上海市浦东新区第四教育署2017.doc
2020年上海市浦东新区第四教育署八年级(上)期中数学试卷
八年级(上)期中数学试卷题号 得分一二三四总分一、选择题(本大题共 6 小题,共 12.0 分) 1. 下列各式中,一定是二次根式的是( )A. B. C.D.2. 下列关于 的方程中,是一元二次方程的为()x A. B. C. D. D.ax 2+bx+c=0 x 2- =1 2x+3y-5=0x 2-1=0 3. 能与 可以合并的二次根式的是()A.B.C.4. 一元二次方程- 2+6 -10=0 的根的情况是( x x )A. C.B. D. 有两个相等的实数根只有一个实数根 有两个不相等的实数根 没有实数根5. 近几年,手机支付用户规模增长迅速,据统计 2016 年手机支付用户约为 4.69 亿人, 连续两年增长后,2018 年手机支付用户达到约 5.83 亿人,如果设这两年手机支付 用户的年均增长率为 ,则根据题意可以列出方程为( x ) A. C. B. D. 4.69(1+x )=5.83 4.69(1+2x )=5.83 4.69(1+x ) =5.83 4.69(1-x ) =5.832 2 6. 下列命题中,真命题的序号为() ①相等的角是对顶角;②在同一平面内,若 ∥ , ∥ ,则∥ ③同旁内角互补; ④ 互为邻补角的两角的角平分线互相垂直 a b b c a c A. B. C. D.②④ ①② ①③ ①②④ 二、填空题(本大题共 12 小题,共 36.0 分)7.× =______. 在实数范围内有意义,则 的取值范围是______. 8. 若x 9. 已知实数 在数轴上的位置如图所示,则化简| -1|-=______.a a 10. 方程 2 2- =0 的根是______. x x 11. 不等式 -2< 的解集是______.xx 12. 已知 = 是关于 的一元二次方程 2+3 -1=0 的根,则 x m x x x=______.13. 把方程 2-2=4 用配方法化为( + )2= 的形式,则 的值是______.x m mnx x n 14. 把命题“平行于同一条直线的两条直线互相平行”改写成“如果…,那么…”的形式为______.,∠=∠,∠∠ACE =30°,则∠AD E =______.∠A CB= EF D =90° B,则17. 如图,点 是△ AB C 三个内角的角平分线的交点,连接 、 、 ,∠AP BP CP = , A CB 60°P 且 = ,则∠C A+AP B C的度数为____. C AB 18. 定义符号 , )的含义为:当 ≥ 时, min{a b}=b a b 的解是______.min{1 -2 =-2 min{-3 -2 =-3 min{x -x}=x -1 , 当 < 时, , , min{a b a bmin{a b}=a如: , ) , , ) ,则方程 ,2 三、计算题(本大题共 小题,共 1 分) 6.0 19. 解方程: 2 .x +2x-1=0 四、解答题(本大题共 小题,共 7 分)46.0 20. +2 -(). - 21. 计算:22. 解方程:(3x-2)( ) () x+1 =2 2-3x 23. 已知,关于 的一元二次方程 2 x -2x+m-1=0有两个不相等的实数根.x ( )求 的取值范围; 1m ( )如果为非负整数,且该方程的根都是整数,求 的值.m2 m24. 某农场要建一个饲养场(矩形 ABC D )两面靠现有墙(A D 位置的墙最大可用长度 AB 15 为 米, 位置的墙最大可用长度为 米),另两边用木栏围成,中间也用木 27 栏隔开,分成两个场地及一处通道,并在如图所示的三处各留 米宽的门(不用木1栏).建成后木栏总长 米.设饲养场(矩形 ABC D )的一边 45长为 米.x AB ( )饲养场另一边 1 米(用含 的代数式表示). B C=______ x ( )若饲养场的面积为 平方米,求 的值. 2 180 x25. 如图,在△AB C 中,, 平分∠BAC 且AB=2AC A D26. 在等腰△OAB 和等腰△O C D 中, , ,连接O A=O B O C=O D、 交于点 . A C B D M ( )如图 ,若∠ ∠A O B= C O D =40°: 1 1 ①AC 与 的数量关系为______; B D ②∠A M B 的度数为______ .( )如图 ,若∠ ∠A O B= C O D =90°: 2 2 ①判断 与 A C B D之间存在怎样的数量关系?并说明理由;②求∠A M B 的度数.答案和解析1.【答案】D【解析】 【分析】本题考查的是二次根式的定义,形如 (a ≥0)的式子叫做二次根式.根据二次根式的 定义判断即可. 【解答】解:A 当 a +1≥0,即 a ≥ 时, 是二次根式,本选项错误; 是二次根式,本选项错误;. -1 B.当 a-1≥0,即 a ≥1 时, C.当 a -1≥0 时, 是二次根式,本选项错误;2 D.a +2a+2=a +2a+1+1=(a+1) +1>0,2 2 2 ∴一定是二次根式,本选项正确;故选 D . 2.【答案】D【解析】解:A 、a ,b ≠0 时,是一元一次方程,故 A 错误;=0B 、是分式方程,故 B 错误;C 、是二元一次方程,故 C 错误;D 、是一元二次方程,故 D 正确.故选:D .根据一元二次方程的定义解答.一元二次方程必须满足四个条件:( )未知数的最高1次数是 ;( )二次项系数不为 ;( )是整式方程;( )含有一个未知数.由这四2 2 03 4个条件对四个选项进行验证.本题考查了一元二次方程的概念,判断一个方程是否是一元二次方程,首先要看是否是 整式方程,然后看化简后是否是只含有一个未知数且未知数的最高次数是 .23.【答案】A【解析】解:A 、 ,能与 合并; =3 B 、 C 、 不能与 合并; 不能与 合并; D 、 不能与 合并;故选:A .根据同类二次根式的概念判断即可.本题考查的是同类二次根式的概念,把几个二次根式化为最简二次根式后,如果它们的 被开方数相同,就把这几个二次根式叫做同类二次根式. 4.【答案】D【解析】解:∵△ ( ) ( ) < ,=6 -4× -1 × -10 =36-40=-4 0 2 ∴方程没有实数根.故选:D .先计算判别式的值,然后根据判别式的意义判断方程根的情况.本题考查了一元二次方程 ax bx c (a ≠0)的根的判别式△ b ac :当△> ,方程有 + + =0 2 = -4 20 两个不相等的实数根;当△ ,方程有两个相等的实数根;当△< ,方程没有实数根.=05.【答案】C【解析】解:设这两年手机支付用户的年平均增长率为x,依题意,得4.69(1+x)=5.83.2故选:C.如果设这两年手机支付用户的年平均增长率为x,那么2017年手机支付用户约为4.69(1+x)亿人,2018年手机支付用户约为4.69(1+x)亿人,根据2018年手机支付用户2达到约5.83亿人列出方程.本题考查的是由实际问题抽象出一元二次方程-平均增长率问题.解决这类问题所用的等量关系一般是:增长前的量×(1+平均增长率)增长的=次增数长后的量.6.【答案】D【解析】解:①相等的角不一定是对顶角,是假命题;②在同一平面内,若a∥b,b∥c,则a∥c,是真命题;③两直线平行,同旁内角互补;是假命题;④互为邻补角的两角的角平分线互相垂直,是真命题;故选:D.根据对顶角的性质、平行线的判定、平行线的性质、角平分线的性质判断即可.本题考查的是命题的真假判断,正确的命题叫真命题,错误的命题叫做假命题.判断命题的真假关键是要熟悉课本中的性质定理.【解析】解:×===.根据二次根式的乘法法则计算,结果要化简.主要考查了二次根式的乘法运算.二次根式的乘法法则=(a≥0,b≥0).8.【答案】x≤【解析】解:根据题意得:1-3x≥0,解得:x≤ .故答案是:x≤ .根据二次根式的性质,被开方数大于或等于0,可以求出x的范围.本题考查的知识点为:二次根式的被开方数是非负数.9.【答案】-2a【解析】解:∵从数轴可知:-1<a<0<1,∴|a-1|-=|a-1|-|a+1|=-a+1-a-1=-2a.故答案为-2a.根据数轴得出-1<a<0<1,根据二次根式的性质得出|a-1|-|a+1|,去掉绝对值符号合并同类项即可.本题考查了二次根式的性质,绝对值以及数轴的应用,利用数轴可以比较任意两个实数 的大小,即在数轴上表示的两个实数,右边的总比左边的大,在原点左侧,绝对值大的 反而小.【解析】解:左边因式分解,得: (2 -1)=0,x x ∴ =0 或 2 -1=0, x x 解得: =0, = , x x 1 2 故答案为: =0, = . x x 1 2将方程的左边分解为两个一次因式的乘积;令每个因式分别为零得到两个一元一次方程;解这两个一元一次方程可得.本题主要考查因式分解法解一元二次方程,通过将方程左边因式分解,把原方程进行了 降次,把解一元二次方程转化为解一元一次方程的问题是因式分解法解一元二次方程的 关键. 11. 【答案】 >-2 -2x 【解析】解: -2< x ,x( -1) >-2,xx >-,x >-2 -2.故答案为: >-2 -2.x不等式移项合并,把 系数化为 1,即可求出解集.x此题考查了解一元一次不等式和分母有理化,熟练掌握运算法则是解本题的关键. 12.【答案】4 【解析】解:把 = 代入 +3 -1=0,得 +3 -1=0. 2x x 2 m mx m 所以 1-3 = . 2 m m 所以==4.故答案是:4.把 = 代入已知方程,得到 1-3 = ,整体代入所求的代数式进行求值即可. 2 m m x m 考查了一元二次方程的解,能使一元二次方程左右两边相等的未知数的值是一元二次方程的解.又因为只含有一个未知数的方程的解也叫做这个方程的根,所以,一元二次方 程的解也称为一元二次方程的根. 13.【答案】-12【解析】解:∵ -2=4, 2 x x ∴ 2-4 =2, x x∴ 2-4 +4=2+4, x x∴( -2)2=6,x ∴ =-2, =6, m n ∴ =-12, mn故答案为:-12根据配方法可求解,值,再代入计算即可求出答案.m n本题考查一元二次方程,解题的关键是熟练运用一元二次方程的解法,本题属于基础题型.14.【答案】如果两条直线平行于同一条直线,那么这两条直线平行【解析】解:命题可以改写为:“如果两条直线平行于同一条直线,那么这两条直线平行”.命题由题设和结论两部分组成,通常写成“如果…那么…”的形式.“如果”后面接题设,“那么”后面接结论.本题考查命题的改写.任何一个命题都可以写成“如果…那么…”的形式.“如果”后面接题设,“那么”后面接结论.在改写过程中,不能简单地把题设部分、结论部分分别塞在“如果”、“那么”后面,要适当增减词语,保证句子通顺而不改变原意.15.【答案】52°【解析】解:∵∠=∠BA C D AE,-∠∴∠AB D=∠2=30°,∵∠1=22°,∴∠3=∠1+∠AB D=22°+30°=52°,故答案为:52°利用全等三角形的性质得出∠AB D=∠2=30°,再利用三角形的外角得出得出即可.本题考查了全等三角形的判定与性质、三角形的外角性质,求出∠1=∠EAC是证明三角形全等的关键.16.【答案】4【解析】解:∵∠=∠=90°,⊥,A CB EF D AB D E∴∠+∠=90°,∠+∠=90°B D B A∴∠=∠,且∠A DA CB EF D=∠=90°,=,AB D E∴△ABC≌△(D E F AAS)∴==6,==8,D F A C EF B C∴=+-=4.由“AAS”可证△AB C≌△DEF,可得==6,==8,即可求A C D F EF BCC F B CD F B D的长.C F本题考查了全等三角形的判定与性质,证明△ABC≌△DEF是本题的关键.17.【答案】80°【解析】【分析】本题考查了全等三角形的判定和性质,角平分线的性质,添加恰当辅助线构造全等三角形是解本题的关键.由角平分线的性质可得∠=60°,由“SAS”可证+∠,可得=,由等腰三角形的性质和外角ABP BAP△ACP≌△ECP,可得=,∠AP PEC AP CEP PE BE=∠性质可得∠PAB=2∠PBA,即可求解.【解答】解:如图,在BC上截取CE=A C,连接PE,∵∠ACB=60°,∴∠CAB+∠AB C=120°∵点P是△AB C三个内角的角平分线的交点,∴∠CAP=∠BAP=∠CAB,∠ABP=∠C BP=∠AB C,∠AC P=∠BCP,∴∠ABP+∠BAP=60°∵CA=CE,∠ACP=∠BCP,CP=CP∴△ACP≌△ECP(SAS)∴AP=PE,∠CAP=∠CEP∵CA+AP=B C,且C B=CE+BE,∴AP=BE,∴BE=PE,∴∠EPB=∠EBP,∴∠PEC=∠EBP+∠EPB=2∠PBE=∠CAP,∴∠PAB=2∠PBA,∵∠ACB=60°,∴∠ABC+∠CAB=120°,∴∠ABP+∠BAP=60°,∴∠PAB=40°,∴∠CAB=80°,故答案为80°.18.【答案】或【解析】【分析】本题考查新定义,一元二次方程,解题的关键是正确理解定义以及熟练运用一元二次方程的解法,本题属于基础题型,根据新定义以及一元二次方程的解法即可求出答案.【解答】解:当x≥-x时,即x≥0,此时-x=x-1,2解得:x=,∵x≥0,∴x=;当x<-x时,即x<0,此时x=x-1,2解得:x=,∵x<0,故答案为:或.19. 【答案】解:∵ +2 =1,x x2 ∴ 2+2 +1=1+1,即( +1)2=2, x x x 则 +1= x , ∴ =-1 x. 【解析】将常数项移到方程的右边,两边都加上一次项系数一半的平方配成完全平方式 后,再开方即可得.本题主要考查解一元二次方程的能力,熟练掌握解一元二次方程的几种常用方法:直接 开平方法、因式分解法、公式法、配方法,结合方程的特点选择合适、简便的方法是解 题的关键. 20.【答案】解: +2 -( =2 +2 -3 + - ) =3 - .【解析】根据二次根式的性质把各个二次根式化简,合并同类二次根式即可.本题考查的是二次根式的加减法,掌握二次根式的性质、二次根式的加减法法则是解题 的关键. 21.【答案】解:原式=5× ×3 .=5 【解析】根据二次根式的乘除法法则计算即可解答.考查了二次根式的乘除法,二次根式的性质与化简.化简后的二次根式中的被开方数中 每一个因数(或因式)的指数都小于根指数2. 22. 【答案】解:∵(3 -2)( +1)=-2(3 -2), x x x ∴(3 -2)( +1)+2(3 -2)=0, x x x 则(3 -2)( +3)=0, x x ∴3 -2=0 或 +3=0, x x 解得 = 或 =-3.x x 【解析】利用因式分解法求解可得.本题主要考查解一元二次方程的能力,熟练掌握解一元二次方程的几种常用方法:直接 开平方法、因式分解法、公式法、配方法,结合方程的特点选择合适、简便的方法是解 题的关键 23. 【答案】解:(1)根据题意得:(-2) -4( -1)>0, 2 m 解得: <2.m故 的取值范围为 <2; m m (2)由(1)得: <2m∵ 为非负整数, m∴ =0 或 1, m把 =0 代入原方程得: -2 -1=0, 2 x xm 解得: =1- , =1+ , x x 1 2 m =0 不合题意舍去;把m=1代入原方程得:x-2x=0,2解得:x=0,x=2.12故m的值是1.【解析】本题主要考查根的判别式及一元二次方程的解,熟练掌握根的判别式及一元二次方程的解的定义是解题的关键.(1)根据方程有两个不相等的实数根知△>0,据此列出关于m的不等式,解之可得;(2)由(1)中m的范围且m为非负整数,且该方程的根都是整数得出m的值即可.24.【答案】解:(1)(48-3x);(2)由题意得:x(48-3x)=180解得x=6,x=10,12∵0≤48-3x≤27,0≤x≤15,∴7≤x≤15,∴x=10.【解析】【分析】考查了一元二次方程的应用.解题关键是要读懂题目的意思,根据题目给出的条件,找出合适的等量关系,列出方程,再求解.(1)用(总长+2个2米的门的宽度)-3x即为所求;(2)由(1)表示饲养场面积计算即可,【解答】解:(1)由题意得:(48-3x)米.故答案是:(48-3x);(2)见答案.25.∵A D=B D∴BE=AE∵AB=2AC∴AE=A C∵A D平分∠BA C∴∠BA D=∠CA D在△AE D和△AC D中∴△EA D≌△CA D(SAS)∴∠C=∠AE D=90°∴C D⊥A C.【解析】本题考查了角平分线的定义,全等三角形的判定与性质,等腰三角形的性质,证明△EA D≌△CAD是本题的关键.过D作DE⊥AB于E,由等腰三角形的性质可得AE=BE=A C,由“SAS”可证△EA D≌△CA D,可得∠C=∠AE D=90°,可得结论.26.【答案】解:(1)①AC=B D;②40°;(2)①AC=B D,理由如下:∵∠A OB=∠C O D=90°,∴∠A OB+∠A O D=∠C O D+∠A O D,∴∠B O D=∠A O C,∴△B O D≌△A O C(SAS),∴B D=A C;②∵△B O D≌△A O C,∴∠OB D=∠OA C,又∵∠OAB+∠OBA=90°,∠AB O=∠AB M+∠O B D,∠M AB=∠M A O+∠O A B,∴∠M AB+∠M B A=90°,又∵在△A M B中,∠A M B+∠AB M+∠BA M=180°,∴∠A M B=180°-(∠AB M+∠BA M)=180°-90°=90°.【解析】【分析】本题考查了等腰直角三角形性质,全等三角形判定和性质,含30°的直角三角形性质,勾股定理等.熟练掌握全等三角形判定和性质是解题关键.(1)①先证明:∠B O D=∠A O C,再证明△B O D≌△A O C(SAS),即可得AC=B D;②由△B O D≌△A O C及三角形内角和定理即可求得∠A M B=40°;(2)①证明△B O D≌△A O C(SAS)即可得B D=A C,②根据全等三角形性质和三角形内角和定理即可求得∠A M B.【解答】解:(1)①∵∠AO B=∠C O D,∴∠A OB+∠A O D=∠C O D+∠A O D,∴∠B O D=∠A O C,在△B O D和△A O C中,,∴△B O D≌△A O C(SAS),∴AC=B D;故答案为:AC=BD;②∵△B O D≌△A O C,∴∠OB D=∠OA C,∵∠A OB=40°,∴∠OAB+∠OBA=180°-∠AOB=180°-40°=140°,又∵∠OAB+∠OBA=∠OAB+∠AB D+∠O B D∴∠OAB+∠OBA=∠O A B+∠AB D+∠OA C=140°,∴∠ M A B AB M + =140°, ∵在△AB M 中,∠ ∴∠A M B =40°; 故答案为:40°; (2)见答案. +∠ + =180°, A M B M A BABM【解析】本题考查了角平分线的定义,全等三角形的判定与性质,等腰三角形的性质,证明△EA D≌△CAD是本题的关键.过D作DE⊥AB于E,由等腰三角形的性质可得AE=BE=A C,由“SAS”可证△EA D≌△CA D,可得∠C=∠AE D=90°,可得结论.26.【答案】解:(1)①AC=B D;②40°;(2)①AC=B D,理由如下:∵∠A OB=∠C O D=90°,∴∠A OB+∠A O D=∠C O D+∠A O D,∴∠B O D=∠A O C,在△B O D和△A O C中,,∴△B O D≌△A O C(SAS),∴B D=A C;②∵△B O D≌△A O C,∴∠OB D=∠OA C,又∵∠OAB+∠OBA=90°,∠AB O=∠AB M+∠O B D,∠M AB=∠M A O+∠O A B,∴∠M AB+∠M B A=90°,又∵在△A M B中,∠A M B+∠AB M+∠BA M=180°,∴∠A M B=180°-(∠AB M+∠BA M)=180°-90°=90°.【解析】【分析】本题考查了等腰直角三角形性质,全等三角形判定和性质,含30°的直角三角形性质,勾股定理等.熟练掌握全等三角形判定和性质是解题关键.(1)①先证明:∠B O D=∠A O C,再证明△B O D≌△A O C(SAS),即可得AC=B D;②由△B O D≌△A O C及三角形内角和定理即可求得∠A M B=40°;(2)①证明△B O D≌△A O C(SAS)即可得B D=A C,②根据全等三角形性质和三角形内角和定理即可求得∠A M B.【解答】解:(1)①∵∠AO B=∠C O D,∴∠A OB+∠A O D=∠C O D+∠A O D,∴∠B O D=∠A O C,在△B O D和△A O C中,,∴△B O D≌△A O C(SAS),∴AC=B D;故答案为:AC=BD;②∵△B O D≌△A O C,∴∠OB D=∠OA C,∵∠A OB=40°,∴∠OAB+∠OBA=180°-∠AOB=180°-40°=140°,又∵∠OAB+∠OBA=∠OAB+∠AB D+∠O B D∴∠OAB+∠OBA=∠O A B+∠AB D+∠OA C=140°,∴∠ M A B AB M + =140°, ∵在△AB M 中,∠ ∴∠A M B =40°; 故答案为:40°; (2)见答案. +∠ + =180°, A M B M A BABM。
2018-2019学年上海市浦东新区第四教育署七年级上学期期末考试英语试题含答案
2018-2019学年上海市浦东新区第四教育署七年级(上)期末英语试卷(五四学制)二、Fill in the blanks with proper words according to the phonetic symbols (根据音标写出正确的单词)(5分)12.(1分)We should collect the used['bætərɪz].13.(1分)Please read the[ɪnˈstrʌkʃn] book carefully.14.(1分)Tom always exercises [ˈregjʊlə(r)li] to keep healthy.15.(1分)The plane [lændz] at Shanghai Pudong International Airport every week.16.(1分)The students in Grade 7 are going to[reɪz] some money for the SPCA.三、Choose the best answer (选择最恰当的答案)(15分)17.(1分)Which of the following underlined parts is different in pronunciation from the others?()A.breakfast B.head C.great D.already18.(1分)John is a kind boy and he often helps _________.()A.others B.another C.the other D.other19.(1分)More than _________ students went to Shanghai Museum yesterday.()A.hundred B.two hundredC.two hundred of D.hundreds of20.(1分)A sign tells us how and where to go.It's a(n)_________ sign.()A.direction B.informationC.warning D.instruction21.(1分)Don't forget to add some eggs _________ the flour.()A.and B.to C.for D.with22.(1分)You _________ buy anything because we have got enough for the picnic.()A.needn't to B.don't need toC.need to D.don't need23.(1分)You must eat _________ sugar,or you will be fatter and fatter.()A.more B.much C.less D.fewer24.(1分)The Thai food pineapple fried rice tastes _________.()A.well B.nicestC.wonderfully D.delicious25.(1分)I think travelling by train is _________ and more exciting than by plane.()A.more cheap B.more cheaperC.much cheap D.much cheaper26.(1分)beautiful the girl is!()A.How B.What a C.What D.How a27.(1分)All the fans are looking forward to _________ their favourite star in the hall.()A.meet B.meeting C.met D.meets28.(1分)Most students like chatting on the Internet._________ Alice.()A.So does B.So doC.Neither does D.Neither do29.(1分)If it _________ rainy tomorrow,we will go for an outing.()A.doesn't B.will be C.is D.isn't30.(1分)Bob _________ up late,but now he gets up very early.()A.use to get B.is used to getC.is used to getting D.used to get31.(1分)﹣Shall we go to see the film on Children's Day?﹣________.()A.I'm sorry to hear that.B.What great fun!C.That's a good idea.D.Thank you very much.四、Complete the following passage with the words or phrases in the box. Each can only be used once (将下列单词或词组填入空格.每空限填一词,每词只能填一次)(6分)32.(3分)将下列单词或词组填入空格.每空限填一词,每词只能填一次A.also B.happy C.ill D.sleepingFood is very important in our daily life.It gives you energy for playing,working,thinking,and watching TV.Even when you're (1),your body needs energy to keep youbreathing.Food (2)helps you grow,keep warm,and get better when you are (3).33.(3分)将下列单词或词组填入空格.每空限填一词,每词只能填一次A.different B.when C.how D.waterYour body can tell you(1)to eat,but it cannot tell you what to eat.People like eating all kinds of food.Ice cream and cola taste good,but they are not good for your health.You need to eat foods with (2)nutrients(营养素).Fish,meat and eggs help you grow.Rice,bread,potatoes give you lots of energy.Fruit and vegetables give your body(3).There are so much delicious food for you to choose.Next time you are in a restaurant,why not try something new?五、Complete the sentences with the given words in their proper forms(用括号中所给单词的适当形式完成下列句子)(6分)34.(1分)This healthy soup is a of different kinds of vegetables.(mix)35.(1分)The terrible earthquake made many children(home).36.(1分)In the fire,the brave fireman put the boy in a place.(safely)37.(1分)We'll put some at home to welcome the Spring Festival.(decorate)38.(1分),the parents found their lost child with the help of the police.(final)39.(1分)Peter is looking forward to in the countryside next year.(hike)六、Rewrite the following sentences as required(根据所给要求,改写下列句子.每空格限填一词)(10分)40.(2分)I'd like to be a singer in the future.(改为否定句)I to be a singer in the future.41.(2分)Her brother has already been to the USA.(改为一般疑问句)her brother been to the USA?42.(2分)There's little food in the fridge.(改为反意疑问句)There's little food in the fridge,?43.(2分)Linda liked the pink dress better than the yellow one.(保持句意不变)Linda the pink dress the yellow one.44.(2分)didn't,any longer,he,pocket money,his,all,spend,(.)(连词成句)七、Reading comprehension (阅读理解): (共23分)A. Choose the best answer. (根据短文内容,选择最恰当的答案) (6分)45.(6分)Florida Hotel Swimming PoolOPENING HOUR :Monday ﹣Saturday 08:00﹣22:00 Sunday 09:00﹣17:00 No diving . No running .No eating or drinking in the pool . Use the steps to go into the pool . Children should be with an adult at all times .times.(1)Where can we find these signs?A.In the restaurant.B.In the park.C.Beside the swimming pool.D.On the road.(2)When is the swimming pool open?A.Monday to Saturday.B.On Sunday.C.On Saturday.D.Every day.(3)What can we do in the swimming pool?A.We can use the steps to go into the pool.B.We can eat the hamburgers.C.We can shout at the lifeguard.D.We can run after each other.(4)What time can we go to the swimming pool on Sundays?A.At 8:00.B.At 10:00.C.At 18:00.D.At 21:00.(5)What does "respect" mean?A.loveB.be afraid ofC.be polite toD.hate(6)What kind of signs are they?A.Direction signs and warning signs.B.Warning signs and instruction signsC.Information signs and instruction signs.D.Warning signs and direction signs.B. Choose the words or expressions and complete the passage.(选择最恰当的单词或词组完成短文)(6分)46.(6分)The American couch﹣potato becomes even a part of the American tradition!In America,many people often(1)their free time sitting on a couch watching TV.As there are many channels to(2),the TV can be quite interesting and very addictive.One may sit and watch TV for hours(3)stop!These people who do(4)all day except watching TV are called couch﹣potatoes.While watching TV,most of these people look so frozen that they almost look dead!Everyone is sitting still with eyes looking right into the same black box.They don't take exercise and usually get(5).So in many ways,these people look like real potatoes!Now,you know (6)couch﹣potato means.Are you couch﹣potatoes?(1)A.take B.pay C.cost D.spend(2)A.like B.choose C.give D.make(3)A.with B.without C.for D.since(4)A.nothing B.everything C.something D.anything(5)A.healthy B.happy C.fat D.clever(6)A.how B.why C.when D.whatC. Read the passage and fill in the blanks with proper words(在短文的空格内填入适当的词,使其内容通顺,每空格限填一词,首字母已给)(5分)47.(5分)在短文的空格内填入适当的词,使其内容通顺,每空格限填一词,首字母已给.Four weeks before my mum's 40th birthday,Dad said to me,Jason,let's do something s(1)for your mother.Let's give her a surprise party.I thought it was a great idea,so we started planning.Mum loves the river,so Dad booked a party boat.It came with aKaraoke machine.That was good,because she loves s(2).Dad invited some friends and n(3),and some people from Mum's work.He even invited some of her old school friends.They all promised to keep it a s.One day,Mum said,It's my birthday soon.Let's go to a restaurant,just the three of us.Dad and I looked at each other and tried not to laugh.Dad said,I'll book a restaurant beside the river.On Mum's birthday,all the people went to the river at 6:30 and got onto the boat.At seven o'clock,the three of us s(5)out for the restaurant.When we walked past the boat,everyone shouted," Surprise!" Mum really was surprised and very happy.She loved every minute.After the party,she said,'Thanks for the best birthday ever!'D. Answer the questions.(根据短文内容回答下列问题)(6分)48.(6分)Pasta is one of the most popular foods in the world.People love pasta because it is cheap,quick and easy to make.You can eat pasta on its own with cheese,or with sauce.Pasta comes in so many different shapes and sizes.There are at least 200 different types of pasta.One of the reasons there are so many types of pasta is that each part of Italy has its own pasta.Pasta has been an important part of Italians' diet since the 13th century.It started as a food for the rich.But by the 17th century,all Italians enjoyed pasta.This is when the first pasta machines were invented.As pasta became easier to make,it became cheaper to buy.Pasta is still an important part of the Italian diet.But now it is an important part of many other people's diet too.Many people buy pasta,but it's easy to make.You only need flour,salt and eggs.It takes some time to mix the pasta dough together.Then you can choose which type of pasta shape to make.A pasta machine is very useful for this,but you can also cut the pasta by hand.(1)Why is pasta so popular?(2)How many types of pasta in the world?(3)Pasta is a food just for the rich,isn't it?(4)What happened after the pasta machine was invented?(5)Would you like to buy pasta or make it by hand?Why?八、Writing (写作)(10分)49.(10分)Write at least 60 words on the topic ‘A/An ______ birthday party' (以"A/An______birthday party" )为题,写一篇不少于60个单词的短文,要求内容适切,意思连贯,标点符号不占格.2018-2019学年上海市浦东新区第四教育署七年级(上)期末英语试卷(五四学制)参考答案与试题解析二、Fill in the blanks with proper words according to the phonetic symbols (根据音标写出正确的单词)(5分)12.【解答】batteries13.【解答】instruction14.【解答】regularly15.【解答】lands16.【解答】raise三、Choose the best answer (选择最恰当的答案)(15分)17.【解答】breakfast的音标是[ˈbrekfəst];head的音标是[hed];great的音标是[greɪt];already 的音标是[ɔ:lˈredi].因此可知great中的划线部分与其他几项不一样.故选:C。
基层单位领导班子及其成员党风廉政建设
许叠云副校长:主要承担学校全体党员教师加强党风廉政建设和职业道德建设相关责任,包括上级文件、政策、规定、要求的学习贯彻,执行落实以及制定与完善学校的规章制度等。
许叠云
本人坚守廉洁制度的相关要求,严于律己。在本人分管领域严格落实:在教材及参考资料征订上严格按照教委文件执行;招生工作公正公开;教学业务考核程序规范,公开正义。
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注:本书一式多份(根据班子成员数),本单位留存一份,班子成员每人留存一份,报所属上级党委二份。
张海
本人严格遵守党风廉政建设各项规定,始终保持共产党员的光荣本色,严格执行签定的廉政责任书各项要求,除了自己以身作则外,还要在班子中开展廉政教育,保证班子成员的廉政建设落到实处,廉政重点主要体现在依法办学,加强民主管理和民主监督,实行校务公开,按规定程序办事。
柳易华
本人严格遵守上级规定的所有廉洁廉政的基本要求。组织支部党员积极参加“廉洁文化进校园”党建示范点建设的相关活动,利用道德讲堂面向全体学生开展社会主义核心价值观建设及“廉洁文化进校园”主题活动。
领导班子成员党风廉政建设责任分工计划
(包括本人廉政承诺、分管领域党风廉政建设的重点项目和具体措施等)
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分工计划概述
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陶宝青
本人严格遵守上级部门有关党风廉政建设的所有规定,主动与上级党委签约,自觉履行廉政承诺,不断深化“廉洁文化进校园”党建工作示范点建设,继续按“四横四纵”模式推进,真正将廉洁自律内化于心,外化于行。
基层单位领导班子及其成员党风廉政建设
2016-2017学年上海市浦东新区第四教育署九年级(上)期中数学试卷(五四学制)
2016-2017学年上海市浦东新区第四教育署九年级(上)期中数学试卷(五四学制)一、选择题:(本大题共6题,每题4分,满分24分)1.(4分)已知两个相似三角形的相似比为1:4,则它们的周长比为()A.1:4 B.4:1 C.1:2 D.1:162.(4分)在△ABC中,∠C=90°,AC=3,CB=4,则cotA的值为()A.B.C.D.3.(4分)在△ABC中,点D,E分别在边AB,AC上,AD:BD=1:2,那么下列条件中能够判断DE∥BC的是()A.B.C.D.4.(4分)已知x:b=c:a,求作x,则下列作图正确的是()A.B. C.D.5.(4分)在梯形ABCD中,AD∥BC,点E,F分别是边AB,CD的中点,AD=BC,=,那么等于()A.B.C.D.6.(4分)如图,在△ABC中,点D、E分别在边AB、AC上,如果DE∥BC,且∠DCE=∠B,那么下列说法中,错误的是()A.△ADE∽△ABC B.△ADE∽△ACD C.△ADE∽△DCB D.△DEC∽△CDB 二、填空题:(本大题共12题,每题4分,满分48分)7.(4分)如果=,那么=.8.(4分)已知线段b是线段a、c的比例中项,且a=2 cm,b=4 cm,那么c=cm.9.(4分)已知点P是线段AB的黄金分割点,AB=4cm,则较长线段AP的长是= cm.10.(4分)计算:sin30°+cos30°•tan60°=.11.(4分)在△ABC中,点D,E分别在线段AB,AC的反向延长线上,DE∥BC,AB=3,AC=2,AD=1,那么CE=.12.(4分)已知等腰△ABC中,AB=AC=5,cos∠B=,则△ABC的面积为.13.(4分)在Rt△ABC中,∠C=90°,∠B=α,AB=m,那么边AC的长为.14.(4分)如图,已知在△ABC中,D是边BC的中点,点E在边BA的延长线上,AE=AB,,那么=.15.(4分)已知菱形ABCD的边长为6,对角线AC与BD相交于点O,OE⊥AB,垂足为点E,AC=4,那么sin∠AOE=.16.(4分)如图,梯形ABCD中,AD∥BC,对角线AC,DB交于点O,如果S△AOD=1,S△BOC=3,那么S△AOB=.17.(4分)新定义:我们把两条中线互相垂直的三角形称为“中垂三角形”.如图所示,△ABC中,AF、BE是中线,且AF⊥BE,垂足为P,像△ABC这样的三角形称为“中垂三角形”,如果∠ABE=30°,AB=4,那么此时AC的长为.18.(4分)将▱ABCD(如图)绕点A旋转后,点D落在边AB上的点D′,点C落到C′,且点C′、B、C在一直线上.如果AB=13,AD=3,那么∠A的余弦值为.三、简答题:(本大题共4题,满分40分)19.(10分)已知:==,x﹣y+z=6,求:代数式3x﹣2y+z的值.20.(10分)已知:如图,在梯形ABCD中,AD∥BC,AD=BC,点M是边BC 的中点,=,=.(1)填空:=,=.(结果用、表示).(2)直接在图中画出向量3+.(不要求写作法,但要指出图中表示结论的向量)21.(10分)如图,已知AD∥BE∥CF,它们依次交直线l1、l2于点A、B、C和点D、E、F.(1)如果AB=6,BC=8,DF=21,求DE的长;(2)如果DE:DF=2:5,AD=9,CF=14,求BE的长.22.(10分)已知:如图,在梯形ABCD中,AD∥BC,点E在边AD上,CE与BD 相交于点F,AD=4,AB=5,BC=BD=6,DE=3.(1)求证:△DFE∽△DAB;(2)求线段CF的长.四、解答题:(本大题共3题,满分38分)23.(12分)如图,点E是四边形ABCD的对角线BD上的一点,∠BAE=∠CBD=∠DAC.(1)求证:DE•AB=BC•AE;(2)求证:∠AED+∠ADC=180°.24.(12分)已知:在平面直角坐标系xOy中,一次函数y=kx+b(k≠0)的图象经过点A(4,0),C(0,﹣4),另有一点B(﹣2,0).(1)求一次函数解析式;(2)联结BC,点P是反比例函数y=的第一象限图象上一点,过点P作y轴的垂线PQ,垂足为Q.如果△QPO与△BCO相似,求P点坐标;(3)联结AC,求∠ACB的正弦值.25.(14分)已知:正方形ABCD的边长为4,点E为BC的中点,点P为AB上一动点,沿PE翻折△BPE得到△FPE,直线PF交CD边于点Q,交直线AD于点G,联接EQ.(1)如图,当BP=1.5时,求CQ的长;(2)如图,当点G在射线AD上时,BP=x,DG=y,求y关于x的函数关系式,并写出x的取值范围;(3)延长EF交直线AD于点H,若△CQE与△FHG相似,求BP的长.2016-2017学年上海市浦东新区第四教育署九年级(上)期中数学试卷(五四学制)参考答案与试题解析一、选择题:(本大题共6题,每题4分,满分24分)1.(4分)已知两个相似三角形的相似比为1:4,则它们的周长比为()A.1:4 B.4:1 C.1:2 D.1:16【分析】直接利用相似三角形的周长比等于相似比,进而得出答案.【解答】解:∵两个相似三角形的相似比为1:4,∴它们的周长比为:1:4.故选:A.2.(4分)在△ABC中,∠C=90°,AC=3,CB=4,则cotA的值为()A.B.C.D.【分析】根据锐角A的余切=邻边:对边可得答案.【解答】解:∵,∠C=90°,AC=2,BC=1,∴cotA==,故选:D.3.(4分)在△ABC中,点D,E分别在边AB,AC上,AD:BD=1:2,那么下列条件中能够判断DE∥BC的是()A.B.C.D.【分析】可先假设DE∥BC,由平行得出其对应线段成比例,进而可得出结论.【解答】解:如图,可假设DE∥BC,则可得==,==,但若只有==,并不能得出线段DE∥BC.故选D.4.(4分)已知x:b=c:a,求作x,则下列作图正确的是()A.B. C.D.【分析】根据第四比例线段的定义列出比例式,再根据平行线分线段成比例定理对各选项图形列出比例式即可得解.【解答】解:∵x:b=c:a,∴=,A、作出的为=,故本选项正确;B、作出的为=,故本选项错误;C、线段x无法先作出,故本选项错误;D、作出的为=,故本选项错误;故选A.5.(4分)在梯形ABCD中,AD∥BC,点E,F分别是边AB,CD的中点,AD=BC,=,那么等于()A.B.C.D.【分析】首先根据梯形的中位线的性质,求得EF=BC,又由,即可求得的值.【解答】解:∵AD∥BC,点E、F分别是边AB、CD的中点,∴EF=(AD+BC),∵AD=BC,∴EF=BC,∵,∴.故选C.6.(4分)如图,在△ABC中,点D、E分别在边AB、AC上,如果DE∥BC,且∠DCE=∠B,那么下列说法中,错误的是()A.△ADE∽△ABC B.△ADE∽△ACD C.△ADE∽△DCB D.△DEC∽△CDB 【分析】由相似三角形的判定方法得出A、B、D正确,C不正确;即可得出结论.【解答】解:∵DE∥BC,∴△ADE∽△ABC,∠BCD=∠CDE,∠ADE=∠B,∠AED=∠ACB,∵∠DCE=∠B,∴∠ADE=∠DCE,又∵∠A=∠A,∴△ADE∽△ACD;∵∠BCD=∠CDE,∠DCE=∠B,∴△DEC∽△CDB;∵∠B=∠ADE,但是∠BCD<∠AED,且∠BCD≠∠A,∴△ADE与△DCB不相似;正确的判断是A、B、D,错误的判断是C;故选:C.二、填空题:(本大题共12题,每题4分,满分48分)7.(4分)如果=,那么=.【分析】根据比例的性质即可得到结论.【解答】解:∵=,∴==﹣,故答案为:﹣.8.(4分)已知线段b是线段a、c的比例中项,且a=2 cm,b=4 cm,那么c=8 cm.【分析】根据比例中项的定义,列出比例式即可求解.【解答】解:根据比例中项的概念结合比例的基本性质,得:比例中项的平方等于两条线段的乘积,所以b2=ac,即42=2c,c=8.故答案为:8.9.(4分)已知点P是线段AB的黄金分割点,AB=4cm,则较长线段AP的长是= 2﹣2cm.【分析】根据黄金分割的概念得到AP=AB,把AB=4cm代入计算即可.【解答】解:∵P是线段AB的黄金分割点,AP>BP,∴AP=AB,而AB=6cm,∴AP=3×=2﹣2.故答案是:2﹣2.10.(4分)计算:sin30°+cos30°•tan60°=2.【分析】分别把特殊角的三角函数值代入,然后再计算即可.【解答】解:原式=+•==2,故答案为:2.11.(4分)在△ABC中,点D,E分别在线段AB,AC的反向延长线上,DE∥BC,AB=3,AC=2,AD=1,那么CE=.【分析】由DE∥BC,则可得其对应线段成比例,进而再结合题干中的条件,即可得出答案【解答】解:∵DE∥BC,∴=,∵AB=3,AC=2,AD=1,∴=,∴AE=,∴CE=AE+AC=+2=,故答案为12.(4分)已知等腰△ABC中,AB=AC=5,cos∠B=,则△ABC的面积为12.【分析】如图作AD⊥BC于D,根据cos∠B=求出BD,再利用勾股定理求出AD,即可解决问题.【解答】解:如图,作AD⊥BC于D,∵AB=AC=5,cos∠B=,∴BD=DC=3,AD===4,∴S=•BC•AD=×6×4=12.△ABC故答案为12.13.(4分)在Rt△ABC中,∠C=90°,∠B=α,AB=m,那么边AC的长为m•sinα.【分析】根据三角函数值的求值可以求得sinα=,故根据AB=m即可求得AC 的值,即可解题.【解答】解:∠C=90°,∠B=α,AB=m,则sinα=,∴AC=AB•sinα=m•sinα.故答案为m•sinα.14.(4分)如图,已知在△ABC中,D是边BC的中点,点E在边BA的延长线上,AE=AB,,那么=2﹣.【分析】根据中点定义可得BD=BC,然后表示出,,再利用向量的三角形法则解答即可.【解答】解:∵D是边BC的中点,∴BD=BC,∵=,∴=,∵AE=AB,=,∴=2,∴=﹣=2﹣.故答案为:2﹣.15.(4分)已知菱形ABCD的边长为6,对角线AC与BD相交于点O,OE⊥AB,垂足为点E,AC=4,那么sin∠AOE=.【分析】菱形对角线互相垂直,故AC⊥BD,根据∠OAE=∠BAO,∠OEA=∠AOB 可以判定△OAE∽△ABO,∴∠AOE=∠BAO,根据AO和AB的值即可求得sin∠AOE的值.【解答】解:∵菱形对角线互相垂直,∴∠OEA=∠AOB,∵∠OAE=∠BAO ,∴△OAE ∽△ABO ,∴∠AOE=∠ABO ,∵AO=AC=2,AB=6,∴sin ∠AOE=sin ∠ABO==. 故答案为:.16.(4分)如图,梯形ABCD 中,AD ∥BC ,对角线AC ,DB 交于点O ,如果S △AOD =1,S △BOC =3,那么S △AOB = .【分析】由AD 与BC 平行,得到三角形AOD 与三角形BOC 相等,由面积比等于相似比的平方求出所求即可.【解答】解:∵AD ∥BC ,∴△AOD ∽△COB ,∵S △AOD =1,S △BOC =3,即S △AOD :S △BOC =1:3,∴OA :OC=1:,∵S △AOB 与S △BOC 高相同,∴S △AOB :S △BOC =1:,则S △AOB =, 故答案为:17.(4分)新定义:我们把两条中线互相垂直的三角形称为“中垂三角形”.如图所示,△ABC 中,AF 、BE 是中线,且AF ⊥BE ,垂足为P ,像△ABC 这样的三角形称为“中垂三角形”,如果∠ABE=30°,AB=4,那么此时AC 的长为 2 .【分析】根据三角形中位线的性质,得到EF∥AB,EF=AB=2,再由勾股定理得到结果.【解答】解:如图,连接EF,∵AF、BE是中线,∴EF是△CAB的中位线,可得:EF=×4=2,∵EF∥AB,∴△PEF~△ABP,∴===,在Rt△ABP中,AB=4,∠ABP=30°,∴AP=2,PB=2,∴PF=1,PE=,在Rt△APE中,∴AE=,∴AC=2,故答案为:.18.(4分)将▱ABCD(如图)绕点A旋转后,点D落在边AB上的点D′,点C落到C′,且点C′、B、C在一直线上.如果AB=13,AD=3,那么∠A的余弦值为.【分析】根据平行四边形的性质得∠DAB=∠D′AB′,AB=AB′=C′D′=13,再由AB′∥C′D′得∠D′AB′=∠BD′C′,加上∠C=∠DAB,则∠C=∠BD′C′,接着由点C′、B、C在一直线上,AB∥CD得到∠C=∠C′BD′,所以∠C′BD′=∠BD′C′,可判断△C′BD′为等腰三角形,作C′H⊥D′B,根据等腰三角形的性质得BH=D′H,由于BD′=10得到D′H=5,然后根据余弦的定义得到cos∠HD′C′=,由此得到∠A的余弦值.【解答】解:∵▱ABCD绕点A旋转后得到▱AB′C′D′,∴∠DAB=∠D′AB′,AB=AB′=C′D′=13,∵AB′∥C′D′,∴∠D′AB′=∠BD′C′,∵四边形ABCD为平行四边形,∴∠C=∠DAB,∴∠C=∠BD′C′,∵点C′、B、C在一直线上,而AB∥CD,∴∠C=∠C′BD′,∴∠C′BD′=∠BD′C′,∴△C′BD′为等腰三角形,作C′H⊥D′B,则BH=D′H,∵AB=13,AD=3,∴BD′=10,∴D′H=5,∴cos∠HD′C′==,即∠A的余弦值为.故答案为.三、简答题:(本大题共4题,满分40分)19.(10分)已知:==,x﹣y+z=6,求:代数式3x﹣2y+z的值.【分析】根据比例的性质,可用设===k,进而解答即可.【解答】解;设===k,可得:x=2k,y=3k,z=4k,把x=2k,y=3k,z=4k代入x﹣y+z=6,可得:2k﹣3k+4k=6,解得:k=2,所以x=4,y=6,z=8,把x=4,y=6,z=8代入3x﹣2y+z=12﹣12+8=8.20.(10分)已知:如图,在梯形ABCD中,AD∥BC,AD=BC,点M是边BC 的中点,=,=.(1)填空:=,=﹣﹣.(结果用、表示).(2)直接在图中画出向量3+.(不要求写作法,但要指出图中表示结论的向量)【分析】(1)由在梯形ABCD中,AD∥BC,AD=,可求得,然后由点M是边BC的中点,求得,再利用三角形法则求解即可求得;(2)利用三角形法则连结AC求解即可.【解答】解:(1)∵在梯形ABCD中,AD∥BC,AD=BC,=,∴=3=3,∵点M是边BC的中点,∴==;∴=﹣=﹣(+)=﹣﹣;故答案为:;﹣﹣;(2)如图所示,连结AC,就是所求作的向量.21.(10分)如图,已知AD∥BE∥CF,它们依次交直线l1、l2于点A、B、C和点D、E、F.(1)如果AB=6,BC=8,DF=21,求DE的长;(2)如果DE:DF=2:5,AD=9,CF=14,求BE的长.【分析】(1)根据三条平行线截两条直线,所得的对应线段成比例可得,再由AB=6,BC=8,DF=21即可求出DE的长.(2)过点D作DG∥AC,交BE于点H,交CF于点G,运用比例关系求出HE及HB的长,然后即可得出BE的长.【解答】解:(1)∵AD∥BE∥CF,∴,∵AB=6,BC=8,DF=21,∴,∴DE=9.(2)过点D作DG∥AC,交BE于点H,交CF于点G,则CG=BH=AD=9,∴GF=14﹣9=5,∵HE∥GF,∴,∵DE:DF=2:5,GF=5,∴,∴HE=2,∴BE=9+2=11.22.(10分)已知:如图,在梯形ABCD中,AD∥BC,点E在边AD上,CE与BD 相交于点F,AD=4,AB=5,BC=BD=6,DE=3.(1)求证:△DFE∽△DAB;(2)求线段CF的长.【分析】(1)AD∥BC,DE=3,BC=6,,.又∠EDF=∠BDA,即可证明△DFE∽△DAB.(2)由△DFE∽△DAB,利用对应边成比例,将已知数值代入即可求得答案.【解答】证明:(1)∵AD∥BC,DE=3,BC=6,∴,∴,∵BD=6,∴DF=2.∵DA=4,∴.∴.又∵∠EDF=∠BDA,∴△DFE∽△DAB.(2)∵△DFE∽△DAB,∴.∵AB=5,∴,∴EF==2.5.∵DE∥BC,∴.∴,∴CF=5.(或利用△CFB≌△BAD).四、解答题:(本大题共3题,满分38分)23.(12分)如图,点E是四边形ABCD的对角线BD上的一点,∠BAE=∠CBD=∠DAC.(1)求证:DE•AB=BC•AE;(2)求证:∠AED+∠ADC=180°.【分析】(1)根据已知条件得到∠BAC=∠EAD,根据三角形额外角的性质得到∠ABC=∠AED,推出△ABC∽△AED,根据根据相似三角形对应边成比例得到结论;(2)根据相似三角形的性质得到,推出△ABE∽△ACD,根据相似三角形的性质得到∠AEB=∠ADC,等量代换即可得到结论.【解答】证明:(1)∵∠BAE=∠DAC,∴∠BAE+∠EAC=∠DAC+∠EAC,即∠BAC=∠EAD,∵∠ABC=∠ABE+∠CBD,∠AED=∠ABE+∠BAE,∵∠CBD=∠BAE,∴∠ABC=∠AED,∴△ABC∽△AED,∴,∴DE•AB=BC•AE;(2)∵△ABC∽△AED,∴,即,∵∠BAE=∠DAC∴△ABE∽△ACD,∴∠AEB=∠ADC,∵∠AED+∠AEB=180°,∴∠AED+∠ADC=180°.24.(12分)已知:在平面直角坐标系xOy中,一次函数y=kx+b(k≠0)的图象经过点A(4,0),C(0,﹣4),另有一点B(﹣2,0).(1)求一次函数解析式;(2)联结BC,点P是反比例函数y=的第一象限图象上一点,过点P作y轴的垂线PQ,垂足为Q.如果△QPO与△BCO相似,求P点坐标;(3)联结AC,求∠ACB的正弦值.【分析】(1)把A、C两点的坐标代入可求得一次函数解析式;(2)可设出P点坐标为(x,),由△POQ和△BCO相似可知有两种情况,当∠BCO=∠POQ时,利用两角的正切值相等,可得到关于x的方程,可求得x的值,可得P点坐标;当∠BCO=∠OPQ时,同理可求得P点坐标;(3)作AD⊥BC于点D,由△ABC的面积可求得AD的长,且可求得AC的长,在Rt△ADC中,可求得∠ACB的正弦值.【解答】解:(1)把A(4,0),C(0,﹣4)代入y=kx+b可得,解得,∴一次函数解析式为y=x﹣4;(2)设P点坐标为(x,),∵,∠PQO=∠BOC=90°,∴当△POQ和△BCO时是有∠BCO=∠POQ或∠BCO=∠OPQ,①当∠BCO=∠POQ时,则tan∠BCO=tan∠POQ,∴=,解得x=2或x=﹣2(舍去),∴P点坐标为(2,);②当∠BCO=∠OPQ时,则tan∠BCO=tan∠OPQ,∴=,解得x=或x=﹣(舍去),∴P点坐标为(,2);综上可得P点坐标为(2,)或(,2);(3)作AD⊥BC交BC于D,如图,∵A(4,0),C(0,﹣4),B(﹣2,0),∴AC=4,BC==2=AB•OC=BC•AD,∵S△ABC∴6×4=2AD,∴AD,∴在Rt△ADC中,sin∠ACB===.25.(14分)已知:正方形ABCD的边长为4,点E为BC的中点,点P为AB上一动点,沿PE翻折△BPE得到△FPE,直线PF交CD边于点Q,交直线AD于点G,联接EQ.(1)如图,当BP=1.5时,求CQ的长;(2)如图,当点G在射线AD上时,BP=x,DG=y,求y关于x的函数关系式,并写出x的取值范围;(3)延长EF交直线AD于点H,若△CQE与△FHG相似,求BP的长.【分析】(1)首先确定∠PEQ=90°,即PE⊥EQ,然后利用△PBE∽△ECQ,列出比例式求出CD的长度;(2)根据△PBE∽△ECQ,求出DQ的表达式;由QD∥AP,列出比例式求解;(3)本问分两种情形,需要分类讨论,避免漏解.【解答】解:(1)由翻折性质,可知PE为∠BPQ的角平分线,且BE=FE.∵点E为BC中点,∴EC=EB=EF,∴QE为∠CQP的角平分线.∵AB∥CD,∴∠BPQ+∠CQP=180°,即2∠EPQ+2∠EQP=180°,∴∠EPQ+∠EQP=90°,∴∠PEQ=90°,即PE⊥EQ.易证△PBE∽△ECQ,∴,即,解得:CQ=.(2)由(1)知△PBE∽△ECQ,∴,即,∴CQ=,∴DQ=4﹣.∵QD∥AP,∴,又AP=4﹣x,AG=4+y,∴,∴y=(1<x<2).(3)由题意知:∠C=90°=∠GFH.①当点G在线段AD的延长线上时,如答图1所示.由题意知:∠G=∠CQE∵∠CQE=∠FQE,∴∠DQG=∠FQC=2∠CQE=2∠G.∵∠DQG+∠G=90°,∴∠G=30°,∴∠BEP=∠CQE=∠G=30°,∴BP=BE•tan30°=;②当点G在线段DA的延长线上时,如答图2所示.由题意知:∠FHG=∠CQE.同理可得:∠G=30°,∴∠BPE=∠G=30°,∴∠BEP=60°,∴BP=BE•tan60°=.综上所述,BP的长为或.。
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上海市浦东新区六年级数学上学期期中试题沪科版五四制
上海市浦东新区第四教育署2016-2017学年六年级数学上学期期中试题(考试时间:90分钟满分:100分)题号一二三四五总分得分一、选择题(本大题共6小题,每小题2分,满分12分)1、2是最小的( )(A)自然数(B)奇数(C)素数(D)合数2、如果两个数互素,那么这两个数( )(A)没有公因数(B)只有公因数1 (C)两个数都是素数(D)都是素因数3、下列说法中正确的是( )(A)任何正整数的因数至少有两个(B)1是所有正整数的因数(C)一个数的倍数总比它的因数大(D)3的因数只有它本身4、大于小于的最简分数有( )(A)1个(B)2个(C)3个(D)无数个5、在分数,,,,,中,与相等的分数共有( )(A)1个(B)2个(C)3个(D)4个6、把一张正方形纸片,连续对折三次,得到的图形面积是这个正方形面积的( )(A) (B)(C)(D)二、填空题(本大题共14小题,每小题2分,满分28分)7、24和16的最大公因数是___________.8、分解素因数:54=__________________.9、一个数的最小倍数是12,这个数的因数有______________________.10、,,则A和B的最小公倍数是__________.11、分数中有______个 .12、比较大小:(填“>”,“<” , “=”).13、一本书看了80页,占全书的,这本书共有________页.14、用带分数填空:200分钟=_________小时.15、一根15米长的绳子,先剪掉它的,再剪掉米,还剩下_______米.16、一个长方形的长是米,宽是长的,则长方形的面积是_________平方米.17、解方程: ,得 .18、计算:.19、如果定义新的运算符号“#”为:,那么(3#2)#2=____________.20、一个正整数n ,若它的所有因数中最小的两个因数的和是4,最大的两个因数的和是100,则n 的值为__________三、简答题(本大题共4小题,每题6分,满分24分) 21、计算: 22、计算:23、计算: 24、解方程:四、应用题:(本大题共3小题,第25题10分,第26题7分,第27题7分,满分24分)25、在数轴上分别画出点A 、B 、C 、D ,点A 表示数,点B 表示数,点C 表示数0,点D 表示数;并将点A 、B 、C 、D 所表示的数用“”连接.26、一张长42厘米,宽30厘米的铁皮,要把它切割成若干个面积相等的正方形铁皮且没有剩余,切成的正方形铁皮至少有几张?求出此时正方形的边长。
2024-2025学年上海市浦东新区第四教育署数学九年级第一学期开学联考试题【含答案】
2024-2025学年上海市浦东新区第四教育署数学九年级第一学期开学联考试题题号一二三四五总分得分A 卷(100分)一、选择题(本大题共8个小题,每小题4分,共32分,每小题均有四个选项,其中只有一项符合题目要求)1、(4分)要了解全校学生的课外作业负担情况,你认为以下抽样方法中比较合理的是()A .调查九年级全体学生B .调查七、八、九年级各30名学生C .调查全体女生D .调查全体男生2、(4分)如图,在△ABC 中,∠C=90°,点D ,E 分别在边AC ,AB 上.若∠B=∠ADE ,则下列结论正确的是()A .∠A 和∠B 互为补角B .∠B 和∠ADE 互为补角C .∠A 和∠ADE 互为余角D .∠AED 和∠DEB 互为余角3、(4分)下列各组数中,不能作为直角三角形的三边长的是()A .1.5,2,3B .6,8,10C .5,12,13D .15,20,254、(4分)如图,△ABC 中,CD ⊥AB 于D ,且E 是AC 的中点.若AD=6,DE=5,则CD 的长等于()A .7B .8C .9D .105、(4分)如图,在正方形ABCD 中,,AC BD 相交于点O ,,E F 分别为,BC CD 上的两点,BE CF =,,AE BF ,分别交,BD AC 于,M N 两点,连,OE OF ,下列结论:①AE BF =;②AE BF ⊥;③2CE CF BD +=;④14ABCD OECF S S =正方形四边形,其中正确的是()A .①②B .①④C .①②④D .①②③④6、(4分)点(1,m)为直线21y x =-上一点,则OA 的长度为A .1B C D .7、(4分)一家鞋店在一段时间内销售了某种男鞋200双,各种尺码鞋的销售量如下表所示:尺码/厘米2323.52424.52525.526销售量/双5102239564325一般来讲,鞋店老板比较关心哪种尺码的鞋最畅销,也就是关心卖出的鞋的尺码组成的一组数据是()A .平均数B .中位数C .众数D .方差8、(4分)在中国集邮总公司设计的2017年纪特邮票首日纪念戳图案中,可以看作中心对称图形的是()A .B .C .D .二、填空题(本大题共5个小题,每小题4分,共20分)9、(4分)函数y =中自变量x 的取值范围是_______.10、(4分)在实数范围内有意义,则x 的取值范围是_____.11、(4分)如图,正方形ABCD 中,对角线AC ,BD 交于点O ,E 点在BC 上,EG OB ⊥,EF OC ⊥,垂足分别为点G ,F ,10AC =,则EG EF +=______.12、(4分)如图,量角器的直径与直角三角板ABC 的斜边AB 重合,其中量角器0刻度线的端点N 与点A 重合,射线CP 从CA 处出发沿顺时针方向以每秒3度的速度旋转,CP 与量角器的半圆弧交于点E ,第24秒时,点E 在量角器上对应的读数是度.13、(4分)如图,一块矩形的土地被分成4小块,用来种植4种不同的花卉,其中3块面积分别是220m ,230m ,236m ,则第四块土地的面积是____2m .三、解答题(本大题共5个小题,共48分)14、(12分)某中学在一次爱心捐款活动中,全体同学积极踊跃捐款.现抽查了九年级(1)班全班同学捐款情况,并绘制出如下的统计表和统计图:捐款(元)2050100150200人数(人)412932求:(Ⅰ)m=_____,n=_____;(Ⅱ)求学生捐款数目的众数、中位数和平均数;(Ⅲ)若该校有学生2500人,估计该校学生共捐款多少元?15、(8分),宽是,他又设计一个面积与其相等的圆,请你帮助张老师求出圆的半径r.16、(8分)如图,△ABC 中,∠ACB =90°,AC =CB =2,以BC 为边向外作正方形BCDE ,动点M 从A 点出发,以每秒1个单位的速度沿着A →C →D 的路线向D 点匀速运动(M 不与A 、D 重合);过点M 作直线l ⊥AD ,l 与路线A →B →D 相交于N ,设运动时间为t 秒:(1)填空:当点M 在AC 上时,BN =(用含t 的代数式表示);(2)当点M 在CD 上时(含点C ),是否存在点M ,使△DEN 为等腰三角形?若存在,直接写出t 的值;若不存在,请说明理由;(3)过点N 作NF ⊥ED ,垂足为F ,矩形MDFN 与△ABD 重叠部分的面积为S ,求S 的最大值.17、(10分)如图,直线2y x m =+与x 轴交于点()2,0A -,直线y x n =-+与x 轴、y轴分别交于B 、C 两点,并与直线2y x m =+相交于点D ,若4AB =.()1求点D 的坐标;()2求出四边形AOCD 的面积;()3若E 为x 轴上一点,且ACE 为等腰三角形,写出点E 的坐标(直接写出答案).18、(10分)某草莓种植大户,今年从草莓上市到销售完需要20天,售价为11元/千克,成本y (元/千克)与第x 天成一次函数关系,当x=10时,y=7,当x=11时,y=6.1.(1)求成本y (元/千克)与第x 天的函数关系式并写出自变量x 的取值范围;(2)求第几天每千克的利润w (元)最大?最大利润是多少?(利润=售价-成本)B 卷(50分)一、填空题(本大题共5个小题,每小题4分,共20分)19、(4分)已知实数m ,n 满足3m 2+6m -5=0,3n 2+6n -5=0,则n m m n +=________20、(4分)如图,BE ,CD 是△ABC 的高,且BD =EC ,判定△BCD ≌△CBE 的依据是“_____”.21、(4分)如图,在平面直角坐标系中,已知OA=4,则点A 的坐标为____________,直线OA 的解析式为______________.22、(4分)如图,在四边形ABCD 中,AB CD ≠,E ,F ,G ,H 分别是AB ,BD ,CD ,AC 的中点,要使四边形EFGH 是菱形,四边形ABCD 还应满足的一个条件是______.23、(4分)端午期间,王老师一家自驾游去了离家170km 的某地,如图是他们离家的距离y(km)与汽车行驶时间x(h)之间的函数图象,当他们离目的地还有20km 时,汽车一共行驶的时间是_____.二、解答题(本大题共3个小题,共30分)24、(8分)关于x 的一元二次方程22(21)10x k x k ++++=有两个不等实根1x ,2x .(1)求实数k 的取值范围;(2)若方程两实根1x ,2x 满足1212x x x x +=-⋅,求k 的值。
2022年上海市浦东新区第四教育署英语九年级第一学期期末质量跟踪监视试题含解析
2022-2023学年九上英语期末模拟试卷注意事项:1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
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Ⅰ. 单项选择1、—I’ll be away for a long time.—Don’t worry. She can look after your pet ________.A.careful enough B.enough carefulC.carefully enough D.enough carefully2、________, the In ternet was only used by the government. But now it’s widely used in every field.A.As usual B.At first C.After all D.So far3、This kind of plant is seen in our city because it lives 4,500m above sea level and is hard to find.A.commonly B.always C.seldom D.easily4、There are ________ teachers in our school, ________of them are women teachers.A.two hundreds ; three fourth B.two hundred ; three fourthsC.two hundred; three forths D.two hundreds; three fourths5、---Everyone knows that knowledge is the most valuable for human beings.---Yes. As a student, we should study hard.A.standard B.treasure C.invention D.instruction6、- It's useless to regret what has been done. Don't make those mistakes again一I won't. That's a(n) _____A.order B.decision C.promise D.agreement7、--Dad, I had a quarrel with my deskmate yesterday. Could you tell me_________ to ask for forgiveness from him?--Of course. You can say sorry to him first.A.what I should do B.how can I speakC.where I can find him D.when I could come back8、—Why did he give up so early?----- Because he did not particulary want to______a competitive sportA.look up B.turn up C.take up D.show up9、Lucy is really brave. She always has the_______ to try again.A.pride B.courage C.praise D.influence10、The T-shirts are all wonderful. But in my son’s eyes, the blue one is ____________.A.nicer B.nicest C.the nicestⅡ. 完形填空11、About ten years ago when I was a student at college, I spent my summer holidays 1 at a museum. Life was hard for me then. Dad had lost his job and Mum was sick in bed. I was 2 if I would be able to go on with my study the next term.One day while I was working, I saw an old man come in with a little girl 3 a wheelchair(轮椅). As I looked at this girl 4 , I found that she had no arms 5 legs. She was wearing a little white dress and she also had a hat on.As the old man pushed the wheelchair up to me, I was busy with my work, I 6 my head toward the girl and gave her a wink(眨眼). When I took the money from her grandfather, I looked back at the girl. She was giving me the prettiest, largest smile I have 7 seen. Suddenly her handicap(生理缺陷)was gone and all I saw was this beautiful girl, whose 8 almost gave me a better understanding of what life is all about. She 9 me into her world of smiles, love and warmth.I’m now a successful businessman and whenever I 10 and think about the trouble of the world, I think about that little girl and the unforgettable lesson about 1ife that she taught me.1.A.visiting B.working C.paining D.cleaning2.A.hoping B.guessing C.wondering D.checking3.A.in B.for C.at D.with4.A.carefully B.hardly C.carelessly D.happily5.A.so B.and C.then D.or6.A.held B.turned C.1ifted D.gave7.A.never B.almost C.ever D.even8.A.face B.smile C.1anguage D.breath9.A.brought B.bought C.taught D.thought10.A.get down B.get up C.get off D.get overⅢ. 语法填空12、根据短文内容及首字母提示写出所缺单词,并将完整单词写在下面对应题号后的横线上。
2019-2020学年上海市浦东新区第四教育署九年级(上)期中数学试卷(五四学制)(解析版)
2019-2020学年上海市浦东新区第四教育署九年级(上)期中数学试卷(五四学制)一、选择题(本大题共6题,每题4分,满分24分)1.下列图形一定是相似图形的是()A.两个矩形B.两个周长相等的直角三角形C.两个正方形D.两个等腰三角形2.如果两个相似的三角形面积之比为4:9,那么它们对应的角平分线之比为()A.2:3B.4:9C.16:81D.9:133.如图,点A、B、C、D、E、F、G、H、K都是7×8方格纸中的格点,为使△DEM∽△ABC,则点M应是F、G、H、K四点中的()A.F B.G C.H D.K4.在Rt△ABC中,已知∠ACB=90°,BC=1,AB=2,那么下列结论正确的是()A.sin A=B.tan A=C.cos B=D.cot B=5.如图,△ABC的顶点是正方形网格的格点,则tan A的值为()A.B.C.D.6.如图,在平行四边形ABCD中,F是边AD上的一点,射线CF和BA的延长线交于点E,如果,那么的值是()A.B.C.D.二、填空题(本大题共12题,每题4分,满分48分)7.已知点P在线段AB上,AP=3PB,那么PB:AB=.8.在1:5000的地图上,某两地间的距离是20cm,那么这两地的实际距离为千米.9.计算:=.10.若α为锐角,已知cosα=,那么tanα=.11.如图,在平行四边形ABCD中,E为CD上一点,联结AE、BD,且AE、BD交于点F,若DE:EC=2:3,则S△DEF:S△ABF=.12.相邻两边长的比值是黄金比的矩形,叫做黄金矩形,从外形看,它最具美感.现在想要制作一张“黄金矩形”的贺年卡,如果较长的一条边长等于20厘米,那么相邻一条边的边长等于厘米.(保留根号)13.如图,AG∥BC,如果AF:FB=3:5,BC:CD=3:2,那么AE:EC=.14.如果α是锐角,且cotα=tan35°,那么α=度.15.如图,正方形DEFG内接于Rt△ABC,∠C=90°,AE=4,BF=9,则tan A=.16.如图,点D在△ABC的边BC上,已知点E、点F分别为△ABD和△ADC的重心,如果BC=12,那么两个三角形重心之间的距离EF的长等于.17.如图,在四边形ABDC中,连接BC,∠A=∠BCD=90°,∠D=30°,∠ABC=45°,如果,那么S四边形ABDC=.18.如图,在Rt△ABC中,∠C=90°,点D在边AB上,线段DC绕点D逆时针旋转,端点C恰巧落在边AC上的点E处.如果=m,=n.那么用含n的代数式表示m 是:m=.三、解答题(本大题共7题,满分78分)19.计算:.20.如图,已知O为△ABC内的一点,点D、E分别在边AB、AC上,且=,,设=,=,试用,表示.21.如图,在Rt△ABC中,∠ACB=90°,BC=6,AC=8.点D是AB边上一点,过点D 作DE∥BC,交边AC于E.过点C作CF∥AB,交DE的延长线于点F.(1)如果=,求线段EF的长;(2)求∠CFE的正弦值.22.如图,在Rt△ABC中,∠C=90°,点D是BC边上的一点,CD=6,cos∠ADC=,tan B=.(1)求AC和AB的长;(2)求sin∠BAD的值.23.已知:梯形ABCD中,AD∥BC,AD=AB,对角线AC、BD交于点E,点F在边BC 上,且∠BEF=∠BAC.(1)求证:△AED∽△CFE;(2)当EF∥DC时,求证:AE=DE.24.如图,线段AB=5,AD=4,∠A=90°,DP∥AB,点C为射线DP上一点,BE平分∠ABC交线段AD于点E(不与端点A、D重合).(1)当∠ABC为锐角,且tan∠ABC=2时,求四边形ABCD的面积;(2)当△ABE与△BCE相似时,求线段CD的长;(3)设CD=x,DE=y,求y关于x的函数关系式,并写出定义域.25.如图,已知线段AB,P是线段AB上任意一点(不与点A、B重合),分别以AP、BP 为边,在AB的同侧作等边△APD和△BPC,连接BD与PC交于点E,连接CD.(1)当BC⊥CD时,试求∠DBC的正切值;(2)若线段CD是线段DE和DB的比例中项,试求这时的值;(3)记四边形ABCD的面积为S,当P在线段AB上运动时,S与BD2是否成正比例,若成正比例,试求出比例系数;若不成正比例,试说明理由.2019-2020学年上海市浦东新区第四教育署九年级(上)期中数学试卷(五四学制)参考答案与试题解析一、选择题(本大题共6题,每题4分,满分24分)1.下列图形一定是相似图形的是()A.两个矩形B.两个周长相等的直角三角形C.两个正方形D.两个等腰三角形【解答】解:A、两个矩形,对应角相等,对应边不一定成比例,故不符合题意;B、两个周长相等的直角三角形的对应角不一定相等,不符合题意;C、两个正方形,形状相同,大小不一定相同,符合相似性定义,故符合题意;D、两个等腰三角形顶角不一定相等,故不符合题意.故选:C.2.如果两个相似的三角形面积之比为4:9,那么它们对应的角平分线之比为()A.2:3B.4:9C.16:81D.9:13【解答】解:两个相似的三角形面积之比为4:9,故它们的相似比为2:3,所以它们对应的角平分线之比为2:3.故选A.3.如图,点A、B、C、D、E、F、G、H、K都是7×8方格纸中的格点,为使△DEM∽△ABC,则点M应是F、G、H、K四点中的()A.F B.G C.H D.K【解答】解:根据题意,△DEM∽△ABC,AB=4,AC=6 DE=2∴DE:AB=DM:AC∴DM=3∴M应是H故选:C.4.在Rt△ABC中,已知∠ACB=90°,BC=1,AB=2,那么下列结论正确的是()A.sin A=B.tan A=C.cos B=D.cot B=【解答】解:如图所示:∵∠ACB=90°,BC=1,AB=2,∴AC=,∴sin A=,故选项A错误;tan A==,故选项B错误;cos B=,故选项C错误;cot B=,正确.故选:D.5.如图,△ABC的顶点是正方形网格的格点,则tan A的值为()A.B.C.D.【解答】解:连接CD.则CD=,AD=2,则tan A===.故选:A.6.如图,在平行四边形ABCD中,F是边AD上的一点,射线CF和BA的延长线交于点E,如果,那么的值是()A.B.C.D.【解答】解:∵在平行四边形ABCD中,∴AE∥CD,∴△EAF∽△CDF,∵,∴,∴,∵AF∥BC,∴△EAF∽△EBC,∴=,故选:D.二、填空题(本大题共12题,每题4分,满分48分)7.已知点P在线段AB上,AP=3PB,那么PB:AB=1:4.【解答】解:如图所示:,∵AP=4PB,那么PB:AB=PB:(AP+PB)=PB:4PB,∴PB:AB=1:4.故答案为:1:4.8.在1:5000的地图上,某两地间的距离是20cm,那么这两地的实际距离为1千米.【解答】解:设两地的实际距离是x厘米,则:1:5000=20:x,∴x=100000,∵100000cm=1千米,∴两地的实际距离是1千米.故答案为1.9.计算:=﹣+5.【解答】解:=2﹣3+5=﹣+5.故答案为:﹣+5.10.若α为锐角,已知cosα=,那么tanα=.【解答】解:由α为锐角,已知cosα=,得sinα==,由正切函数等于正弦值与与余弦的比,得tanα===,故答案为:.11.如图,在平行四边形ABCD中,E为CD上一点,联结AE、BD,且AE、BD交于点F,若DE:EC=2:3,则S△DEF:S△ABF=4:25.【解答】解:∵四边形ABCD是平行四边形,∴AB∥CD,AB=CD,∴△DEF∽△BAF,∴=()2,∵DE:EC=2:3,∴DE:CD=DE:AB=2:5,∴S△DEF:S△ABF=4:25.故答案为:4:25.12.相邻两边长的比值是黄金比的矩形,叫做黄金矩形,从外形看,它最具美感.现在想要制作一张“黄金矩形”的贺年卡,如果较长的一条边长等于20厘米,那么相邻一条边的边长等于(10﹣10)厘米.(保留根号)【解答】解:设相邻一条边的边长为x厘米,由黄金分割的定义可知,=,解得,x=10﹣10,故答案为:10﹣10.13.如图,AG∥BC,如果AF:FB=3:5,BC:CD=3:2,那么AE:EC=3:2.【解答】解:∵AG∥BC,∴△AGF∽△BDF,∴==,设AG=3k,BD=5k,∵=,∴=∴CD=2k,∵AG∥CD,∴△AGE∽△CDE,∴===,故答案为3:2.14.如果α是锐角,且cotα=tan35°,那么α=55度.【解答】解:∵cotα=tan35°,∴cot(90°﹣35°)=tan35°.∴α=55°.故答案是:55.15.如图,正方形DEFG内接于Rt△ABC,∠C=90°,AE=4,BF=9,则tan A=.【解答】解:∵四边形DEFG为正方形,∴∠DEA=∠GFB=90°,DE=GF,∵∠C=90°,∴∠A+∠B=∠A+∠ADE=90°,∴∠ADE=∠B,∴△ADE∽△GFB,∴=,即=,解得DE=6,∴tan A===,故答案为:.16.如图,点D在△ABC的边BC上,已知点E、点F分别为△ABD和△ADC的重心,如果BC=12,那么两个三角形重心之间的距离EF的长等于4.【解答】解:如图,连接AE并延长交BD于G,连接AF并延长交CD于H,∵点E、F分别是△ABD和△ACD的重心,∴DG=BD,DH=CD,AE=2GE,AF=2HF,∵BC=12,∴GH=DG+DH=(BD+CD)=BC=×12=6,∵AE=2GE,AF=2HF,∠EAF=∠GAH,∴△EAF∽△GAH,∴==,∴EF=4,故答案为:4.17.如图,在四边形ABDC中,连接BC,∠A=∠BCD=90°,∠D=30°,∠ABC=45°,如果,那么S四边形ABDC=.【解答】解:如右图,在Rt△ABC中,BC=,∠ABC=45°,∴∠ACB=45°,∴AB=AC=1,∴S△ABC=×1×1=;在Rt△BCD中,∠D=30°,BC=,∴BD=2,∴CD==,∴S△BCD=××=,∴S四边形ABCD=S△ABC+S△BCD=+=.故答案是.18.如图,在Rt△ABC中,∠C=90°,点D在边AB上,线段DC绕点D逆时针旋转,端点C恰巧落在边AC上的点E处.如果=m,=n.那么用含n的代数式表示m 是:m=2n+1.【解答】解:作DH⊥AC于H,如图,∵线段DC绕点D逆时针旋转,端点C恰巧落在边AC上的点E处,∴DE=DC,∴EH=CH,∵=n,即AE=nEC,∴AE=2nEH=2nCH,∵∠C=90°,∴DH∥BC,∴=,即m===2n+1.故答案为:2n+1.三、解答题(本大题共7题,满分78分)19.计算:.【解答】解:原式==.20.如图,已知O为△ABC内的一点,点D、E分别在边AB、AC上,且=,,设=,=,试用,表示.【解答】解:∵=,=,∴=﹣=﹣,∵=,∴=.又∵,∴DE∥BC∴=,∴DE=BC,∴=(﹣).21.如图,在Rt△ABC中,∠ACB=90°,BC=6,AC=8.点D是AB边上一点,过点D 作DE∥BC,交边AC于E.过点C作CF∥AB,交DE的延长线于点F.(1)如果=,求线段EF的长;(2)求∠CFE的正弦值.【解答】解:(1)∵DE∥BC,∴△ADE∽△ABC,∴==,又∵BC=6,∴DE=2,∵DF∥BC,CF∥AB,∴四边形BCFD是平行四边形,∴DF=BC=6,∴EF=DF﹣DE=4;(2)∵四边形BCFD是平行四边形,∴∠B=∠F,在Rt△ABC中,∠ACB=90°,BC=6,AC=8,利用勾股定理,得AB===10,∴sin B===,∴sin∠CFE=.22.如图,在Rt△ABC中,∠C=90°,点D是BC边上的一点,CD=6,cos∠ADC=,tan B=.(1)求AC和AB的长;(2)求sin∠BAD的值.【解答】解:(1)如图,在Rt△ACD中,∵∠ACD=90°,CD=6,cos∠ADC=,∴=,即=,则AD=10,∴由勾股定理知,AC===8.又∵tan B=,∴=,即=,则BC=12.∴在Rt△ABC中,利用勾股定理知,AB===4.综上所述,AC=8,AB=4;(2)如图,过点D作DE⊥AB于点E.由(1)易知,BD=6.∵tan B=,∴=.则BE=DE.则由勾股定理得到:62=DE2+DE2,解得DE=,∴sin∠BAD===.23.已知:梯形ABCD中,AD∥BC,AD=AB,对角线AC、BD交于点E,点F在边BC 上,且∠BEF=∠BAC.(1)求证:△AED∽△CFE;(2)当EF∥DC时,求证:AE=DE.【解答】证明:(1)∵∠BEC=∠BAC+∠ABD,∠BEC=∠BEF+∠FEC,又∵∠BEF=∠BAC,∴∠ABD=∠FEC,∵AD=AB,∴∠ABD=∠ADB,∴∠FEC=∠ADB,∵AD∥BC,∴∠DAE=∠ECF,∴△AED∽△CFE;(2)∵EF∥DC,∴∠FEC=∠ECD,∵∠ABD=∠FEC,∴∠ABD=∠ECD,∵∠AEB=∠DEC.∴△AEB∽△DEC,∴,∵AD∥BC,∴,∴.即AE2=DE2,∴AE=DE.24.如图,线段AB=5,AD=4,∠A=90°,DP∥AB,点C为射线DP上一点,BE平分∠ABC交线段AD于点E(不与端点A、D重合).(1)当∠ABC为锐角,且tan∠ABC=2时,求四边形ABCD的面积;(2)当△ABE与△BCE相似时,求线段CD的长;(3)设CD=x,DE=y,求y关于x的函数关系式,并写出定义域.【解答】解:(1)过C作CH⊥AB与H,由∠A=90°,DP∥AB,得四边形ADCH为矩形,在△BCH中,CH=AD=4,∠BHC=90°,tan∠CBH=2,得HB=CH÷2=2,所以CD=AH=5﹣2=3,则四边形ABCD的面积=.(2)由BE平分∠ABC,得∠ABE=∠EBC,①∠BCE=∠BAE=90°,由BE=BE,得△BEC≌△BEA,得BC=BA=5,于是在△BCH中,BH=,所以CD=AH=5﹣3=2.②∠BEC=∠BAE=90°,延长CE交BA延长线于T,由∠ABE=∠EBC,∠BEC=∠BET=90°,BE=BE,得△BEC≌△BET,得BC=BT,且CE=TE,又CD∥AT,得AT=CD.令CD=x,则在△BCH中,BC=BT=5+x,BH=5﹣x,∠BHC=90°,所以BC2=BH2+CH2,即(5+x)2=(5﹣x)2+42,解得.综上,当△ABE∽△EBC时,线段CD的长为2或.(3)延长BE交CD延长线于M.由AB∥CD,得∠M=∠ABE=∠CBM,所以CM=CB.在△BCH中,.则DM=CM﹣CD=,又DM∥AB,得,即,解得y=(0<x<4.1).25.如图,已知线段AB,P是线段AB上任意一点(不与点A、B重合),分别以AP、BP 为边,在AB的同侧作等边△APD和△BPC,连接BD与PC交于点E,连接CD.(1)当BC⊥CD时,试求∠DBC的正切值;(2)若线段CD是线段DE和DB的比例中项,试求这时的值;(3)记四边形ABCD的面积为S,当P在线段AB上运动时,S与BD2是否成正比例,若成正比例,试求出比例系数;若不成正比例,试说明理由.【解答】解:(1)∵等边△APD和△BPC,∴PC=BC,∠CPD=60°,∠DPA=∠CBP=60°,∴PD∥BC,∴∠DPC=∠PCB=60°,∵BC⊥CD,∴∠DCB=∠PDC=90°,∴∠DCP=30°,∴tan∠DBC===cos30°=;(2)由已知,CD2=DE•DB,即,又∵∠CDE=∠CDE,∴△DCE∽△DBC,∴,又∵CP=BC,,∵PD∥BC,∴,∴,∴CD=BE,∴,即点E是线段BD的黄金分割点.∴,又∵PC∥AD,∴,(3)设AP=a,PB=b,∴,,因为AD∥PC,PD∥BC,∴,,∴,∴,∴,作DH⊥AB,则,,∴BD2=DH2+BH2=(a)2+(a+b)2=a2+ab+b2,∴,∴S与BD2成正比例,比例系数为.。
上海市浦东新区第四教育署2019-2020学年九年级上学期期中数学试卷(五四学制) (含答案解析)
上海市浦东新区第四教育署2019-2020学年九年级上学期期中数学试卷(五四学制)一、选择题(本大题共6小题,共24.0分)1.下列两个图形一定相似的是()A. 两个菱形B. 两个矩形C. 两个正方形D. 两个等腰梯形2.若两个相似三角形的面积之比为1:4,则它们的周长之比为()A. 1︰2B. 1︰4C. 1︰5D. 1︰163.如图,在方格纸中,△ABC和△EPD的顶点均在格点上,若△ABC∽△EPD,则点P所在的格点为下列各点中的()A. P1B. P2C. P3D. P44.已知Rt△ABC中,∠C=90°,AC=4,BC=6,那么下列各式中,正确的是()A. sinA=23B. cosA=23C. tanA=23D. tanB=235.如图所示,△ABC的顶点是正方形网格的格点,则sin A的值为()A. 12B. √55C. √1010D. 2√556.如图,四边形ABCD为平行四边形,E,F为CD边的两个三等分点,连接AF,BE交于点G,则S△EFG:S△ABG=()A. 1:3B. 3:1C. 1:9D. 9:1二、填空题(本大题共12小题,共48.0分)7.线段a=4,线段b=9,线段c是线段a与线段b的比例中项,则线段c=________8.比例尺为1:50000的地图上,量得两地相距4厘米,则两地的实际距离为___ 千米.a⃗−b⃗ )=______ .9.化简:2a⃗−3(1310.已知α是锐角且tanα=3,则sinα+cosα=___________.411.如图,在平行四边形ABCD中,E为边BC上一点,AC与DE相交于点F,若CE=2EB,S△AFD=27,则S△EFC等于______.12.已知一本书的宽与长之比为黄金比,且这本书的长是20cm,则它的宽为______(结果保留根号).13.如图,在△ABC中,点D,E分别在边AB,AC上,DE//BC,∠BDC=∠CED,如果DE=4,CD=6,那么AD:AE等于______ .14.若sinα=cos40°,则锐角α=______.15.如图,正方形CDEF内接于Rt△ABC,点D、E、F分别在边AC、AB和BC上,当AD=2,BF=3时,正方形CDEF的面积是______ .16.如图,G为△ABC的重心,DE过点G,且DE//BC,交AB、AC,分别于D、E两点,若△ADE的面积为5,则四边形BDEC的面积为______.17.在△ABC中,已知AB=2,∠B=30°,AC=√2.则S△ABC=.18.如图,△ABC绕点A逆时针旋转得到△ADE,点D落在BC上,若∠B=65°,则∠CAE=______°.三、计算题(本大题共1小题,共10.0分)19.计算:sin223°+2sin60°+tan45°−tan60°+cos223°.四、解答题(本大题共6小题,共68.0分)20. 如图,在△ABC 中,点D 、E 分别在边AB 、AC 上,DE//BC ,且DE =23BC . (1)如果AC =6,求AE 的长;(2)设AB ⃗⃗⃗⃗⃗ =a ⃗ ,AC⃗⃗⃗⃗⃗ =b ⃗ ,求向量DE ⃗⃗⃗⃗⃗⃗ (用向量a ⃗ 、b ⃗ 表示).21. 已知Rt △ABC 中,∠B =90°,AC =20,AB =10,P 是边AC 上一点(不包括端点A 、C),过点P 作PE ⊥BC 于点E ,过点E 作EF//AC ,交AB 于点F.设PC =x ,PE =y .(1)求y 与x 的函数关系式;(2)是否存在点P 使△PEF 是Rt △?若存在,求此时的x 的值;若不存在,请说明理由.22.如图,在Rt△ABC中,∠C=90°,sinB=3,点D在BC边上,且∠ADC=545°,DC=6,求BD的长和tan∠BAD的值.23.如图,已知梯形ABCD中,AD//BC,AB=CD,点E在对角线AC上,且满足∠ADE=∠BAC.(1)求证:CD⋅AE=DE⋅BC;(2)以点A为圆心,AB长为半径画弧交边BC于点F,联结AF.求证:AF2=CE⋅CA.24.已知在菱形ABCD中,AB=4,∠BAD=120°,点P是直线AB上任意一点,联结PC.在∠PCD内部作射线CQ与对角线BD交于点Q(与B、D不重合),且∠PCQ=30°.(1)如图,当点P在边AB上时,如果BP=3,求线段PC的长;(2)当点P在射线BA上时,设BP=x,CQ=y,求y关于x的函数解析式及定义域;(3)联结PQ,直线PQ与直线BC交于点E,如果△QCE与△BCP相似,求线段BP的长.25.如图(1),已知点G在正方形ABCD的对角线AC上,GE⊥BC,垂足为点E,GF⊥CD,垂足为点F.(1)证明与推断:①求证:四边形CEGF是正方形;②推断:AG的值为______:BE(2)探究与证明:将正方形CEGF绕点C顺时针方向旋转α角(0∘<α<45∘),如图(2)所示,试探究线段AG与BE之间的数量关系,并说明理由;(3)拓展与运用:正方形CEGF在旋转过程中,当B,E,F三点在一条直线上时,如图(3)所示,延长CG交AD 于点H.(1)求证:△AHG∽△CHA;(2)若AG=6,GH=2√2,则BC=______.-------- 答案与解析 --------1.答案:C解析:解:A、两个菱形,对应边成比例,对应角不一定相等,不符合相似的定义,故不符合题意;B、两个矩形,对应角相等,对应边不一定成比例,不符合相似的定义,故不符合题意;C、两个正方形,对应角相等,对应边一定成比例,一定相似,故符合题意;D、两个等腰梯形同一底上的角不一定相等,对应边不一定成比例,不符合相似的定义,故不符合题意;故选:C.根据相似图形的定义:对应角相等,对应边成比例的两个图形一定相似,结合选项,用排除法求解.本题考查相似形的定义,熟悉各种图形的性质是解题的关键.2.答案:A解析:此题考查了相似三角形的性质.注意相似三角形的面积比等于相似比的平方,相似三角形的周长的比等于相似比.根据相似三角形的面积比等于相似比的平方,即可求得其相似比,又由相似三角形的周长的比等于相似比,即可求得答案.解:∵两个相似三角形的面积之比为1:4,∴它们的相似比为1:2,∴它们的周长之比为1:2.故选A.3.答案:B解析:本题考查了相似三角形的判定:两组对应边的比相等且夹角对应相等的两个三角形相似.利用两个三角形都为直角三角形,则根据两组对应边的比相等且夹角对应相等的两个三角形相似,当PDBC =DEAC时,△ABC∽△EPD,然后利用比例性质计算出PD后可判断P点的位置.解:∵∠EDP=∠ACB=90°,∴当PDBC =DEAC时,△ABC∽△EPD,即PD3=42,∴PD=6,∴点P在格点P2的位置.故选:B.4.答案:D解析:解:∵∠C=90°,BC=6,AC=4,∴AB=√62+42=2√13,A、sinA=BCAB =3√1313,故此选项错误;B、cosA=ACAB =2√1313,故此选项错误;C、tanA=BCAC =32,故此选项错误;D、tanB=ACBC =23,故此选项正确.故选:D.本题可以利用锐角三角函数的定义以及勾股定理分别求解,再进行判断即可.此题主要考查了锐角三角函数的定义以及勾股定理,熟练应用锐角三角函数的定义是解决问题的关键.5.答案:B解析:解:连接DC,由网格可得:CD⊥AB,则DC=√2,AC=√10,故sinA=DCAC =√2√10=√55.故选:B.直接连接DC,得出CD⊥AB,再结合勾股定理以及锐角三角函数关系得出答案.此题主要考查了锐角三角函数关系,正确构造直角三角形是解题关键.6.答案:C解析:本题考查平行四边形的性质、相似三角形的性质等知识,解题的关键是灵活运用所学知识解决问题,属于中考常考题型.利用相似三角形的性质面积比等于相似比的平方即可解决问题;解:∵四边形ABCD是平行四边形,∴CD=AB,CD//AB,∵DE=EF=FC,∴EF:AB=1:3,∴△EFG∽△BAG,∴S△EFGS△BAG =(EFAB)2=19,故选:C.7.答案:6解析:本题考查了比例线段,理解比例中项的概念,这里注意线段不能是负数.根据比例中项的定义,列出比例式即可得出线段c的值.解:根据比例中项的概念结合比例的基本性质,得比例中项的平方等于两条线段的乘积.即c2=ab,则c2=4×9,解得c=±6,(线段是正数,负值舍去).故答案为6.8.答案:2解析:此题考查比例的性质,解决的关键在于掌握比例的性质.解:比例尺为1:50000的地图上,量得两地相距4厘米,则两地的实际距离为50000×4=200000厘米=2千米,故答案为2.9.答案:a⃗+3b⃗解析:解:2a⃗−3(13a⃗−b⃗ ),=2a⃗−a⃗+3b⃗ ,=a⃗+3b⃗ .故答案为:a⃗+3b⃗ .根据向量的加减运算法则进行计算即可得解.本题考查了平面向量,熟记向量的加减运算法则是解题的关键.10.答案:75解析:本题考查同角的三角函数的关系,关键是掌握sin2α+cos2α=1,tanα=sinαcosα.设sinα=3k,cosα=4k,k>0,求出k值,即可求出sinα+cosα的值.解:∵tanα=34=sinαcosα,∴设sinα=3k,cosα=4k,k>0,∵sin2α+cos2α=1,∴9k2+16k2=1,∴k=15,∴sinα=35 ,cosα=45,∴sinα+cosα=75.故答案为75.11.答案:12解析:根据题意可知△EFC∽△DFA,根据相似比CE:AD即可求出面积比,从而得到△EFC的面积.本题考查的是相似三角形的判定与性质,利用相似三角形面积比是相似比的平方进行解题是关键.解:在平行四边形ABCD中,CE//AD,∴△EFC∽△DFA,,又∵CE=2EB,∴CECB =23,而CB=DA,∴CEDA =23,∴S△EFC27=49,∴S△EFC=12,故答案为12.12.答案:(10√5−10)cm解析:本题考查的是黄金分割的概念和性质,把一条线段分成两部分,使其中较长的线段为全线段与较短线段的比例中项,这样的线段分割叫做黄金分割,他们的比值√5−12叫做黄金比.根据黄金比值和题意列出关系式,计算即可得到答案.解:设宽为xcm,由题意得,x:20=√5−12,解得x=10√5−10.故答案为:(10√5−10)cm.13.答案:3:2解析:解:∵DE//BC,∴∠EDC=∠BCD,ADAE =BDEC∵∠BDC=∠DEC,∴△BDC∽△CED,∴BDCE =DCDE=64=32,∴ADAE =32.故答案为3:2.由DE//BC,推出∠EDC=∠BCD,ADAE =BDEC,由△BDC∽△CED,推出BDCE=DCDE=64=32,由此即可解决问题.本题考查相似三角形的判定和性质、平行线分线段成比例定理等知识,解题的关键是灵活运用相似三角形的性质,属于中考常考题型.14.答案:50°解析:本题考查互余两角三角函数的关系,解题的关键是明确题意,熟练掌握三角函数的意义.根据锐角三角函数的意义解答即可.解:∵sinα=cos40°,∴锐角α=90°−40°=50°,故答案为:50°.15.答案:6解析:本题考查了相似三角形的判定和性质,正方形的性质,熟练掌握相似三角形的性质定理是解题的关键.根据正方形的性质得到DE//BC,由平行线的性质得到∠AED=∠B,∠ADE=∠EFB=90°,推出△ADE∽△EFB,根据相似三角形的性质得到ADEF =DEBF,代入数据即可得到结论.解:∵四边形CDEF是正方形,∴DE//BC,∴∠AED=∠B,∠ADE=∠EFB=90°,∴△ADE∽△EFB,∴ADEF =DEBF,即2EF =DE3,∴DE⋅EF=2×3=6,∴正方形CDEF的面积是6.故答案为:6.16.答案:254解析:本题考查的是三角形的重心的性质、相似三角形的判定和性质,三角形的重心是三角形三条中线的交点,且重心到顶点的距离是它到对边中点的距离的2倍.连接AG并延长交BC于H,根据重心的概念得到AG=2GH,根据平行线的性质、相似三角形的性质计算即可.解:连接AG并延长交BC于H,∵G为△ABC的重心,∴AG=2GH,∵DE//BC,∴ADAB =AGAH=23,∵DE//BC,∴△ADE∽△ABC,相似比为23,∴△ADE与△ABC的面积之比为49,∵△ADE的面积为5,∴四边形BDEC的面积=254,故答案为:254.17.答案:√3±12解析:本题考查了含30°直角三角形的性质,勾股定理、三角形的面积,利用数形结合与分类讨论是解题的关键.分△ABC是锐角三角形与钝角三角形两种情况进行讨论,然后分别解直角△ABD与直角△ACD,求出AD、BD、CD的长,再根据S△ABC=12BC⋅AD,代入数值计算即可.解:当△ABC是锐角三角形时,过点A作AD⊥BC于点D,∵AB=2,∠B=30°,∴AD=12AB=1,∴由勾股定理可知:BD=√AB2−AD2=√22−12=√3,∵AC=√2,∴由勾股定理可知:CD=√AC2−AD2=√2−12=1,∴BC=BD+DC=√3+1,∴S△ABC=12BC⋅AD=12×(√3+1)×1=√3+12;当△ABC是钝角三角形时,同理可得:BD=√3,CD=1,∴BC =BD −DC =√3−1, ∴S △ABC =12BC ⋅AD =12×(√3−1)×1=√3−12. 故答案为√3±12.18.答案:50解析:解:∵将△ABC 绕点A 逆时针旋转得到△ADE ,∴AB =AD ,∠BAD =∠CAE ,∴∠B =∠ADB =65°∴∠BAD =∠CAE =180°−2×65°=50°故答案为:50由旋转的性质可得AB =AD ,∠BAD =∠CAE ,由等腰三角形的性质和三角形内角和性质可得∠CAE 的度数.本题考查了旋转的性质,熟练运用旋转的性质是本题的关键.19.答案:解:原式=sin 223°+ cos 223°+2×√32+1−√3 =1+√3+1−√3=2.解析:本题考查了特殊角的三角函数值和实数的运算,能熟记特殊角的三角函数值是解题的关键.根据sin 2a +cos 2a =1和特殊三角函数值代入计算即可.20.答案:解:(1)如图,∵DE//BC ,且DE =23BC ,∴AE AC =DE BC =23.又AC =6,∴AE =4.(2)∵AB ⃗⃗⃗⃗⃗ =a ⃗ ,AC ⃗⃗⃗⃗⃗ =b ⃗ ,∴BC ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ =b ⃗ −a ⃗ .又DE//BC ,DE =23BC ,∴DE ⃗⃗⃗⃗⃗⃗ =23BC ⃗⃗⃗⃗⃗ =23(b ⃗ −a ⃗ ).解析:考查了平行线分线段成比例定理,平面向量,需要掌握平面向量的三角形法则和平行向量的定义.(1)由平行线截线段成比例求得AE的长度;(2)利用平面向量的三角形法则解答.21.答案:解:(1)在Rt△ABC中,∠B=90°,AC=20,AB=10,∴sinC=12,∵PE⊥BC于点E,∴sinC=PEPC =12,∵PC=x,PE=y,∴y=12x(0<x<20);(2)存在点P使△PEF是Rt△,①如图1,当∠FPE=90°时,四边形PEBF是矩形,BF=PE=12x,四边形APEF是平行四边形,PE=AF=12x,∵BF+AF=AB=10,∴x=10;②如图2,当∠PFE=90°时,Rt△APF∽Rt△ABC,∠ARP=∠C=30°,AF=40−2x,平行四边形AFEP中,AF=PE,即:40−2x=12x,解得x=16;③当∠PEF=90°时,此时不存在符合条件的Rt△PEF.综上所述,当x=10或x=16,存在点P使△PEF是Rt△.解析:(1)在Rt△ABC中,根据三角函数可求y与x的函数关系式;(2)分三种情况:①如图1,当∠FPE=90°时,②如图2,当∠PFE=90°时,③当∠PEF=90°时,进行讨论可求x的值.考查了相似三角形的判定与性质,平行四边形的性质,矩形的性质,解直角三角形,注意分类思想的运用,综合性较强,难度中等.22.答案:解:∵∠ADC=45°,∴AC=DC=6.又,∴AB=10,根据勾股定理,得:BC=√AB2−AC2=8,∴BD=BC−CD=8−6=2,过点D作DE⊥AB,垂足为E,如图,在Rt△BDE中,,∴DE=BD×sinB=1.2,∴BE=√BD2−DE2=1.6,AE=AB−BE=8.4,,∴BD=2,tan∠BAD的值为1.7解析:本题考查了解直角三角形的知识,解答本题的关键是作出辅助线,构造直角三角形,注意熟练掌握锐角三角函数的定义.根据已知条件得到AC=DC=6,由锐角三角函数的定义得到,求出AB=10,根据勾股定理求出BC的长,进而得到BD的长;过点D作DE⊥AB,垂足为E,在Rt△BDE中由锐角三角函数的定义得,求出DE,根据勾股定理得到BE,进而求出AE,即可得到tan∠BAD的值.23.答案:证明:(1)∵AD//BC,∴∠DAE=∠ACB,∵∠ADE=∠BAC,∴△ADE∽△CAB,∴DEAB =AEBC,∴AB⋅AE=DE⋅BC,∵AB=CD,∴CD⋅AE=DE⋅BC;(2)∵AD//BC,AB=CD,∴∠ADC=∠DAB,∵∠ADE=∠BAC,又∵∠ADC=∠ADE+∠CDE,∠DAB=∠BAC+∠CAD,∴∠CDE=∠CAD,∴△CDE∽△CAD,∴CDCA =CECD,∴CD2=CE⋅CA,由题意,得AB=AF,AB=CD,∴AF=CD,∴AF2=CE⋅CA.解析:此题考查相似三角形的判定和性质,关键是根据相似三角形的判定得出相似三角形.(1)根据相似三角形的判定得出△ADE∽△CAB,再利用相似三角形的性质证明即可;(2)根据相似三角形的判定得出△CDE∽△CAD,再利用相似三角形的性质证明即可.24.答案:解:(1)如图1中,作PH⊥BC于H.∵四边形ABCD是菱形,∴AB=BC=4,AD//BC,∴∠A+∠ABC=180°,∵∠A =120°,∴∠PBH =60°,∵PB =3,∠PHB =90°,∴BH =PB ⋅cos60°=32,PH =PB ⋅sin60°=3√32, ∴CH =BC −BH =4−32=52,∴PC =√PH 2+CH 2=(3√32)(52)=√13; (2)如图1中,作PH ⊥BC 于H ,连接PQ ,设PC 交BD 于O . ∵四边形ABCD 是菱形,∴∠ABD =∠CBD =30°,∵∠PCQ =30°,∴∠PBO =∠QCO ,∵∠POB =∠QOC ,∴△POB∽△QOC , ∴PO QO =BOCD ,∴OP BO =QO CD ,∵∠POQ =∠BOC ,∴△POQ∽△BOC ,∴∠OPQ =∠OBC =30°=∠PCQ ,∴PQ =CQ =y , ∴PC =√3y ,在Rt △PHB 中,BH =12x ,PH =√32x , ∵PC 2=PH 2+CH 2, ∴3y 2=(√32x)2+(4−12x)2,∴y =√3x 2−12x+483(0≤x <8);(3)①如图2中,若直线QP 交直线BC 于B 点左侧于E .此时∠CQE=120°,∵∠PBC=60°,∴△PBC中,不存在角与∠CQE相等,此时△QCE与△BCP不可能相似;②如图3中,若直线QP交直线BC于C点右侧于E.则∠CQE=∠B=QBC+∠QCP=60°=∠CBP,∵∠PCB>∠E,∴只可能∠BCP=∠QCE=75°,作CF⊥AB于F,则BF=2,CF=2√3,∠PCF=45°,∴PF=CF=2√3,此时PB=2+2√3,③如图4中,当点P在AB的延长线上时,∵△CBE与△CBP相似,∴∠CQE=∠CBP=120°,∴∠QCE=∠CBP=15°,作CF⊥AB于F.∵∠FCB=30°,∴∠FCB=45°,∴BF=12BC=2,CF=PF=2√3,∴PB=2√3−2.综上所述,满足条件的PB的值为2+2√3或2√3−2.解析:本题考查相似形综合题,考查了菱形的性质,解直角三角形,相似三角形的判定和性质等知识,解题的关键是学会用分类讨论的思想思考问题,属于中考压轴题.(1)如图1中,作PH⊥BC于H.解直角三角形求出BH,PH,在Rt△PCH中,理由勾股定理即可解决问题.(2)如图1中,作PH⊥BC于H,连接PQ,设PC交BD于O.证明△POQ∽△BOC,推出∠OPQ=∠OBC=30°=∠PCQ,推出PQ=CQ=y,推出PC=√3y,在Rt△PHB中,BH=12x,PH=√32x,根据PC2=PH2+CH2,可得结论.(3)分两种情形:①如图2中,若直线QP交直线BC于B点左侧于E.②如图3中,若直线QP交直线BC于C点右侧于E.分别求解即可.25.答案:(1)①∵四边形ABCD是正方形,∴∠BCD=90°,∠BCA=45°,∵GE⊥BC、GF⊥CD,∴∠CEG=∠CFG=∠ECF=45°,∴四边形CEGF是矩形,∠CGE=∠ECG=45°,∴EG=EC,∴四边形CEGF是正方形;②√2;(2)连接CG,由旋转性质知∠BCE=∠ACG=α,在Rt△CEG和Rt△CBA中,∵CECG =cos45°=√22、CBCA=cos45°=√22,∴CGCE=CACB=√2∴△ ACG∽△BCE,∴AGBE =CACB=√2,∴线段AG与BE之间的数量关系为AG=√2BE;(3)①∵∠CEF=45°,点B、E、F三点共线,∴∠BEC=135°,∵△ACG∽△BCE,∴∠AGC=∠BEC=135°,∵∠AGH=∠CAH=45°,∵∠CHA=∠AHG,∴△AHG∽△CHA;②3√5.解析:本题主要考查相似形的综合题,解题的关键是掌握正方形的判定与性质、相似三角形的判定与性质等知识点.(1)①由GE⊥BC、GF⊥CD结合∠BCD=90°可得四边形CEGF是矩形,再由∠ECG=45°即可得证;②由正方形性质知∠CEG=∠B=90°、∠ECG=45°,据此可得CGCE=√2、,利用平行线分线段成比例定理可得;(2)连接CG,只需证△ACG∽△BCE即可得;(3)证△AHG∽△CHA得AGAC =GHAH=AHCH,设BC=CD=AD=a,知AC=√2a,由AGAC=GHAH得AH=23a、DH=13a、CH=√103a,由AGAC=AHCH可得a的值.解:(1)①见答案;②由①知四边形CEGF是正方形,∴∠CEG=∠B=90°,∠ECG=45°,∴CGCE=√2,,∴AGBE =CGCE=√2,故答案为:√2;(2)见答案;(3)①见答案;②由①知∽△CHA,∴AGAC =GHAH=AHCH,设BC=CD=AD=a,则AC=√2a,则由AGAC =GHAH得√2a=2√2AH,∴AH=23a,则DH=AD−AH=13a,CH=√CD2+DH2=√103a,∴AGAC =AHCH得√2a=23a√103a,解得:a=3√5,即BC=3√5,故答案为:3√5.。
2017-2018年上海市第四教育署六上(五四制)第一次月考英语试题 含答案
预备年级 2018年 9 月英语语质量监测(满分 100 分 时间 75 分钟)第一部分 听力 (25%)Ⅰ. Listen and choose the right picture. (6%)II. Listen to the dialogue and choose the right answer. (8 分) ( ) 7. A. Cook for his family.B. Go to the supermarket.C. Play with his family.D. Go swimming.( ) 8. A. To the zoo.B. To the park.C. To the school.D. To the museum.( ) 9. A. No, she doesn’t. B. Yes, she d oes.C. No, she isn’t.D. Yes, she is.( ) 10. A. To leave rubbish everywhere.B. Not to reuse shopping bags.C. Not to pick up rubbish.D. To keep the environment clean. () 11. A. At Eight o’clock.B. At Nine o’clock.C. At Ten o’clock.D. At Eleven o’clock.( ) 12. A. The Space Museum.B. Ocean Park.C. North City Park.D. Water World.( ) 13. A. Because her mother is ill.B. Because she is too young.C. Because she likes her sister.D. Because she doesn’t want to go to school.( ) 14. A. Her mother.B. Her brother.C. Her sister.D. Her cousin.III.Listen to the passage and say whether the following sentences are true (T) or false (F). (6 分)( ) 15. There are five people in Tom’s family.( ) 16. Tom’s mum usually washes the dishes for them.( ) 17. Tom’s grandparents like going shopping.( ) 18. Dad waters the flowers in the garden.( ) 19. Peter often goes to the Football Club at weekends.( ) 20. Tom likes looking for things on the Internet.IV.Listen and fill in the blanks with the words you hear. (5 分)21.Today is Mary’s.22.Mary’s mother buys a nice for her.23.The party begins at a.m.24.Mary’s friends bring some for her—a basketball, some comic books and toys.25.After lunch they go to the .第二部分词汇与语法(44%)Ⅰ. Complete the sentences according to the given phonetic transcriptions.(根据所给音标,完成句子)5%1.Where ./els/can Allen buy a cheaper computer ?2.The /ˈəunli/ way to the bus stop is to walk along this road.3.The Pacific /ˈəʊʃn/ is in the west of America.4.We must not /pəˈlu:t/ the water and air.5.Hurry up! It is /ˈɔ:lməʊst/ time for school.Ⅱ. Choose the best answer.(选择最恰当的答案, 用A,B,C 或 D 表示):20%( )1. Mr Backer, is my new Maths teacher, Miss Gao.A.sheB. heC. itD. this( )2. Alice goes shopping with .A. he and IB. I and heC. me and himD. him and me( )3. Tom often plays football but Tim always plays piano after school.A. the, theB. the, /C. /, /D. /, the( )4. My parents always tell me out at night.A. goB. not goC. not to goD. not going( )5. people are there in your family.A. How manyB. How muchC. How oldD. How long( )6. He is never late for school because he gets up early in the morning.A. oftenB. neverC. sometimesD. always( )7. I have American friend. He is from USA.A.an…/B. a …/C. an….theD. /….the( )8. How about to see a film with me this Sunday?A.goB. goingC. goesD. to go( )9. Alice has got a lot of presents her family.A. fromB. toC. inD. on( )10. Kate hasn’t been to England yet. I haven’t yet, either.A. been thereB. been to thereC. gone thereD. gone to there( )11. We always together.A. go to the schoolB. go to school on footC. walk schoolD. walk to the school( )12 . I always with my brother. We can eat a lot of delicious food.A. see a filmB. go to the libraryC. go to the restaurantD. play basketball( )13. Jack sometimes _ with his cousin.A. play footballB. plays gamesC. cookingD. go shopping( )14. Kitty promises rubbish into the river.A. not throwB. don’t throwC. not to throwD. to not throw( )15. Alice is very helpful. She likes to help .A. otherB. others peopleC. othersD. another( )16. We often play football Saturday and Sunday.A.inB. onC. atD. by( )17. Jack’s younger sister is ill. So he must stay at home and her.A.look atB. look forC. look upD. look after( )18. There is some rubbish on the ground. Please .A.pick up itB. pick it upC. pick itD. pick out it( )19. Danny late for school and hard.A.never is… always worksB. is never…. always workC.is never ….always worksD. be never …. works always( )20. Mr. White teaches Chinese and we all like very much....him B. us.. he C. our... him D. our (i)III Complete the following passage with the words in the box. Each word can only be used once. (5 分)A. thirteenB. smallC. bothD. worksE. looksF. largeKate White has a 1 family. There are nine people in her family. Her grandfather’s name is James White. He 2 as a doctor. Her grandmother’s name is Alice White. She is a h ousewife. They have a son and a daughter. Her father’s name is Thomas White and he is a policeman. Her mother’s name is Jenny White and she is a nurse. Kate is 3 and she is a middle school student. She has a little brother, Jimmy. Jimmy is only three and 4 very cute. Charles is her uncle, and Casey is her aunt. They are 5 teachers. They have a little baby. His name is Tommy. Kate loves her family very much.I V.Fill in the blanks with the w ords in their proper forms (用所给词的适当形式填空, 一空一词):5%1.How many has Mr. Wang got? (child)2.We all know air is bad for our health. ( pollute)3.It’s fun to go in the park. (cycle)4.John usually helps me with my study and he is very to me. ( help)5.He is always the to get to school. (one)V.Rewrite the following sentences.(按要求改写下列句子,每空一词) : 9%1.I have already been to Ocean Park. (改为一般疑问句)you been to Ocean Park ?2.Lily has got a lot of birthday presents from her friends. (改否定句)Lily got birthday presents from her friends3.Alice sometimes go to the park with her parents.( 对划线部分提问)does Alice go to the park with her parents?4.My father doesn’t watch TV at all.( 保持原意)My father TV .5.school,friend, is, for, never, late ,my(连词成句)第二部分阅读与写作(31%)Ⅰ. Reading comprehension.(A)Choose the best answer to the questions. (根据短文内容,选出最恰当的答案,用A, B, C…等表示) 5%A young man went to a car shop. He was wearing old boots and a dirty jacket. His hair was long and untidy(不整洁的). The young man came up to an expensive (昂贵的) car. He looked at it carefully and then asked the shop owner ( 物主), “How much is this car?” “Nine thousand two hundred dollars,” the owner answered. “I want six of them,” the young man said. The shop owner looked at him in surprise and then he smiled coldly. He found it hard to be polite and showed the door to the young man.At that moment, the young man took money out of his pocket and paid for the cars. The shop owner felt sorry for what he said. The young man said one car was for himself and the rest were for his friends. Now each of them would have one. He also said that they were all working on a fishing-boat. “We have got much money this season,” the young man said. “When we are free, we will go camping in our new cars.”( ) 1. The young man was very .A.poorB. kindC. richD. clean( ) 2. There were friends of the young man’s who wanted new cars.A. fourB. fiveC. sixD. seven( ) 3. The young man was a .A. shop ownerB. fishermanC. factory workerD. teacher( ) 4. “The shop owner showed the door to the young man.” means .A.The shop owner wanted to tell him the way to the door.B.The shop owner wanted to tell him to enjoy the door of the car.C.The shop owner asked him to go away.D.The shop owner wanted to ask him to look at the door of the car.( ) 5. What’s the main idea of the passage?A.Cars are very expensive.B.We can’t show the door to others.C.We can’t tell whether a man is rich by his clothes.D.People wearing old boots and jackets may be rich.(B)Read the passage and choose the best answer. (阅读短文并选择最恰当的答案,用A,B,C…等表示):6%Alice and Eric are good friends. They are in the same class and sit next to each other. After school, they1 go home together. However(尽管如此), they have many2 habits. On weekends, Alice likes to stay at home, watch TV or sleep, but Eric often goes out3 . Alice eats candy every day, although she knows it’s4 for her health. Eric also likes eating candy. But he only eats it once a week,5 he thinks eating less candy can help him to keep in good health.Today is Alice’s birthday. Alice thinks Eric will buy a bag of candy for her. But in fact, Eric buys her a book. It’s about how to form (养成)good habits. Eric thinks it’s really 6 to keep good habits. And it’s helpful to have a healthy lifestyle.( ) 1. A. never B. unusually C. often D. hardly( ) 2. A. same B. different C. right D. wrong( ) 3. A. exercise B. exercises C. at exercising D. to exercise( ) 4. A. good B. bad C. well D. badly( ) 5. A. so B. but C. although D. because( ) 6. A. funny B. interesting C. exciting D. important(C)Read the passage and fill in the blanks with proper words.(在短文空格上填入恰当的词,使其内容通顺,每空格限填一词,首字母已给出): 7%I have a happy family. There are five m 1 in my family: grandfather, grandmother, father, mother and me. My g 2 live in the countryside. They have six rooms. The house is big. There are two apple trees in the yard. They have sixteen sheep and three cows. The sheep are w 3 . The cows are yellow. They are healthy. My parents and I live in the c 4 . My father is a worker. He works hard in a car factory. My mother is a high school’s Chinese teacher, she u 5 rides her bicycle to go to work at seven o’clock. In the evening, she m 6 supper for us. The food she cooks is delicious. I am a student. My school is not far from my home. I w 7 to school on sunny days. I have many friends at school After class, I like p 8 games with them. We are tired, but we are very happy. I love my family.(D)Read the passage and answer the questions.(回答问题)(5%)When you are in England, you must be very careful in the street because the traffic drives on the left. Before you cross the street, you must look to the right first and then the left.In the morning and in the evening when people go to or come from work, the streets are very busy. Traffic is the most dangerous then.When you go by bus in England, you have to be careful, too. Always remember the traffic moves on the left, so you must be careful. Have a look first, or you will go the wrong way.In many English cities there are big buses with two floors. They are called double-decker buses.(双层车) You can sit on the second floor. From there, you can see the city very well. It’s very interesting.1.In England, why must you be very careful in the street?Because the traffic .2.What must you do before crossing the street?We must .3.Is traffic the most dangerous when people go to or come from work?, .4.What should you always remember?We should always remember .5.What kinds of buses are there in many English cities?There are big buses with .Ⅱ. Writing: Write at least 50 words about the topic “my good friend” (根据题目“我的好朋友” 写一篇不少于50 个单词的短文,要求意思贴切,内容连贯,无重大语法错误): 8%以下提示仅供参考:1.w ho’s your good friend?2.W hat do you like about your good friend?3 What do you usually do together?.Part 1 听力部分录音原文及参考答案:I. 1. I have been to Yu Garden once.2.Lucy promises to reuse shopping bags.3.Mike often plays with his dog.4.Look! This is my good friend Kitty.5.Lily often helps pick up the rubbish.6.Tom and Peter often eat their lunch together.(1—6 BADCFE)II.7.M: What do you often do at weekends?W: I often cook for my family. What about you, Tom? M: Aha, I often go to the supermarket.Q: What does Tom often do at weekends?8.W: Hi, Bill. What do you want to do tomorrow?M: I have no idea. What about you?W: Shall we go to the zoo?M: That sounds great.Q: Where will they go tomorrow?9.W: Do you know about Alice, Ben?M: Of course. She is my good friend.W: What do you think about her?M: She is friendly and helpful. And she never gets angry. Q: Does Alice often get angry?10.W: Do you want to be a Friend of the Earth, Joe? M: Of course, I do.W: What do you promise to do?M: I promise to keep the environment clean.Q: What does a Friend of the Earth promise to do?11.W: Let’s plan a visit to Shanghai.M: That’s a good idea. Shall we visit the Bund?W: That’s great! What time?M: How about ten o’clock in the morning?W: That’s a good time.Q: When are they going to visit the Bund?12.W: Where have you been in Garden City, Nick?M: I have been to Ocean Park.W: Have you been to North City Park yet?M: No, not yet. I am going there this afternoon.Q: Where has Nick been in Garden City?13.M: Let’s go to school, Lily. It’s seven o’clock.W: But I have to look after my sister. My mother is ill. Q: Why does the girl have to look after her sister?14.M: Sally, do you have a computer in your room? W: Yes. It is on my desk.M: Who do you share your bedroom with?W: I share my bedroom with my sister.Q: Who does Sally share her bedroom with?(7—10 BAAD 11—14 CBAC)III.My name is Tom. I have a big family. There are six people in it. They are my dad and mum, my grandpa and my grandma, my little brother Peter and me. We often have a good time at weekends. Mum cooks for us, but she doesn’t wash the dishes. Dad washes the dishes and he also waters the flowers in the garden. My grandpa and grandma often go shopping. They like shopping very much. My little brother Peter likes playing football. At weekends, he often goes to the Football Club. I like looking for things on the Internet. Sometimes, I also chat with my friends on it. We all enjoy our weekends.(15—20 FFTTTT)IV.Mary is 14 years old. Today is Sunday. It’s her birthday. Her mother buys a nice dress for her. She has a birthday party at home. The party begins at 8:00 a.m. Her friends come to the party. And they bring some presents for her—a basketball, some comic books and toys. After lunch they go to the park. First they climb the hills. Then the girls go boating and the boys go swimming. They have a good time.(21. birthday 22. dress 23. 8:00 24. Presents 25. park参考答案:1. else2. only3. ocean4. pollute5. almost 1-5 DDDCA 6-10 DCBAA 11-15 BCBCC 16-20BDBCA III1. F2. D3. A4. E5. CIV1. children 2. pollution 3. cycling 4. helpful 5. firstV1. Have , yet 2. doesn’t, many 3. How often 4. never watches5.My friend is never late for school.(A)1-5 CBBCC(B)1-6 CBDBCD(C)1. members 2. grandfather 3. white 4. center 5. usually 6. makes 7. walk 8. playing(D)1. drives on the left.2.look to the right first and then the left.3.Yes, it is4.the traffic moves on the left, so you must be careful.5.with two floors.6.。
上海浦东第四教育署2024届中考英语模拟精编试卷含答案
上海浦东第四教育署2024届中考英语模拟精编试卷含答案注意事项1.考生要认真填写考场号和座位序号。
2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。
第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。
3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。
Ⅰ. 单项选择1、The _______ of staying healthy is to eat healthy food and exercise more.A.place B.secret C.time D.game2、–Would you mind if I borrow your digital camera?--___________. Because I stored plenty of important pictures in it.A.Y es, I’d be glad to.B.Sorry, I’m afraid I can’t.C.No problem.D.Y es, of course.3、All the in our school enjoyed themselves on March 8th because it was their own holiday.A.man teachers B.women teachers C.woman teacher D.man teacher4、Peter fell off his bike this morning, and his knees were hurt ________.A.quietly B.slowly C.carefully D.badly5、We must be more careful and try to /ə'vɔɪd/ making the same mistakes.A.aloud B.allow C.avoid D.afford6、—It’s said that the TV program Readers was quite popular.—Yes. _______ my parents _______ my little sister likes watching it very much.A.Neither; nor B.Both; andC.Either; or D.Not only; but also7、I have some trouble with my English writing. Can you give me some ______ ?A.advice B.decisions C.messages D.suggestion8、China’s first home-built aircraft carrier(航空母舰) hit the water in Dalian _______the morning of April 26, 2017.A.in B.on C.at D.to9、----Mike wants to know if ________ a picnic tomorrow.---- Yes. But if it ________, we will visit the museum instead.A.you have; will rain B.you will have; will rainC.you will have; rains D.will you have; rains10、—Who is in the classroom?—________________. Look! The students are playing outside.A.None B.NobodyC.Nothing D.AnybodyⅡ. 完形填空11、完形填空(10小题,每小题1分,共10分)A traveler came to a village carrying nothing but an empty pot. He wa 1 but nobody would share their food with him. The traveler then 2 the pot with water, dropped a large stone in it, and placed it 3 a fire in the village square.One of the villagers asked 4 he was doing. The traveler answered: “I’m making ‘stone soup’.” The vi llager seemed 5 .The traveler continued: “The soup tastes wonderful but I need some carrots to make 6 more tasty.” The villager then 7 him some carrots and put them in the pot.Another villager walked by, asking about the pot, 8 he traveler again told him about his stone soup. This time he really needed some meat. The villager brought him some 9 to help him out.More and more villagers walked by. Each added 10 ingredient. Finally, the traveler had a delicious pot of soup. Of course, he shared it with everyone.1.A.thirsty B.tired C.hungry D.angry2.A.compared B.filled C.offered D.provided3.A.over B.above C.into D.onto4.A.why B.what C.where D.whose5.A.worried B.interested C.tired D.moved6.A.them B.their C.it D.its7.A.took B.bought C.brought D.sold8.A.but B.or C.until D.and9.A.meat B.stone C.carrots D.cabbage10.A.other B.another C.others D.the otherⅢ. 语法填空12、The Value of a DollarWhen I was nineteen, I had a job at a local bookstore. One night, a young couple came in and walked here and there looking for a book. They seemed 1.(be) the usual type to give mall 2.(worker) a hard time. When they came to the register (收银处), she was a dollar or two short of 3.(buy) the book she wanted.She looked 4.(disappoint). I had a customer discount (折扣) card and it was still active. I told the girl5.(gentle), “Hang on, don’t worry, you’ll have the money to buy the book.” I put in my pas sword.She gave me a 6.(thank) smile. With a dollar left, she and her boyfriend left the store. I believed that I would never see 7.(they)again.I don’t remember 8.long it was, but they did return later that evening just to bring me 9.beautiful card that said “It’s people like you who make the world a 10.(good) place to live in.”I’m deeply moved by the little girl and her boyfriend. In fact, they have also made the world a little brighter.Ⅳ. 阅读理解A13、Are you still using text messages or QQ to chat with your family or friends? If your answer is “yes”, we have to say you are out. A new way of communicating called “WeChat” has been more and more popular in the past three years. WeChat is a mobile text and voice messaging communication service developed by Tencent in China, first released in January 2011. With WeChat, we can communicate with friends by sending text messages, voice messages, we can share our photos or videos, share our location, we can even chat with lots of friends at the same time. More important, it is free for the users.In May 2011, WeChat had four to five million users, and by the end of 2011, it had 50 million users. WeChat had more than 100 million users by March 2012. In April 2012, Weixin changes its name as WeChat for the international market. In September 2012, WeChat had more than 200 million users according to Tencent CEO, Ma Huateng. Today WeChat has influence not only in Mainland China, HK, Taiwan, South-East Asia, but Chinese People areas around the world. In July 2013, Tencent announced WeChat had 70 million users outside China and Lionel Messi, the famous Argentine football player became the latest product ambassador of WeChat.In a word, WeChat is a lifestyle now. Do you WeChat today?1.According to the passage, we can do many things with WeChat EXCEPT ______.A.text messagesB.share locationC.shop for free2.The underline word “release” here means ________ in Chinese.A.发布B.设置C.释放3.How many users did WeChat have outside China in July 2013?A.70 million.B.50 million.C.100 million.4.Who is the CEO of Tencent?A.Lionel Messi. B.Ma Yun. C.Ma Huateng.5.Where do you probably read this message?A.National Geography.B.Trend & Health.C.New Science and Technology.B14、Within IOC, Samaranch was a man much admired and sometimes loved, the Spanish who made the organization into one that is rich and so powerful.No one doubted Samaranch’s ability to make something impossible come true. In 1981 he ensured(确保) that Seoul was elected to be the host of the 1988 Olympics. Remember, that was a time when the two Koreas were at war with each other. And because Samaranch helped swing(使改变) Spanish votes(投票) across to London, rather than to Paris, London was made host for the 2012. To China and the Chinese people, Samaranch was considered a great friend. Thanks to his efforts(努力) during his last term as IOC president, Beijing and China won the Olympics for the first time.Samaranch led the IOC from 1980 to 2001. Certainly, throughout his IOC presidential term, he had a soft spot for Moscow, for it was there that he was elected president of the IOC in 1980, and it was there that he managed to reach his final wishes as IOC president. Very early on in that 2001 Moscow session he wanted the Belgian surgeon, Jacques Rogge, a newcomer to the Olympic Movement, to replace him as president, for China to host the 2008 Olympic Games and for his son, Juan Antonio Samaranch junior, to become an IOC member. He got all three wishes.Samaranch’s death came three months before his 90th birth day on April 21st, 2010.1.Samaranch became the president of the IOC in __________.A.Spain B.London C.Paris D.Moscow2.Samaranch was born _________.A.in June, 1921 B.in July, 1921C.in July, 1920 D.in June, 19203.Which of the following is NOT one of his final wishes?A.London was made host for the 2012.B.Rogge replaced Samaranch to be IOC president.C.His son became an IOC member.D.The 29th Olympic games was held in Beijing.4.Which of the following is FALSE about Samaranch?A.Samaranch ruled IOC for 21 years.B.Samaranch was good at communicating.C.Samaranch was born in Russia.D.Samaranch was regarded as a great friend to China.C15、阅读下列材料,从A、B、C、D四个选项中选出最佳答案,并把答题卡上对应题目的答案标号涂黑。
八年级英语上学期期试题 牛津上海版五四制
上海市浦东新区第四教育署2016-2017学年八年级英语上学期期中试题(满分100分完卷时间 100 分钟)Part I Listening (第一部分听力共25分)A.Listen to the sentence and choose the right picture.(听句子,选出与内容符合的图片,用A,B,C,D,E,F等表示)(6分)A B C DE F G1._______2.________3._______4.________5._________6._________B. Listen to the dialogue and choose the best answer to the question you hear.(根据你听到的对话和问题,选出最恰当的答案,用A,B, C,D表示)(6分)7.A) Swimming. B) Running. C) Jogging. D) Badminton.8.A) In a shop. B) In a library. C) In a book store. D) In a computer room9.A) Sports. B) Stories. C) Reports. D) Advertisements.10.A) 9:45. B) 11:05. C) 10:15. D) 10:50.11.A) Because she had a headache.B) Because she had a fever.C) Because she watched a wonderful film.D) Because she did her homework too late.12.A) To drink a cup of milk. B) To cook a meal at home.C) To have lunch at a restaurant. D) To finish their work.C. Listen to the passage and tell whether the following statements are true or false.(根据你所听到的短文内容判断下列句子是否正确,符合的用“T”表示,不符合的用“F”表示)(5分)13. We usually go out with our school or work friends.14. We often talk about something important with our fun f riends.15. Fun friends can make us happy and forget our troubles.16. One kind of friend is rich enough to pay for us when we go out.17. You would have some problems spending the whole day with your best friends.D. Listen to the passage and fill in the blanks.(听短文,完成下列句子,每空格限填一词)(8分)18.Scientist may be able to explain why children _______ _______ so much.19.Children who loved the _______ ______ drinks grow the fastest.20.The researchers gave more than _______ _______ six drinks to taste.21.A certain chemical helps bone grow, however, it may ____ ____ bring tooth and weight problems.Part II Vocabulary and Grammar(第二部分词汇和语法共37分)A. Choose the best answer. (选择最恰当的答案,用A,B, C,D表示))(15分)22. Which of the following words matches the sound /feil/?A) fall B) fell C) fail D) fear23.Eddie’s brother is studying in ________ European country.A) a B) an C) the D)/24.We had a welcome party for the new students _____ the afternoon of September 1st.A) in B) at C) to D) on25.There are more ________ in my school than in yours.A) woman teachers B) women teachersC) women teacher D) woman teacher26.There are about three ________ books in our school library.A) thousand B) thousands C) thousands of D) thousand of27.The Yangtze River is one of ________ in China.A) longer rivers B) the longest riversC) the longest river D) the long river28.Linda gave me two story books, but ________ of them is interesting.A) either B) both C) neither D) none29.My little brother often asks me _________ basketball with her.A) playing B) to play C) to not play D) play30.Jenny ________ a dance club in her school twice a week.A) takes part in B) joins C) attends D) enters31.My ambition is _______ a pliceman like my father in the future.A) become B) to become C) became D) becomes32.________ important news you’ve told me!A) What B) What a C) What an D) How33.You ________ make a decision right now. You can think about it for a few days.A) can’t B) mustn’t C) needn’t D) shouldn’t34.Bill spends a lot of money on books _________ he is n ot rich.A) though B) if C) when D) because35.-_________?-I don’t feel quite well. My back hurts.A) What was going on? B) What’s wrong with you?C) How do you like it? D) Where is the hopital?36.-I’m very sorry for being late. I was caught in a traffic jam.-_________.A) That’s right. B) That’s all right.C) You’re welcome. D) Well done!B. Complete the following passage with the words or phrases in the box. Each word or phrasecan only be used once.(将下列单词或词组填入空格。
浦东新区教育系统通讯录-第四教育署幼儿园
上南三村幼儿园竺月华第四教育署幼儿园上南路1251弄24号200126021-58838279紫叶幼儿园池奕桦第四教育署幼儿园上海市浦东新区紫叶路246号20120458440827周东幼儿园陈瑛第四教育署幼儿园上海市浦东新区周浦镇关岳路258弄20131868135283周浦镇幼儿园吴炯第四教育署幼儿园浦东新区周浦镇公元新村15号(分园:康桥镇秀康路536号)201318021-20903996杨思幼儿园郭萍第四教育署幼儿园南园:西营南路75号北园:杨新路82号200126南园:58420385北园海阳之星幼儿园第四教育署幼儿园上海市浦东新区海阳路480弄1号20012638760966雪野幼儿园王建英第四教育署幼儿园大道站路496号20012361069229小天鹅幼儿园顾群英第四教育署幼儿园凌兆路531弄68号200124021-58491372小浪花幼儿园彭根娣第四教育署幼儿园三林路1466弄83号20012350633731下沙幼儿园施惠萍第四教育署幼儿园浦东新区航头镇下沙新街51弄2号20131758141692七色花幼儿园瞿红梅第四教育署幼儿园浦东下南路551弄38号20012558531209小天使幼儿园王淑琴第四教育署幼儿园上海市浦东新区上浦路36弄60号20012458499509未来之星幼儿园王洁第四教育署幼儿园西泰林路375号20012350656904瓦屑幼儿园孙艳第四教育署幼儿园浦东新区周浦镇瓦屑建设路14号20132158152293童心幼儿园第四教育署幼儿园浦东新区南汇周浦镇瑞浦路631号20131868096183(转分机8000天天乐幼儿园柳纹第四教育署幼儿园浦东新区浦建路211弄13号20012758890157欧风幼儿园盛晴第四教育署幼儿园上海市浦东新区周浦镇瑞意路339号20131833891591塘桥幼儿园杭君第四教育署幼儿园塘桥路223弄7号20012758811537齐河幼儿园潘丽君第四教育署幼儿园上海浦东新区昌里东路651弄44号20012550777882兰亭幼儿园徐红明第四教育署幼儿园蓝村路471弄32号20012758810248康桥镇二幼赵婵第四教育署幼儿园浦东新区康桥拯安路75号(总园)浦东新区康桥康弘路528弄6号(分园)20131958131414济阳一村幼儿园唐夏君第四教育署幼儿园德州路420弄21号20012658743563绿川幼儿园黄金娣第四教育署幼儿园浦东博华路999弄9号201204021-50611392临沂一村幼儿园沈洁珍第四教育署幼儿园临沂一村56号20012538810231临沂五村幼儿园陈爱娟第四教育署幼儿园临沂路381弄30号(总部)南码头路551弄20号(分部)20012558705497、58812954临沂八村幼儿园曹湘瑜第四教育署幼儿园浦东新区临沂路81弄71号20012558753092澧溪幼儿园顾剑英第四教育署幼儿园上海市浦东新区周浦镇年家浜路353号20131858111619康桥镇一幼第四教育署幼儿园上海市浦东新区康桥镇梓康路17号20131502158111402恒宇幼儿园赵燕第四教育署幼儿园杨南路1039号200124021-68529885航头幼儿园施惠萍第四教育署幼儿园航头镇航梅路525弄5支弄5号20131668221872东方锦绣幼儿园周敏莉第四教育署幼儿园浦东新区华春路180号50123568上钢新村幼儿园干叶敏第四教育署幼儿园昌里东路80弄62号20012658838741昌里幼儿园朱艺晓第四教育署幼儿园南码头路1621弄37号20012550781853贝贝星幼儿园赵青第四教育署幼儿园御山路391号20120468931975北蔡幼儿园朱幸嫣第四教育署幼儿园上海市浦东新区北中路455号20120458910246小叮当幼儿园朱伟利第四教育署幼儿园上海市浦东新区环林西路608弄89号20012350832347上南一村幼儿园陈蕾第四教育署幼儿园上南路929弄3号20012668586280百合花幼儿园021-50651808*800第四教育署幼儿园浦东新区西泰林路926号200123021-50651808春之声幼儿园汪绿绮第四教育署幼儿园东书房路411号200123021-61343488鹤琴幼儿园第四教育署幼儿园浦东新区合庆镇庆利路58号20120158971872红苹果幼儿园何佳丽第四教育署幼儿园南码头路1136弄35号甲200125021-50775536红蜻蜓幼儿园陈蕾第四教育署幼儿园上海市浦东新区德州路255弄58号200126021-58836118南码头幼儿园朱剑萍第四教育署幼儿园南码头路193弄10号200125021-58892281三林幼儿园袁雪明第四教育署幼儿园三新路1号20012458410379上钢九村幼儿园徐宝华第四教育署幼儿园西营路33弄10号200126021-50560688上南五村幼儿园陈艺第四教育署幼儿园上南五村8号200126021-58833312天池幼儿园胡健第四教育署幼儿园上海市浦东新区北艾路227弄25号20120458435527小太阳幼儿园胡健第四教育署幼儿园浦东上浦路510弄34号200124021-58497490。
浦东新区教育系统通讯录-第四教育署中学
江镇中学凌慧良第三教育署中学浦东新区东亭路631号20120268785217香山中学潘超炜第二教育署中学浦东新区灵山路1672弄36号20013668561542五三中学柴建荣第三教育署中学浦东新区川沙镇东河浜路26号2012058922712老港中学严福明第二教育署中学上海市浦东新区老港镇建中路1001号20130258051045上南中学张海第四教育署中学浦东新区昌里路93弄1号20012658475643致远中学赵世强第二教育署中学灵山路2100号20013650757977云台中学陈炜第四教育署中学昌里东路190弄13号20012658808448育民中学孙建良第一教育署中学上海市浦东新区高桥镇西街217号20013758678402侨光中学黄爱娟第三教育署中学浦东新区川沙镇新德路463号2012058981710*浦东中学吴原第四教育署中学上海市浦东新区浦三路648号200125021-50779448川沙中学北校黄冰洁第三教育署中学浦东新区华夏东路2475号20120021-68650060平和学校任国芳第二教育署中学浦东黄杨路261号20120650315621高行中学谢建初第一教育署中学浦东行泰路210号20120858651132高桥-东陆学校高崇华第一教育署中学浦东新区巨峰路712弄18号20012950256531高东中学顾建荣第一教育署中学浦东新区高东镇光明路476号20013758480918长岛中学马丽秦第一教育署中学上海市浦东新区长岛路555号200129北蔡中学马淑颖第四教育署中学莲园路246号(莲园路校区)北中路493号(北中路校区)20120458912358东方世纪学校乐秀峻第二教育署中学上海市浦东新区龙东大道4328号201201021-58588888第二工业大学附属龚路中学倪瑞明第一教育署中学浦东新区曹路镇龚路北街50号20120938923288绿川学校赵金荣第四教育署中学上海市浦东新区北蔡镇绿林路409号20120450610613上中东校薛建平第二教育署中学上海市临港新城环湖西三路1398号20130620943302华东师范大学张江实验中学陈胜庆第三教育署中学浦东张江江东路54号2012138950082华师大二附中何晓文第三教育署中学上海市浦东新区晨晖路555号20120350801890沪新中学刘永和第一教育署中学上海市浦东新区莱阳路224号20012958710139民远高级中学陈鹏第二教育署中学浦东新区商城路1288号(商城路源深路口)2001258400162洋泾中学李海林第三教育署中学浦东新区潍坊路111号20012258306444。
浦东新区第四教育署六年级数学上学期第一次阶段考试试题(无答案)沪教版五四制(2021学年)
考试试题(无答案)沪教版五四制编辑整理:尊敬的读者朋友们:这里是精品文档编辑中心,本文档内容是由我和我的同事精心编辑整理后发布的,发布之前我们对文中内容进行仔细校对,但是难免会有疏漏的地方,但是任然希望(上海市浦东新区第四教育署2017-2018学年六年级数学上学期第一次阶段考试试题(无答案)沪教版五四制)的内容能够给您的工作和学习带来便利。
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段考试试题(90分钟完成 满分100分)一、选择题(每小题3分,满分共18分)1.在数18,—24,0,2。
5,43,2005,3.14,—10中,整数有……………( )(A) 2个 (B) 3个 (C ) 4个 (D) 5个2.已知正整数a 能整除17,那么 a是…………………………………………( )(A) 任何正整数 (B) 1或17 (C ) 17 (D ) 34 3.下列分解素因数的正确算式是………………………………………………( )(A) 2×3×3=18 (B) 8=2×4 (C )12=2×2×3 (D)9=1×3×3 4.在1552551532515,,,中,和31相等的分数是…………………………………( ) (A )2515 (B )153 (C)255 (D )1555.铺一条2千米长的管道要9天完成,平均每天铺设的管道长度是全长的…( ) (A)29; (B )29千米; (C )19; (D) 19千米. 6.下列叙述错误的有 …………………………………………………………( )(1)没有最小的自然数.(2)如果两个数都是奇数,那么这两个数互素。
2017-2018上海市浦东新区第四教育署2017-2018学年七年级(五四学制)上学期期中质量调研英语试题
2017学年第一学期英语学科质量调研初一年级试卷(满分100分完卷时间75 分钟)考生注意:本卷有8 大题,共81小题。
试题均采用连续编号,所有答案务必按照规定在答题纸上完成,做在试卷上不给分。
Part 1 Listening(第一部分听力)共25分I. Listening Comprehension (听力理解):A. Listen and choose the right picture(根据你听到的内容,选出相应的图片)(5分)A B CD E F1._______2._______3._______4._______5.________B. Listen to the dialogue and choose the best answer to the question you hear (根据你听到的对话和问题,选出最恰当的答案):(8分)6. A. Basketball. B. Football. C. V olleyball. D. Badminton.7. A. Six yuan. B. Fifty-six yuan. C. Sixty yuan. D. Fifty yuan.8. A. To Beijing. B. To Sanya. C. To Xinjiang. D. To Nanjing.9. A. Monday. B. Friday. C. Saturday. D. Sunday.10. A. An engineer. B. A teacher. C. An architect. D. A driver.11. A. A shirt. B. A toy. C. A computer. D. A bicycle.12. A. His friend. B. His grandmother.C. His parents.D. His pet dog.13. A. Because he had no time. B. Because he had a cold.C. Because he was hurt.D. Because he got up late.C. Listen to the passage and tell whether the following statements are true or false (判断下列句子是否符合你听到的短文内容,符合的用“T”表示,不符合的用“F”表示)(6分)14.Mike lives in a big town and works in a zoo.15.He often saves money to buy food for his clever dog.16.One of his friends asked him and his dog to dinner one day.17.His friend gave the dog ten dollars at dinner time .18.The dog came back with the newspapers an hour later.19.Mike thinks his dog may go to see a film.D. Listen to the passage and complete the sentence(听短文,完成下列句子。
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上海市浦东新区第四教育署2017 上海市浦东新区第四教育署2017-2018学年八年级物理下学期期中试题(完卷时间60分钟,满分100分)一.选择题(每题2分,共20分) 1.下列关于杠杆的说法中正确的是()A、杠杆必须是一根直棒B、有力作用在杠杆上不一定能使杠杆平衡C、作用在杠杆上的力其力臂不可能为0 D、力臂就是力的作用点到支点的距离2.如图1所示中属于费力杠杆的是()图1 3.下列温度中,约在36℃~37℃之间的是()A、人体的正常体温B、标准气压下沸水的温度C、冰箱冷藏室的温度D、人感觉舒适的环境的温度 4.升旗的旗杆顶上有一个滑轮,如图2升旗时往下拉动绳子,旗子就会上升,对该滑轮的说法中,正确的是()A、这是一个定滑轮,费力B、这是一个定滑轮,可省力C、这是一个定滑轮,不能改变力的方向D、这是一个定滑轮,可改变用力的方向图2 5.甲机器的功率为1000瓦,乙机器的功率为2000瓦,两台机器正常工作,下列说法中正确的是()A、甲机器做功一定多B、乙机器做功一定多C、甲机器做功一定快D、乙机器做功一定快 6.如图3所示,杂技表演者在离板后的上升过程中,她的A.重力势能增大,动能减小B.重力势能增大,动能增大C.重力势能减小,动能减小D.重力势能减小,动能增大图3 7. 下列四种情况中,人对物体做功的是()A、提着水桶在水平地面上匀速前进B、扛着米袋慢慢爬上楼梯C、用力推汽车,汽车没动D、举着杠铃原地不动8. 下列物体,即具有动能又具有势能的是()A、静止在树上的小鸟B、静止在高架上的列车C、正在下降的飞机D、在地面上跑动的运动员9. 在图4中,O是杠杆PQ的支点,在P点挂一重物G,在Q点分别作用不同方向的力F。
其中能使杠杆在水平位置平衡且最省力的是图()图4 A B C D Q O P Q O P Q O P Q O P 10.如图5所示的轻质杠杆,AO小于BO,在A、B两端悬挂重物G1和G2后杠杆平衡,若在此现象中,同时在G1和G2下面挂上相同质量的钩码,则()A.杠杆仍保持平衡B.杠杆A端向下倾斜C.杠杆B端向下倾斜D.无法判断图5 二.填空(每空1分,共31分图6 A F 11.温度是表示物体_____________的物理量。
温度计上的符号℃表示采用的是______温标,它规定在标准大气压下,冰水混合物的温度为_____℃,________的温度为100℃,常用的温度计是根据_______(选填“煤油”或“水”)热胀冷缩的原理制成的12.“酒香不怕巷子深”说明分子在做__________________;固体很难被压缩或拉伸说明分子间存在______________。
13.若不计摩擦及滑轮重,用如图6所示的滑轮竖直匀速向上提升重为100牛的物体A,则拉力F的大小为牛,物体匀速上升过程中,物体的动能_________,(选填“变大”,“不变”或“变小”)重力势能________(选填“变大”,“不变”或“变小”),此过程选填“是”或“不是”物体A的动能与势能间的转化。
14.小明用200 牛的水平推力,推动重为100 牛的物体在水平地面上匀速前进 4 米,所用的时间10秒,在这个过程中,重力做功为__________焦,推力做功__________焦,推力做功的功率为_________瓦,该值表示__________________________。
15.意大利科学家伽利略发明了世界上第一支气体温度计,如图7所示,该温度计的工作原理是利用气体的热胀冷缩,由此判断当气温上升时,玻璃泡内的气体_________,该温度计的液柱面将_______。
图8 16.图8所示轻质杠杆OA始终在水平位置保持静止,手对细线需施加的拉力物体的重力G(选填“小于”、“等于”或“大于”)。
若保持图中细线的悬挂点不变,将物体逐渐水平移至A 端的过程中,手对细线需施加的拉力,物体具有的重力势能。
(后两空均选填“变小”、“不变”或“变大”)图9 17.如图9所示,物体所受重力是80牛,不计摩擦及滑轮的重,当用力匀速提升物体时,力F为牛,它相当于一个选填“省力”、“费力”或“等臂”杠杆。
若物体在10秒内被提高2米,则拉力F做的功为_________焦,拉力F的功率为_________瓦18.如图10所示,轻质杠杆AC可绕A点转动,AB0.2米,AC0.6米。
B点处挂一个质量为3千克的物体,C点处加一个竖直向上的力F,杠杆在水平位置平衡,则物体的重力大小为_________牛,力F大小为________牛,这是一个________杠杆。
(选填“省力”或“费力”)若拉力F向左倾斜,此杠杆_________(选填“有”或“没有”)可能变成等臂杠杆。
19.某同学为了“探究影响重力势能大小的因素”,利用质量不同的实心铜块A和B、刻度尺、相同的小桌和沙面进行实验。
如图11(a)和(b)所示,他先将小桌放在沙面上,然后让铜块A从一定的高度下落到小桌的桌面上。
接着他按图11(c)、(d)所示,重新实验。
请仔细观察图中下落的铜块与小桌陷入沙面的情况,然后归纳得出初步结论。
A A A AB B a b c d 图11 ①实验中,该同学认为小桌下陷深度越深,说明物体具有的重力势能越大,这样判断的依据是______________________________ ②比较图11中的(a)、(b)和(c)实验过程及相关条件可知当物体质量相同时,_____________________,物体具有的重力势能越大。
③比较图11中的(a)、(c)和(d)实验过程及相关条件可知当物体_______________时,质量越大,物体具有的重力势能越大。
三.作图题(每题3分,共9分)20.如图12所示,轻质杠杆OA在力F1、F2作用下处于静止状态,l2是力F2的力臂,请在图中画出力F1的力臂l1和力F2的示意图。
21.如图13所示,O为杠杆AC的支点,在B处挂一小球,AOOBBC,为使杠杆在水平位置平衡,请画出施加在杠杆上最小的动力F。
22.如图14所示,是温度计的一部分,其中图a所示是某一天的最低温度-5℃,图b所示的是这一天的最高温度10℃。
请在图中将这两个温度表示出来。
F1 O l2 A 图14 图13 图12 四. 计算题(4分4分6分6分23.杠杆平衡时,动力的大小为2牛,动力臂为0.5米,阻力臂为0.1米,求阻力的大小。
24. 物体在20牛的水平拉力作用下沿拉力方向做匀速直线运动,5秒内前进了10米。
求此过程中拉力做的功和功率。
F1 A F2 B O 25.如图15所示,一轻质杠杆可绕O点转动,已知OA=1.6米,OB=0.4米,在杠杆的B点挂一重为600牛的物体,若使杠杆在水平位置平衡,求(1)作用在杠杆上B点的力F2大小为多少牛(2)竖直作用在A点的力F1大小为多少牛图15 图16 26.工人用如图16所示的滑轮装置,将物体匀速向上提起,不计绳子和滑轮之间的摩擦及滑轮重,工人作用在绳子上的力为100牛,求①物体的重力。
②若工人拉动绳子做功100焦,则求绳子自由端上升的距离。
五.实验题(每空1分,共20分)27.如图17所示温度计的测量范围是_____________,最小刻度为_______,因此此温度计更适合测量________的温度,此时体温计的读数为___________。
图17 28.如图18是使用温度计测量水的温度,其中操作正确的是________________。
图18 29. 如图19所示,小陈和小蔡同学做“探究杠杆平衡的条件”实验,他们首先将杠杆的_______固定在铁架台上,然后调节平衡螺母向(选填“左”或“右”)移动使杠杆在位置上平衡,这样是为了直接读出_____________。
接着他们改变钩码的数量和位置做了两次实验,实验中获得的数据记录在下表中。
图19 实验序号F1 (牛)L1 (厘米)F2 (牛)L2 (厘米)1 2 4 4 2 2 3 5 5 3 小陈分析了表中的数据后得出杠杆平衡时,F1 L1 F2 L2。
小蔡分析了表中的数据后得出杠杆平衡时,F1L1 F2L2。
就实验数据来看,小陈同学的结论_______,小蔡同学的结论_______(选填“正确”或“错误”),为了进一步判断结论的普遍性,可采取的方法是__________________。
实验中_________用弹簧测力计(选填“需要”或“不需要”),这主要是为了改变杠杆受到力的__________。
(a)图20 (b)(c)C O A B C D O A O A C θ 30. 某小组同学在学习了支点、动力和阻力的概念后,想研究“用杠杆提起物体,能够提起物体的重力大小与哪些因素有关”,他们用如图20所示的装置进行实验,用钩码代替物体悬挂在杠杆的C点(每只钩码重0.5牛),用测力计对杠杆施加向上的动力F,每次都将杠杆拉到水平位置保持静止,然后将图13(a)、(b)和(c)的实验数据分别记录在表一、表二和表三中。
表一(动力F1牛)表二(d40厘米)表三(动力F1牛)实验序号支点到力F作用点的距离d(厘米)钩码重(牛)实验序号动力F (牛)钩码重(牛)实验序号动力方向θ(度)钩码重(牛)支点到力F作用线的距离l(厘米)1 40 2.0 4 0.5 1.0 7 左偏60 1.0 20 2 30 1.5 5 1.0 2.0 8 左偏40 1.5 30 3 10 0.5 6 1.5 3.0 9 竖直向上2.0 40 10 右偏40 1.5 30 11 右偏60 1.0 20 ①分析比较表一的实验数据及相关条件可知该小组同学想利用图20(a)的实验装置研究杠杆上同一点C提起物体的重力大小与______________的关系。
②分析比较表二的实验数据及相关条件可得到的初步结论是当动力F的方向和d的大小不变时,杠杆上同一点C 提起物体的重力大小_______________________________。
③分析比较表三中第二列和第三列的实验数据及相关条件发现当动力F和d的大小不变时,杠杆上同一点C提起物体的重力大小与动力F __________有关。
④该小组同学在以上结论的基础上,又测量了支点到动力F作用线的垂直距离l,并将测量结果填写在表三中的最后一列,分析表三中最后两列数据,归纳得出的结论是动力大小不变,杠杆上同一点C提起的物体重力大小随__________变大而变大。
⑤该小组同学还想探究杠杆提起物体的重力大小与重物在杠杆上的位置的关系,则应该保持动力的大小、作用点、__________不变,改变_____________进行探究。
2017年第四教育署初二第二学期期中试卷答案题号答案及评分标准一、选择题(共20分)1.B。