△ABC为直角三角形
又∵BC=AB=4
又∵平移的距离为3······································2分
∠B=∠A′B′C=90°································3分
∴∠A=∠ACB=∠A′=∠C′=45°························4分
BB′=CC′=3cm·····································5分
∴∠BOC=∠A′B′B-∠ACB=90°-45°=45°·················6分
∴B′O=B′C············································7分
∴B′C=BC-BB′=4-3=1(cm)·······························8分
∴B′O=1cm·············································9分
∴△DB′C′=B′O·B′C/2=1/2······························10分
11、∵∠A+∠C+∠E=∠AOE································2分
又∵∠B+∠D+∠F=∠BOD··································4分
又∵∠BOD=∠AOE········································6分
又∵∠AOE=60°·········································8分
∴∠A+∠B+∠C+∠D+∠E+∠F=60°×2=120°···············10分
12、∵∠EBC=80°,∠CDF=60°···························2分
∴∠FBE=100°,∠FDE=120°·····························4分
∴∠FGE=
(∠FBE+∠FDE)·······························7分
=
×(100°+120°)=110°····························10分
13、
(1)小车在送前4人的同时,剩下的人也同时步行不停的往前走,
小车送到火车站后立即再返回接剩下的人。
·················2分
设步行的4个人步行了x小时,
则有:
解得x=
·············································3分
所以共用时间:
÷5+(15-
)÷60=
小时=
(分钟)<42(分钟),故此方案可行;································4分
(2)先用小汽车把第一批人送到离火车站较近的某一处,让第一批人步行,与此同时第二批人也在步行中;接着小汽车再返回接第二批人,使第二批人与第一批同时到火车站,在这一方案中,每个人不是乘车就是在步行,没有人浪费时间原地不动,所以两组先后步行相同的路程。
···············································7分
设这个路程为y千米,
则
=
·······································8分
解得:
y=2(千米),·····································9分
所用时间为:
+
=
小时=37(分钟)<42(分钟),故此方案也可行。
···············································10分