2020年广东省汕头市中考数学试卷及答案初中数学2018年广东省汕头市初中毕业生学业考试数 学讲明:1.全卷共4页，考试用时100分钟，总分值为150分.2.答题前，考生必须将自己的姓名、准考证号、学校按要求填写在答卷密封线左边的空格内；并填写答卷右上角的座位号，将姓名、准考证号用2B 铅笔写、涂在答题卡指定的位置上。

3.选择题的答题必须用2B 铅笔将答题卡对应小题所选的选项涂黑．4．非选择题可用黑色或蓝色字迹的钢笔、签字笔按各题要求写在答卷上，不能用铅笔和红笔．写在试卷上的答案无效．姓名5．必须保持答卷的清洁．考试终止时，将试题、答卷、答题卡交回。

一、选择题〔本大题8小题，每题4分，共32分〕在每题列出的四个选项中，只有一个是正确的，请将答题卡上对应的小题所选的选项涂黑． 1．4的算术平方根是〔 〕A ．2±B ．2C ．D2．运算32()a 结果是〔 〕 A ．6aB ．9aC ．5aD ．8a3．如下图几何体的主〔正〕视图是〔 〕A ． C ．4．«广东省2018年重点建设项目打算〔草案〕»显示，港珠澳大桥工程估算总投资726亿元，用科学记数法表示正确的选项是〔 〕A ． 107.2610⨯元 B ．972.610⨯元 C ．110.72610⨯元 D ．117.2610⨯元 5.满足2〔x-1〕≤x+2的正整数x 有多少个〔 〕 A ．3 B.4 C.5 D.66.数据3,3,4,5,4,3,6的众数和中位数分不是( ) A.3,3 B.4,4 C.4，3 D.3，47.菱形ABCD 的边长为8,∠A=120°，那么对角线BD 长是多少〔 〕 A ．12 B.123 C.8 D.838.如下图的矩形纸片，先沿虚线按箭头方向向右对折，接着将对折后的纸片沿虚线剪下一个小圆和一个小三角形，然后将纸片打开是以下图中的哪一个二、填空题〔本大题5小题，每题4分，共20分〕 9．分解因式2x 3-8x= ．10．O ⊙的直径8cm AB C =，为O ⊙上的一点，30BAC ∠=°，那么BC = cm ．11．一种商品原价120元，按八折〔即原价的80%〕出售，那么现售价应为 元．12．在一个不透亮的布袋中装有2个白球和n 个黄球，它们除颜色不同外，其余均相同．假设从中随机摸出一个球，摸到黄球的概率是45，那么n =_____________．13．用同样规格的黑白两种颜色的正方形瓷砖，按以下图的方式铺地板，那么第〔3〕个图形中有黑色瓷砖 块，第n 个图形中需要黑色瓷砖________块〔用含n 的代数式表示〕．……〔1〕 〔2〕 〔3〕 三、解答题〔一〕〔本大题5小题，每题7分，共35分〕 14．〔此题总分值7分〕运算：19sin 30π+32-+-0°+()． 15．〔此题总分值7分〕解方程22111x x =---16. 〔此题总分值7分〕如下图，在平面直角坐标系中，一次函数y=kx+1的图象与反比例函数y=x9的图象在第一象限相交于点A 。

过点A 分不作x 轴、y 轴的垂线，垂足为点B 、C 。

假如四边形OBAC 是正方形，求一次函数的关系式。

第10题CBO第13题图17．〔此题总分值7分〕如下图，ABC △是等边三角形， D 点是AC 的中点，延长BC 到E ，使CE CD =，〔1〕用尺规作图的方法，过D 点作DM BE ⊥，垂足是M 〔不写作法，保留作图痕迹〕； 〔2〕求证：BM EM =．18．〔此题总分值7分〕如下图，A 、B 两都市相距100km ，现打算在这两座都市间修建一条高速公路〔即线段AB 〕，经测量，森林爱护中心P 在A 都市的北偏东30°和B 都市的北偏西45°的方向上，森林爱护区的范畴在以P 点为圆心，50km 为半径的圆形区域内，请咨询打算修建的这条高速公路会可不能穿越爱护区，什么缘故？〔参考数据：1.732 1.414〕四、解答题〔二〕〔本大题3小题，每题9分，共27分〕 19．〔此题总分值9分〕某种电脑病毒传播专门快，假如一台电脑被感染，通过两轮感染后就会有81台电脑被感染．请你用学过的知识分析，每轮感染中平均一台电脑会感染几台电脑？假设病毒得不到有效操纵，3轮感染后，被感染的电脑会可不能超过700台？ 20．〔此题总分值9分〕某中学学生会为了解该校学生喜爱球类活动的情形，采取抽样调查的方法，从足球、乒乓球、篮球、排球等四个方面调查了假设干名学生的爱好爱好，并将调查的结果绘制成如下的两幅不完整的统计图〔如图1，图2要求每位同学只能选择一种自己喜爱的球类；图中用乒乓球、足球、排球、篮球代表喜爱这四种球类中的某一种球类的学生人数〕，请你依照图中提供的信息解答以下咨询题： 〔1〕在这次研究中，一共调查了多少名学生？〔2〕喜爱排球的人数在扇形统计图中所占的圆心角是多少度？ 〔3〕补全频数分布折线统计图．A C D 第17题图30° A BFE P45°第18题图 图2乒乓球 20% 足球排球 篮球40%图1第20题图21．〔此题总分值9分〕如下图，在矩形ABCD 中，12AB AC =，=20，两条对角线相交于点O ．以OB 、OC 为邻边作第1个平行四边形1OBB C ，对角线相交于点1A ，再以11A B 、1A C 为邻边作第2个平行四边形111A B C C ，对角线相交于点1O ；再以11O B 、11O C 为邻边作第3个平行四边形1121O B B C ……依次类推． 〔1〕求矩形ABCD 的面积；〔2〕求第1个平行四边形1OBB C 、第2个平行四边形111A B C C 和第6个平行四边形的面积．五、解答题〔三〕〔本大题3小题，每题12分，共36分〕 22、〔此题总分值12分〕〔1〕如图1，圆心接ABC △中，AB BC CA ==，OD 、OE 为O ⊙的半径，OD BC ⊥于点F ，OE AC ⊥于点G ，求证：阴影部分四边形OFCG 的面积是ABC △的面积的13．〔2〕如图2，假设DOE ∠保持120°角度不变， 求证：当DOE ∠绕着O 点旋转时，由两条半径和ABC △的两条边围成的图形〔图中阴影部分〕面积A 1O 1A 2B 2 B 1C 1 B C 2A OD 第21题图 C 第22题图D 图1 图2始终是ABC △的面积的13． 23．〔此题总分值12分〕小明用下面的方法求出方程30=的解，请你仿照他的方法24．〔此题总分值12分〕正方形ABCD 边长为4，M 、N 分不是BC 、CD 上的两个动点，当M 点在BC 上运动时，保持AM 和MN 垂直，〔1〕证明：Rt Rt ABM MCN △∽△；〔2〕设BM x =，梯形ABCN 的面积为y ，求y 与x 之间的函数关系式；当M 点运动到什么位置时，四边形ABCN 面积最大，并求出最大面积； 〔3〕当M 点运动到什么位置时Rt Rt ABM AMN △∽△，求x 的值．2018年广东省汕头市初中毕业生学业考试数学试题参考答案及评分建议一、选择题〔本大题8小题，每题4分，共32分〕1．B 2．A 3．B 4．A 5.C 6.D 7.D 8．C 二、填空题〔本大题5小题，每题4分，共20分〕 9．2x 〔x+2〕〔x-2〕 10．4 11．96 12．8 13．10，31n + 三、解答题〔一〕〔本大题5小题，每题7分，共35分〕14．解：原式=113122+-+ ··················································································· 4分 =4． ······························································································· 7分NDA CB M第24题图15．解：方程两边同时乘以(1)(1)x x +-， ······························································· 2分 2(1)x =-+， ···································································································· 4分 3x =-， ··········································································································· 5分 经检验：3x =-是方程的解． ················································································ 7分 16.依题意可得：xy =9=OB ·OC ，……………………2分 又四边形ABCD 为正方形，因此 OC=OB=3因此有 A 〔3，3〕， ……………………3分 直线y ＝kx ＋1过点A ，因此得3＝3k ＋1，因此 k ＝32……………………5分 故有直线 y ＝32x ＋1 ……………………7分17．解：〔1〕作图见答案17题图，··························································· 2分 〔2〕ABC △是等边三角形，D 是AC 的中点，BD ∴平分ABC ∠〔三线合一〕， 2ABC DBE ∴∠=∠． ························································································· 4分 CE CD =，CED CDE ∴∠=∠．又ACB CED CDE ∠=∠+∠，2ACB E ∴∠=∠． ····························································································· 5分 又ABC ACB ∠=∠， 22DBC E ∴∠=∠， DBC E ∴∠=∠， BD DE ∴=． 又DM BE ⊥，BM EM ∴=． ·································································································· 7分18．解：过点P 作PC AB ⊥，C 是垂足， 那么30APC ∠=°，45BPC ∠=°， ·································· 2分A BFEPC 答案17题图AC BDE Mtan30AC PC =°，tan 45BC PC =°，AC BC AB +=， ························································ 4分 tan30tan 45100PC PC ∴+=°°，11003PC ⎛⎫∴+= ⎪ ⎪⎝⎭， ··················································· 6分50(350(3 1.732)63.450PC ∴=⨯->≈≈，答：森林爱护区的中心与直线AB 的距离大于爱护区的半径，因此打算修建的这条高速公路可不能穿越爱护区． ···························································································· 7分 四、解答题〔二〕〔本大题3小题，每题9分，共27分〕 19．解：设每轮感染中平均每一台电脑会感染x 台电脑， ············································ 1分 依题意得：1(1)81x x x +++=， ··········································································· 4分2(1)81x +=，19x +=或19x +=-，12810x x ==-，〔舍去〕，··················································································· 6分 33(1)(18)729700x +=+=>． ············································································ 8分答：每轮感染中平均每一台电脑会感染8台电脑，3轮感染后，被感染的电脑会超过700台． ························································································································ 9分 20．解：〔1〕2020%100÷=〔人〕． ····································································· 1分〔2〕30100%30%100⨯=， ··················································································· 2分 120%40%30%10%---=，36010%36⨯=°°． ···························································································· 4分 〔3〕喜爱篮球的人数：40%10040⨯=〔人〕， ························································ 5分喜爱排球的人数：10%10010⨯=〔人〕． ································································ 7分······················· 9分答案20题图21．解：〔1〕在Rt ABC △中，16BC =，1216192ABCD S AB BC ==⨯=矩形． ······································································ 2分〔2〕矩形ABCD ，对角线相交于点O ，4ABCD OBC S S ∴=△． ···························································································· 4分四边形1OBB C 是平行四边形，11OB CB OC BB ∴∥，∥，11OBC B CB OCB B BC ∴∠=∠∠=∠，．又BC CB =，1OBC B CB ∴△≌△，112962OBB C OBC ABCD S S S ∴===△， ······································································· 6分 同理，111111148222A B C C OBB C ABCD S S S ==⨯⨯=， ························································ 8分第6个平行四边形的面积为6132ABCD S =． ······························································· 9分五、解答题〔三〕〔本大题3小题，每题12分，共36分〕 22．证明：〔1〕如图1，连结OA OC ，， 因为点O 是等边三角形ABC 的外心，因此Rt Rt Rt OFC OGC OGA △≌△≌△． ····························· 2分2OFCG OFC OAC S S S ==△△，因为13OAC ABC S S =△△，因此13OFCGABC S S =△． ························································································ 5分 〔2〕解法一：连结OA OB ，和OC ，那么AOC COB BOA △≌△≌△，12∠=∠， ························ 6分 不妨设OD 交BC 于点F ，OE 交AC 于点G ，3412054120AOC DOE ∠=∠+∠=∠=∠+∠=°，°，35∴∠=∠． ······································································· 8分 在OAG △和OCF △中， 1235OA OC ∠=∠⎧⎪=⎨⎪∠=∠⎩，，，答案22题图〔1〕 AE O G FBCD 答案22题图〔2〕A E O GFB C D 1 2 3 45OAG OCF ∴△≌△， ························································································· 10分 13OFCG AOC ABC S S S ∴==△△． ··············································································· 12分 解法二： 不妨设OD 交BC 于点F ，OE 交AC 于点G ， 作OH BC OK AC ⊥⊥，，垂足分不为H K 、， ·················· 6分 在四边形HOKC 中，9060OHC OKC C ∠=∠=∠=°，°， 360909060120HOK ∴∠=-︒-︒=︒°-?， ························ 8分 即12120∠+∠=°．又23120GOF ∠=∠+∠=°，13∴∠=∠． ····································································································· 8分 AC BC =， OH OK ∴=，OGK OFH ∴△≌△， ························································································ 10分 13OFCG OHCK ABC S S S ∴==△． ················································································ 12分24．解：〔1〕在正方形中，，AM MN ⊥，90AMN ∴∠=°，90CMN AMB ∴∠+∠=°．在Rt ABM △中，90MAB AMB ∠+∠=°， CMN MAB ∴∠=∠，Rt Rt ABM MCN ∴△∽△． ··········································· 3分 〔2〕Rt Rt ABM MCN △∽△，44AB BM xMC CN x CN∴=∴=-，， 244x x CN -+∴=， ···························································································· 5分答案第22题图〔3〕 A EO GF B C D 1 3 2H K N DA CBM答案24题图22214114428(2)102422ABCNx x y S x x x ⎛⎫-+∴==+=-++=--+ ⎪⎝⎭梯形， 当2x =时，y 取最大值，最大值为10． ································································· 7分 〔3〕90B AMN ∠=∠=°，∴要使ABM AMN △∽△，必须有AM ABMN BM=， ··················································· 9分 由〔1〕知AM ABMN MC=， BM MC ∴=，∴当点M 运动到BC 的中点时，ABM AMN △∽△，现在2x =．····························· 12分〔其它正确的解法，参照评分建议按步给分〕。