广东省三水八年级上学期学业水平考试数学试卷有答案

广东省三水市八级学学水平考试数学试卷说明：．卷共4页，试用时分钟，满分为120分．2．答卷前，考生务必用黑色字的签字笔或钢笔在答题卡填写自己的准考证号、姓名、试室号、座位号．用2B铅把对应该号码的标号涂黑．3题小题选出答案后2B铅笔把答题卡上对应题目选项的答案信息点涂黑，如需改动，用橡皮擦干净后，再选涂其他答案，答案不能答在试题上．4．非选择题必须用黑色字迹钢或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液．不按以上要求作答的答案无效．5．考生务必保持答题卡的整洁考试结束时，将试卷和答题卡一并交回．一、选择题（本大题10小题每小题3分共分）在每小题列出的四个选项中只有一个是正确的，请把答题卡上对应目所选的选项涂黑．1．的绝值是A．-6B．6C．±D．2．下列各图中，经过折叠能围一个立方体的是A．B．．D．3．中国航母辽宁舰（如题3图）中国人民海军第一艘可以搭载固定翼飞机的航空母舰，满载排水量为67500吨，这个数据用科学记数法表示为

A．6.75×10

B．6.75×10C．6.75×10

吨D．6.75×10

吨4．若()

mn

的值是

题3图A．-1B．1C2

D．35．一组数据2，4，x，，，的众数是2则这组数据的平均数、中位数分别为A．3，

B．3.5，C．3，3.5

D．4，6．如题6图直线l∥l，∠1=140°∠2=70°则3的度数是A．60°C．70°

B．65°D．80°7．如果ab，那么下列判断正确的是A．a＜，b＜0B．a＞0，＞C．a≥，b≤0D．a＜0，＞或＞，b＜8．不等式组的集数轴上表示正确的是x

题6图

＝＝ABCD9．下列标志中，可以看作是中对称图形的是ABCD10．次函数y

的图象如题图示，那么一函数

y

与反比例函数

在同一平面直角坐标系中的大致图象为题10图ABCD二、填空题（本大题6小题，每题4分共24）请将下列各题的正确答案填写在答题卡相应的位置上．11．解因式：2x

x

．12．知

，则

__________．13．∠=80°，则α的角为__________．14．题14图在山的东侧点一个热气球，由于受西风的影响，以30米分速度沿与地面成75°角的方向飞行25分钟后到达C处此时热气球上的人测得小山西侧B点的角为30°，则小山东西两侧、两间的距离__________．15．题15图，在△中，∥BCAD=1AB，=2，则BC=__________．16．题16图，AB是半圆O直径，且AB，C为圆上的点．将此半圆沿所在的直线折叠，若圆弧恰过圆心，则图中阴影部分的面积（果保留π题14图

题图

题图三、解答题（一本大题3小，每小题分，15分17．方程：

52(8

．18．化简，再求值：

3xx2()x

，再选择一个使原式有意义的代求值．19．题19图，已知线段AB

（）出线段AB的直分线l用尺规作图，保留作图痕迹，不要求写出作法（）（1）中所作的直线l上意取两点，（线段AB的上方结AM，ANBM，BN．求证：∠=∠MBN题19图四、解答题（二本大题3小，每小题分，24分20．商场以每件280元价购进一批商品，当每件商品售价为360元，每月可售出60件为了扩大销售，商场决定采取适当降价的方式促销，经调查发现，如果每件商品降价1元那么商场每月就可以多售出件（）价前商场每月销售该商品的利润是多少元？（）使商场每月销售这种商品的利润达到7200元，且更有利于减少库存，则每件商品应降价多少元？21．障房建设是民心工程，广省某市从年开始加快保障房建设进程，现统计了该市2009年2013年年新保障房情况，绘制成如题22图示的折线统计图和不完整的条形统计图．某市2009～2013年新建保障房数年增长率折线统计图

某市2009～年新建保障房套数条形统计图20092010201120122013年20092010201120122013年份题22图（）丽看了统计图后说市2012年建保障房的套数比2011年了认小丽说法正确吗？请说明理由；（）补全条形统计图；（）这5年均每年新建保障房的套数．22．题22图．点A、、、⊙上，⊥于点，过点作OFBC于F求证）△AEB△OFC；（）=2FO．题22图五、解答题（三本大题3小，每小题分，27分

23．题23图，抛物线2（a＞0）经过原点O和点A（，（）接写出抛物线的对称轴与轴的交点坐标；（）（x，y，y）抛物线上，若＜＜，比较y的小；（）B-1，）该抛物线上，点C与B关抛物线的对轴对称，求直线的函数关系式．题23图24．题24图直线直平分⊙的径OA于点B，交⊙O于点、，是⊙的线为点，连结AE，交于点F（）⊙的径为8，的长；（）证PE；（）PF=13，A=，的长．题24图25．知：如图①，在平行四边中，=6AD．AD为斜边在平行四边形ABCD的内部作eq\o\ac(△,Rt)，∠=30°∠°．（）△的长；（2）△每秒2个单位长度的速度沿向平行移动，得到△D，AD与重合时停止移动运动时间为t秒eq\o\ac(△,，)ED与△重的面积为，请直接写出与t之间函数关系式，并写出t的取范围；（）图②，在），当停止移动后得到△，BEC绕点按时针方向旋转（°＜α＜180°旋过程中的应点为B，的应点为E，设直线B与线交于点P与线CB交于Q是否存在这样的使BPQ为等腰三角形？若存在，求出度数；若不存在，请说明理由．题25图

题25图

参考答案及评分标准数

学一、选择题（本大题小题每小题分共分）题号答案

B

A

A

D

B

C

D

D

D

B二、填空题（本大题小题，每小题4分，共分）．x(x

．

．14750

．

．

三、解答题（一本大题3小题，小题分，共15分17解：x+2=8+··························································································x=6······························································································x=3．···························································································18解：原=

2(x(x(x2)(2)

····························································=2+8·························································································3分当x=1时原式=2×1+8=10的值不能为2或）······························5分19）：如图所示：················································································1分·······················································2分（2证明：∵l是垂平分线∴=BMAN=BN，·································································∴∠=∠MBA，∠NAB∠NBA·············································4分∴∠∠NAB∠MBA∠NBA，即：∠MAN=∠．································································四、解答题（二本大题3小题，小题分，共24分20解）题意，得60-280）元（2设要使商场每月销售这种商品的利润达到7200，且更有利于减少库存，则每件商品应降价元，由题意，得（360-x+60），解得：x，x12∵有利于减少库存，∴x=60．答）价商场每月销售该商品的利润是4800元）使商场每月销售这种商品的利润达到元且更有利于减少库存，则每件商品应降价60．

AE，∴AE，∴21解）市2012年建保障房的增长率比的增长率减少了，但是保障房的总数在增加，故小丽的说法错误．···················································（2年障房的套数为：750×(1+20%)=900套·····························3分年保障房的套数为：x(1+20%)=600，则x=500···························4分如图所示：··········································6分（3（）·············································答：这年平均每年新建保障房的套数为784套······························8分22证明）图，连接，则=∠，··········································1分∵⊥，∴∠∠，∴∠BAE=∠COF，······················又∵AC，⊥，∴∠∠AEB，···························3分∴△∽△；···································································（2∵∽△，∴，················································BEFC由圆周角定理，=∠，∠DAE=∠CBE∴△∽，∴

AEFOADBC

，····························∵⊥，∴BC，∴=FOFO，即FO············································································8分五、解答题（三本大题3小题，小题分，共27分23解）物线的对称轴与轴的交点坐标为，0···································（2抛物线的对称轴是直线x．由图可知，当x＜1时随的增大而减小，····································

kk3∴当x＜＜1时y＞；····························································12（3∵对称轴是x=1，（1）在该抛物线上，点与关于抛物线的对称轴对称，∴点C坐标是，····························································设直线的系式为=kx+b（k≠0，·························6分解得．··············································································∴直线的数关系式是：=24．··············································24解）接OD，··················································································∵直线PD垂直平分O的径于B⊙O的径为8，∴OB=4，BC=BDCD，························································∴在eq\o\ac(△,Rt)OBD中BDOD

，∴=3

；········································································（2∵PE⊙O的线∴PEO，···············································∴∠PEF-∠，=∠-∠，································∵OE，∴A∠AEO∴∠PEF=∠，∴PE；·························································6分（3过点P作PG⊥于，····························································∴∠==90°，∵∠=，∴∠∠，∴=A=13×···························································∵PE=PF∴=10．····························································9分25解）△的长为3；····························································1分（2①S＝t2（0t································································2

33