黑龙江省哈尔滨市萧红中学2019-2020学年 2021届八年级(五四制 )9月份教学素养监测数学
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黑龙江省哈尔滨市萧红中学2020-2021学年八学年下学期3月份学科素养监测数学试题(wd无答案)一、单选题1. 下列几组数中,能作为直角三角形三边长度的是()A.2,3,4B.5,8,9C.8,15,17D.13,14,152. 顺次连接平行四边形各边的中点得到的四边形是()A.平行四边形B.菱形C.矩形D.正方形3. 下列给出的条件能判定四边形ABCD为平行四边形的是()A.AB//CD,AD=BC B.∠A=∠B,∠C=∠D C.AB=CD,AD=BC D.AB=AD,CB=CD4. 已知a、b、c为的三边,且满足,则是()A.等腰三角形B.直角三角形C.等腰直角三角形D.等腰三角形或直角三角形5. 如图,小贤为了体验四边形的不稳定性,将四根木条用钉子钉成一个矩形框架ABCD,B与D两点之间用一根橡皮筋拉直固定,然后向右扭动框架,观察所得四边形的变化,下列判断错误的是()A.四边形ABCD由矩形变为平行四边形B.BD的长度增大C.四边形ABCD的面积不变D.四边形ABCD的周长不变6. 已知直角三角形中30°角所对的直角边长是cm,则另一条直角边的长是()A.4cm B. cm C.6cm D. cm7. 菱形的周长为,一条对角线长为,则菱形的面积为().A.B.C.D.8. 如图,在Rt △ABC中,∠ACB=90°,CD为AB边上的高,CE为AB边上的中线,AD=2,CE=5,则CD=()A.2B.3C.4D.29. 下列命题中正确的是()A.一组对边相等,另一组对边平行的四边形是平行四边形B.对角线相等的四边形是矩形C.对角线互相垂直的四边形是菱形D.对角线互相垂直平分且相等的四边形是正方形10. 如图,矩形纸片ABCD中,AB=4,BC=8,将纸片沿EF折叠,使点C 与点A重合,则下列结论错误的是()A.AF=AE B.△ABE≌△AGF C.EF=D.AF=EF二、填空题11. 已知平行四边形ABCD中,∠A=65°,则∠D的度数是 _______ .12. 已知等腰三角形的一条腰长是5,底边长是6,则底边上的高为 _____ .13. 菱形的对角线长分别是10、16,则它的面积是 _______ .14. 已知等边三角形的边长为2cm,则它的面积为 ________ .15. 已知,如图,一轮船以16海里/时的速度从港口A出发向东北方向航行,另一轮船以12海里/时的速度同时从港口A出发向东南方向航行,离开港口2小时后,则两船相距 _____ .16. 如图,所有的四边形都是正方形,所有的三角形都是直角三角形,其中最大的正方形的边长为2cm,则图中7个正方形的面积之和为 _______ cm 2.17. 如图,将正方形OEFG放在平面直角坐标系中,O是坐标原点,点E的坐标为(2,3),则点F的坐标为 _____ .18. 如图,将一根25㎝长的细木棒放入长、宽、高分别为8㎝、6㎝和10㎝的长方体无盖盒子中,则细木棒露在盒外面的最短长度是 ____ ㎝.19. 已知平行四边形ABCD中,AB=10,BC边上的高为6,AC=3 ,则平行四边形ABCD的周长 ______ .20. 矩形ABCD与CEFG如图放置,点B,C,E共线,点C,D,G共线,连接AF,取AF的中点H,连接GH.若BC=EF=2 ,CD=CE=1 ,则GH= _____ .三、解答题21. (1)如图1,在Rt△ABC中,∠C=90°,BC=2,AB=4,求AC的长.(2)如图2,在△ABC中,AB=2,AC=4,∠A=120°,求BC的长.22. 如图,图1、图2分别是10×8的网格,网格中每个小正方形的边长均为1,A、B两点在小正方形的顶点上,请在图1取一点C、在图2中取一点D(点C、点D必须在小正方形的顶点上),使以A、B、C和A、B、D为顶点的三角形分别满足以下要求:(1)在图1中画一个△ABC,使△ABC为面积为5的直角三角形( AB为斜边);(2)在图2中画一个△ABD,使△ABD为钝角等腰三角形且其面积为............... 1023. 如图,正方形ABCD的边长为5,点E,F分别在AD,DC上,AE=DF=2,BE与AF相交于点G,H为BF的中点,连接GH,求GH的长.24. 如图,在△ABC中,D、E分别是AB、AC的中点,BE=2DE,过点C作CF∥BE交DE的延长线于F,连接CD.(1)求证:四边形BCFE是菱形;(2)在不添加任何辅助线和字母的情况下,请直接写出图中与△BEC面积相等的所有三角形(不包括△BEC).25. 如图,一艘渔船正以30海里/时的速度由西向东追赶鱼群,在A处看见小岛C在船的北偏东60°方向上,40分钟后,渔船行至B处,此时看见小岛C在渔船的北偏东30°方向上,(1)求A处与小岛C之间的距离;(2)渔船到达B处后,航向不变,继续航行多少时间与小岛C的距离恰好为10 海里.26. 如图1,在菱形ABCD中,E、F分别在AB、CD上,DE平分∠AEF.(1)求证:EF= DF;(2)如图2,连接BD交EF于点N,若AE= CF,求证:点N在AC上;(3)在(2)的条件下,∠ANE+∠BDE=45°,在DE上取一点M,连接AM,AD= DM,AM= ,求EM的长.27. 已知如图,平面直角坐标系内的矩形OABC,点A在x轴上,点C在y轴上,点B坐标为(),D为AB边上一点,将△BCD沿直线CD折叠,得到△ECD,点B的对应点E落在线段OA上.(1)求OE的长;(2)点P从点C出发,以每秒2个单位长度的速度沿射线CD方向运动,设运动时间为t,△PBD的面积为S,求S关于t的关系式;(3)在(2)的条件下,点Q为直线DE上一点,是否存在t,使得以点A、B、Q、P为顶点的四边形为平行四边形?若存在,请求出t的值,并直接写出点P、点Q的坐标;若不存在,请说明理由.。
绝密★启用前 黑龙江省哈尔滨市萧红中学2019-2020学年八年级上学期10月月考数学试题 试卷副标题 注意事项: 1.答题前填写好自己的姓名、班级、考号等信息 2.请将答案正确填写在答题卡上 第I 卷(选择题) 请点击修改第I 卷的文字说明 一、单选题 1.方程23x x =-+的解是( ) A .1 B .13 C .–1 D .13- 2.下列方程中是一元一次方程的是( ) A .23x y = B .2222(1)3x x x -+=+ C .()21112x x +-= D .234x x -= 3.下列说法中,正确的是( ) A .若ca=cb ,则a=b B .若a b c c =,则a=b C .若a 2=b 2,则a=b D .由4-532x x =+,得到4352x x -=-+ 4.解方程 时,去分母正确的是( ) A. B. C. D.5.根据“x 的3倍与5的和比x 的13少2”列出的方程是( ) A .3523x x +=- B .3523x x +=+C .()3523x x +=-D .()3523x x +=+ 6.甲队有32人,乙队有28人,现从乙队抽x 人到甲队,使甲队人数是乙队人数的2倍,据题意,可列方程为( ) A .32+x=56 B .32=2(28-x ) C .32+x=2(28-x ) D .2(32+x )=28-x 7.一套仪器由1个A 部件和3个B 部件构成,1立方米钢材可做40个A 部件或240个B 部件,现要用6立方米钢材制作这种仪器,设应用x 立方米钢材做B 部件,其他钢材做A 部件,恰好配套,则可列方程为( )A .340240(6)x x ⨯=-B .324040(6)x x ⨯=-C .403240(6)x x =⨯-D .240340(6)x x =⨯-8.某个体商贩在一次买卖中,同时卖出两件上衣,售价都是135元,若按成本计,其中一件盈利25%,另一件亏本25%,在这次买卖中他( )A .不赚不赔B .赚9元C .赔18元D .赚18元 9.日历上横向相邻三个数的和为57,则三个数中最大的数是( )A .26B .20C .19D .1810.商店将进价2400元的彩电标价3200元出售,为了吸引顾客进行打折出售,售后核算仍可获利20%,则折扣为( )A .九折B .八五折C .八折D .七五折第II 卷(非选择题) 请点击修改第II 卷的文字说明 二、填空题 11.一元一次方程360x -+=,方程的解是______。
萧红中学2021届八年级6月份教学质量监测数学学科A. a =1, b = 2, c = 3B. a =b = 1, c =3C. a= 4, b = 5, c = 62.下列各曲线表示的y与x之间的关系中,y不是x的函数的是()3.下列方程是关于x的一元二次方程的是();A.ax 2 +bx +c = 0B. 1+1= 2 x2xC.x2 +2x =(x+1)(x-1)D.3(x+1)2 =2(x+1)4.顺次连接四边形ABCD各边的中点,得到四边形EFGH,在下列条件中,可使四边形EFGH成为菱形的是( )A.AB=CDB. AC=BDC. AC⊥BDD. AD∥BC5.如图,以Rt△ABC的三边为边长向外作正方形,三个正方形的面积分别为S1 、S2 、S3 ,若S 1=13,S2=12,则S3的值为().A.1B.5C.25D.1446.如图,正方形ABCD的边长为3,∠ABE=15 ,且AB =AE,则DE=()A.3B.4C.6D.9第5题图第6题图7.已知正比例函数y=(m-1)x的图象上两点A(x1,y1),B(x2,y2),当x1<x2时,有y1>y2,那时间:120分钟总分:120分一.选择题(每题 3 分,共 30 分)1.由下列线段 a,b,c 可以组成直角三角形的是( ).么m的取值范围是( )A.m<1 B.m>1 C.m<2 D.m>08.下列四个命题中是假命题的是( )A.对角线相等且互相平分的四边形是矩形;B.一组对边平行一组对角相等的四边形是平行四边形;C.一条对角线平分一组对角的平行四边形是菱形;D.对角线相等且互相垂直的四边形是菱形;9.如图,函数y=2x 和 y=ax+4 的图象相交于点A(m,3),则不等式2x≥ax+4的解集为()10.甲乙两同学从A地出发,骑自行车在同一条路上行驶到B地,他们离出发地的距离s(千米)和行驶时间t(时)之间的函数关系的图象,如图所示。
萧红中学2023-2024学年度八年级下学期语文学科寒假学情测试一、积累与运用(54分)一个对读书着迷的孩子,一捧起书,他的心中便觉得有了着落,有了依靠。
他抚摸一本书,不仅用手指和眼睛,还用心灵。
在温暖的阳光下,他翻开一本书仔细阅读,与一段优美的文字邂①。
这些文字不端架子,如行云流水,又如云端紫燕,显得气象万千。
此刻,他就沉浸在雅洁从容的文字里,仿佛和一位智者相对而坐,听智者娓娓道来,他的心就如长出翅膀,飞②千古。
通过读书能使他成长为精神富足、知识渊博、思维敏捷的少年。
上面是一位同学在读书活动中写的一段文字,请阅读这段文字回答下面的(1-3)题。
1.(5分)依次给这段文字中加点的字注音,全部正确的一项是……………………………()A.着迷(zhāo) 着落(zhuó)B.着迷(zhe) 着落(zháo)C.着迷(zháo) 着落(zhuó)D.着迷(zhuó) 着落(zhe)2.(5分)在这段文字横线处填入汉字,全部正确的一项是…………………………………()A.①逅②跃B.①逅②越C.①垢②跃D.①垢②越3.(5分)在文段中,画横线的句子有语病,下列修改最恰当的一项是……………………()A.读书能使他成长为精神富足、知识渊博、思维敏捷的少年。
B.通过读书,他能成长为精神富足、知识渊博、思维敏捷的少年。
C.读书能使他成长为知识渊博、思维敏捷、精神富足的少年。
D.通过读书能使他成长为知识渊博、思维敏捷、精神富足的少年。
4.(5分)《钢铁是怎样炼成的》一书的主人公保尔在谁的影响下走向革命道路………()A.阿维尔巴赫教授B.朱赫来C.巴扎诺娃D. 母亲5. (5分) 传说,有一天,苏东坡去金山寺给诗僧老友佛印禅师祝寿,席间,即兴为佛印禅师写了这副对联:花甲一周尚余半百岁月,古稀双庆犹欠春秋。
请问,横线上填写多少岁?( )A.二十B.三十C.四十D.五十6.(5分)关于新闻知识,下列表述不恰当的一项是…………………………………………………()A.新闻作品包括消息、新闻特写、新闻通讯、新闻评论等不同体裁。
2023-2024学年黑龙江省哈尔滨市南岗区萧红中学八年级(下)期中数学试卷(五四学制)一、选择题:本题共10小题,每小题3分,共30分。
在每小题给出的选项中,只有一项是符合题目要求的。
1.下列图形中,不是轴对称图形的是()A.矩形B.菱形C.平行四边形D.等腰三角形2.已知三角形的边长分别为1,,,那么这个三角形一定是()A.锐角三角形B.直角三角形C.钝角三角形D.等腰直角三角形3.下列式子中,能表示两个变量y与x是正比例函数关系的是()A. B. C. D.4.下列各曲线中表示y是x的函数图象的是()A. B.C. D.5.已知:三角形的各边分别为6cm,8cm,10cm,则连结各边中点所成三角形的周长为()A.24cmB.18cmC.14cmD.12cm6.在平面直角坐标系中,一次函数的图象经过()A.第一、二、三象限B.第一、三、四象限C.第一、二、四象限D.第二、三、四象限7.如图,在矩形ABCD中,,,将矩形ABCD沿EF折叠,使点C与点A重合,则折痕EF的长为()A.6B.12C.D.8.下列命题中,正确的是()A.对角线相等的四边形是矩形B.对角线互相平分的四边形是平行四边形C.对角线互相垂直的四边形是菱形D.对角线互相垂直且相等的四边形是正方形9.10.“五一节”期间,乐乐老师一家自驾游去了离家260千米的某目的地,下面是她们离家的距离千米与汽车行驶时间小时之间的函数图象,她们出发小时时,离目的地还有千米.A.22B.32C.238D.228二、填空题:本题共10小题,每小题3分,共30分。
11.在函数中,自变量x的取值范围是______.12.在▱ABCD中,::2,则______.13.已知正比例函数,且y随x的增大而减小,则k的取值范围是______.14.在中,,斜边,若,则______.15.如图,在矩形ABCD中,于E,BE::3,则______.16.如图,▱ABCD的周长为18,对角线AC,BD相交于点O,,E为CD边上的中点,则______.17.如图,在菱形ABCD中,,,则的周长是______18.如图,已知正方形ABCD的面积为64,点F在CD上,点E在CB的延长线上,且,,则BE的长为______.19.将直线沿着y轴平移,平移后所得直线经过点,并且交y轴于点B,若点O为坐标原点的面积为2,则k的值为______.20.如图,在四边形ABCD中,,,垂足为点E,连接AC交DE于点F,点G为AF的中点,,若,,则DE的长为______.三、解答题:本题共7小题,共60分。
黑龙江省哈尔滨市南岗区萧红中学【最新】八年级下学期期末化学试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.图片中的内容不属于化学对人类贡献的是()A.医务人员佩戴的口罩B.消毒用75%的酒精C.工厂排出的有害烟尘D.超塑延展性的纳米铜2.下列过程中没有发生化学变化的是()A.炸药爆破拆除楼房B.发酵制取沼气C.生成炭黑的实验D.稀有气体填充霓虹灯3.下列实验操作错误的是()A.用橡胶塞塞住试管B.收集氧气C.往试管里放入固体粉末D.液体的量取4.下列有关物质的应用错误的是()A.氧气用于气焊B.稀有气体用于磁悬浮列车C . 活性炭用于净水D . 二氧化碳用作气体肥料5.下列有关叙述正确的是()A .家用电器短路着火,应该立即用水基型灭火器灭火B .日常生活中用蒸馏的方法将硬水软化C .我国科学家张青莲教授为相对原子质量的测定做出了卓越贡献D .当燃气灶的火焰呈黄色或橙色时,调大炉具的进气口 6.下列实验现象描述正确的是() A .甲烷燃烧:产生淡蓝色火焰,放热B .铁钉与硫酸铜溶液反应:银白色固体表面有紫红色固体析出,溶液由浅绿色变为蓝色C .电解水实验:两个电极表面产生气泡,一段时间后正负极产生的气体体积比约为1∶2D .镁条燃烧:白色固体燃烧,发出耀眼的白光7.下列应用以及化学方程式的书写和基本反应类型都正确的是() A .拉瓦锡测定空气中氧气含量;2254P+5O 2P O ↑点燃氧化反应B .用过氧化氢溶液制氧气:22222H O 2H O+O =↑ 分解反应C .铁丝在氧气中燃烧:2343Fe+2O Fe O 点燃化合反应D .一氧化碳和氧化铜的反应:2CuO+CO Cu+CO = 还原反应8.下列有关粒子结构示意图的说法,正确的是()A.①和②的化学性质相似B.③④⑥对应的元素位于周期表的同一周期C.①③④具有相对稳定结构D.③和⑤可以形成A2B型化合物9.这一阶段我们经常会用75%的酒精溶液灭菌消毒,此浓度的酒精能较快渗入细胞内,使蛋白质凝固变性,从而起到杀菌的作用。
2021-2022学年黑龙江省哈尔滨市南岗区萧红中学八年级(上)月考数学试卷(10月份)(五四学制)一、选择题(本大题共10小题,共30.0分)1.下列图形中是轴对称图形的是()A. B.C. D.2.已知点P(2,1),那么点P关于x轴对称的P′的坐标是()A. P′(−2,−1)B. P′(2,−1)C. P′(−1,2)D. P′(2,1)3.如图,已知△ABC和△A′B′C′关于直线l对称,下列选项中的结论不正确的是()A. △ABC≌△A′B′C′B. ∠BAC=∠B′A′C′C. 直线l垂直平分CC′D. 直线BC和B′C′的交点不一定在直线l上4.如图是屋架设计图的一部分,其中∠A=30°,点D是斜梁AB的中点,BC、DE垂直于横梁AC,AB=16m,则DE的长为()A. 8mB. 4mC. 2mD. 6m5.小强在车站要从黑龙江鸡西去山东寿光,从镜子中看到镜子对面墙上A. 15:01B. 10:51C. 10:21D. 12:016.如果一个三角形有两个角分别为80°,50°,则这个三角形是()三角形.A. 等腰B. 等边C. 不等边D. 直角7.如图,∠EAF=15°,AB=BC=CD=DE=EF,则∠DEF等于()A. 90°B. 75°C. 70°D. 60°8.若等腰三角形的周长为26cm,一边为11cm,则腰长为()A. 11cmB. 7.5cmC. 11cm或7.5cmD. 以上都不对9.如图,在△ABC中,BC=8cm,AB的垂直平分线交AB于点D,交边AC于点E,△BCE的周长等于18cm,则AC的长等于()A. 6cmB. 8cmC. 10cmD. 12cm10.如图所示,在△ABC中,AB=AC,∠A=36°,BD、CE分别为∠ABC与∠ACB的角平分线且相交于点F,则图中的等腰三角形有()A. 6个B. 7个C. 8个D. 9个二、填空题(本大题共11小题,共37.0分)11.已知点P在线段AB的垂直平分线上,PA=6,则PB=______ .12.等腰三角形的一底角等于30°,则其它两个内角各为______.13.等腰三角形的两边分别为6和3,则它的第三边为______.14.已知点A(x,−4)与点B(3,y)关于x轴对称,那么x+y的值为______.15.如图,在△ABC中,AB=AC,AD是BC边上的高,点E、F是AD的三等分点,若△ABC的面积为12cm2,则图中阴影部分的面积是______cm2.16.如图,在Rt△ABC中,∠ACB=90°,AB的垂直平分线DE交AC于E,交BC的延长线于F,若∠F=30°,DE=1,则BE的长是______.17.Rt△ABC中,∠C=90°,∠B=2∠A,AB=4,则BC=______.18.如图,在等边△ABC中,D、E分别是AB、AC上的点,且AD=CE,则∠BCD+∠CBE=______度.19.一个等腰三角形的一个外角等于100°,则这个三角形的顶角应该为______.20.如图,在平面直角坐标系中,对△ABC进行循环往复的轴对称变换,若原来点A的坐标是(a,b),则经过第2021次变换后所得A点坐标是______.21.如图,已知点M、N和∠AOB,求作一点P,使P到点M、N的距离相等,且到∠AOB的两边的距离相等.(要求用尺规画图,保留作图痕迹)三、解答题(本大题共6小题,共53.0分)22.等腰三角形一腰上的中线把这个三角形的周长分成12cm和21cm两部分,求这个等腰三角形的底边长.23.已知:如图,已知△ABC,分别画出与△ABC关于x轴、y轴对称的图形△A1B1C1和△A2B2C2.24.如图,AC和BD相交于点O,且AB//DC,OC=OD,求证:OA=OB.25.如图:已知等边△ABC中,D是AC的中点,E是BC延长线上的一点,且CE=CD,DM⊥BC,垂足为M.求证:(1)DE=2DM;(2)M是BE的中点.26.如图,已知∠MAN=120°,AC平分∠MAN.B、D分别在射线AN、AM上.(1)在图(1)中,当∠ABC=∠ADC=90°时,求证:AD+AB=AC.(2)若把(1)中的条件“∠ABC=∠ADC=90°”改为∠ABC+∠ADC=180°,其他条件不变,如图(2)所示.则(1)中的结论是否仍然成立?若成立,请给出证明;若不成立,请说明理由.27.如图,在平面直角坐标系中,直线AB与y轴的正半轴交于点A,与x轴交于点B(2,0),三角形△ABO的面积为2.点Q的坐标是(4,0).动点P从点O出发,以每秒1个单位长度的速度在射线OB上运动,过P作PM⊥x轴交直线AB于M.(1)求点A的坐标;(2)当点P在线段OB上运动时,设△MBQ的面积为S,点P运动的时间为t秒,请用含t的式子来表示s;(3)当点P在线段OB延长线上运动时,是否存在某一时刻t(秒),使△MBQ是以QM为腰的等腰三角形?若存在,求出时间t值.答案和解析1.【答案】C【解析】解:选项A、B、D均不能找到这样的一条直线,使图形沿一条直线折叠,直线两旁的部分能够互相重合,所以不是做轴对称图形;选项C能找到这样的一条直线,使图形沿一条直线折叠,直线两旁的部分能够互相重合,所以是做轴对称图形;故选:C.根据轴对称图形的概念对各选项分析判断即可得解.如果一个图形沿一条直线折叠,直线两旁的部分能够互相重合,这个图形叫做轴对称图形.本题考查了轴对称图形的概念,轴对称图形的关键是寻找对称轴,图形两部分折叠后可重合.2.【答案】B【解析】解:∵点P(2,1),∴点P关于x轴对称的P′的坐标是(2,−1).故选B.根据“关于x轴对称的点,横坐标相同,纵坐标互为相反数”解答.本题考查了关于x轴、y轴对称点的坐标,解决本题的关键是掌握好对称点的坐标规律:(1)关于x轴对称的点,横坐标相同,纵坐标互为相反数;(2)关于y轴对称的点,纵坐标相同,横坐标互为相反数;(3)关于原点对称的点,横坐标与纵坐标都互为相反数.3.【答案】D【解析】解:∵△ABC和△A′B′C′关于直线l对称,∴△ABC≌△A′B′C′,∠BAC=∠B′A′C′,直线l垂直平分线段CC′,直线BC和直线B′C′的交点在对称轴l上,故A,B,C正确,不符合题意;D不正确,故符合题意.根据轴对称的性质一一判断即可.本题考查轴对称的性质,全等三角形的性质等知识,解题的关键是掌握轴对称的性质,属于中考常考题型.4.【答案】B【解析】解:∵∠A=30°,AB=16m,∴BC=12AB=12×16=8m,∵BC、DE垂直于横梁AC,∴BC//DE,∵点D是斜梁AB的中点,∴DE=12BC=12×8=4m.故选:B.先根据30°角所对的直角边等于斜边的一半求出BC的长度,再根据三角形的中位线平行于第三边并且等于第三边的一半进行求解.本题考查了30°角所对的直角边等于斜边的一半的性质以及三角形的中位线定理,熟记定理是解题的关键.5.【答案】C【解析】解:电子表的实际时刻是10:21.故选:C.镜子中看到的数字与实际数字是关于镜面成垂直的线对称.注意镜子的5实际应为2.此题主要考查了镜面对称,可以把数据抄下来,反过来看看,这样最直观.6.【答案】A【解析】解:第三个角=180°−80°−50°=50°,故这个三角形是等腰三角形.故选:A.根据三角形的内角和可知第三个角是50度,由等腰三角形的判定得到这个三角形是等腰三角形,本题主要考查了三角形的内角和是180度及等腰三角形的判定;利用三角形内角和求得角的度数是证明等腰三角形常用方法之一,要熟练掌握.7.【答案】D【解析】【分析】此题主要考查了等腰三角形的性质及三角形内角和外角之间的关系.(1)三角形的外角等于与它不相邻的两个内角和;(2)三角形的内角和是180度,属于基础题.根据已知条件,利用等腰三角形的性质及三角形的内角和外角之间的关系进行计算.【解答】解:∵AB=BC=CD=DE=EF,∠EAF=15°,∴∠BCA=∠EAF=15°,∴∠CBD=∠BDC=∠BCA+∠EAF=15°+15°=30°,∴∠BCD=180°−(∠CBD+∠BDC)=180°−60°=120°,∴∠ECD=∠CED=180°−∠BCD−∠BCA=180°−120°−15°=45°,∴∠CDE=180°−(∠ECD+∠CED)=180°−90°=90°,∴∠EDF=∠EFD=180°−∠CDE−∠BDC=180°−90°−30°=60°,∴∠DEF=180°−(∠EDF+∠EFD)=180°−120°=60°.故选:D.8.【答案】C【解析】解:①11cm是腰长时,腰长为11cm,(26−11)=7.5cm,②11cm是底边时,腰长=12所以,腰长是11cm或7.5cm.故选:C.分边11cm是腰长与底边两种情况讨论求解.本题考查了等腰三角形的性质,难点在于要分情况讨论.9.【答案】C【解析】解:∵DE是边AB的垂直平分线,∴AE=BE.∴△BCE的周长=BC+BE+CE=BC+AE+CE=BC+AC=18.又∵BC=8,∴AC=10(cm).故选:C.AC=AE+EC=BE+EC,根据已知条件易求.此题主要考查线段的垂直平分线的性质等几何知识.线段的垂直平分线上的点到线段的两个端点的距离相等.10.【答案】C【解析】解:∵在△ABC中,AB=AC,∠A=36°,=72°,∴∠ABC=∠ACB=180°−36°2∵BD、CE分别为∠ABC与∠ACB的角平分线,∴∠ABD=∠CBD=∠ACE=∠BCE=∠A=36°,∴AE=CE,AD=BD,BF=CF,∴△ABC,△ABD,△ACE,△BFC是等腰三角形,∵∠BEC=180°−∠ABC−∠BCE=72°,∠CDB=180°−∠BCD−∠CBD=72°,∠EFB=∠DFC=∠CBD+∠BCE=72°,∴∠BEF=∠BFE=∠ABC=∠ACB=∠CDF=∠CFD=72°,∴BE=BF,CF=CD,BC=BD=CE,∴△BEF,△CDF,△BCD,△CBE是等腰三角形.∴图中的等腰三角形有8个.故选C.由在△ABC中,AB=AC,∠A=36°,根据等边对等角,即可求得∠ABC与∠ACB的度数,又由BD、CE分别为∠ABC与∠ACB的角平分线,即可求得∠ABD=∠CBD=∠ACE=∠BCE=∠A=36°,然后利用三角形内角和定理与三角形外角的性质,即可求得∠BEF=∠BFE=∠ABC=∠ACB=∠CDF=∠CFD=72°,由等角对等边,即可求得答案.此题考查了等腰三角形的判定与性质、三角形内角和定理以及三角形外角的性质.此题难度不大,解题的关键是求得各角的度数,掌握等角对等边与等边对等角定理的应用.【解析】解:∵点P在线段AB的垂直平分线上,PA=6,∴PB=PA=6.故答案为:6.直接根据线段垂直平分线的性质进行解答即可.本题考查的是线段垂直平分线的性质,熟知垂直平分线上任意一点,到线段两端点的距离相等是解答此题的关键.12.【答案】30°和120°【解析】解:∵等腰三角形的一底角等于30°,∴等腰三角形的另一底角也等于30°,∴等腰三角形的顶角等于180°−30°−30°=120°,故其它两个内角各为30°和120°,故答案为:30°和120°.根据等腰三角形的性质即可得到答案.本题考查的是等腰三角形的性质及三角形内角和定理;熟练掌握等腰三角形的性质是解答本题的关键.13.【答案】6【解析】解:由题意得,当腰为3时,则第三边也为腰,为3,此时3+3=6.故以3,3,6不能构成三角形;当腰为6时,则第三边也为腰,此时3+6>6,故以3,6,6可构成三角形.故答案为:6.题目给出等腰三角形有两条边长为3和6,而没有明确腰、底分别是多少,所以要进行讨论,还要应用三角形的三边关系验证能否组成三角形.本题考查了等腰三角形的性质和三角形的三边关系;已知没有明确腰和底边的题目一定要想到两种情况,分类进行讨论,还应验证各种情况是否能构成三角形进行解答,这点非常重要,也是解题的关键.【解析】解:∵点A(x,−4)与点B(3,y)关于x轴对称,∴x=3,y=4,∴x+y=7,故答案为:7.根据关于x轴对称点的坐标特点:横坐标不变,纵坐标互为相反数可得x、y的值,进而可得x+y的值.此题主要考查了关于x轴对称点的坐标,关键是掌握点的坐标的变化规律.15.【答案】6【解析】解:∵△ABC中,AB=AC,AD是BC边上的高,∴△ABC是轴对称图形,且直线AD是对称轴,∴△CEF和△BEF的面积相等,=S△ABD,∴S阴影∵AB=AC,AD是BC边上的高,∴BD=CD,∴S△ABD=S△ACD=1S△ABC,2∵S△ABC=12cm2,∴S阴影=12÷2=6cm2.故答案为:6.由图,根据等腰三角形是轴对称图形知,△CEF和△BEF的面积相等,所以阴影部分的面积是三角形面积的一半.本题考查了等腰三角形的性质及轴对称性质;利用对称发现并利用△CEF和△BEF的面积相等是正确解答本题的关键.16.【答案】2【解析】解:∵∠ACB=90°,FD⊥AB,∴∠ACB=∠FDB=90°,∵∠F=30°,∴∠A=∠F=30°(同角的余角相等).又∵AB的垂直平分线DE交AC于E,∴∠EBA=∠A=30°,∴直角△DBE中,BE=2DE=2.故答案是:2.根据同角的余角相等、等腰△ABE的性质推知∠DBE=30°,则在直角△DBE中由“30度角所对的直角边是斜边的一半”即可求得线段BE的长度.本题考查了线段垂直平分线的性质、含30度角的直角三角形.解题的难点是推知∠EBA=30°.17.【答案】2【解析】解:在Rt△ABC中,∵∠C=90°,∴∠A+∠B=90°,∵∠B=2∠A,∴3∠A=90°,∴∠A=30°,∴BC=12AB=12×4=2,故答案为:2.首先可求出∠A=30°,再根据含30°角的直角三角形的性质即可解答.本题主要考查了含30°角的直角三角形的性质,根据题意求出∠A=30°是解题的关键.18.【答案】60【解析】【试题解析】【分析】此题考查了等边三角形的性质及全等三角形的判定方法,常用的判定方法有SSS,SAS,AAS,HL等.根据等边三角形的性质,得出各角相等各边相等,已知AD=CE,利用SAS判定△ADC≌△CEB,从而得出∠ACD=∠CBE,所以∠BCD+∠CBE=∠BCD+∠ACD=∠ACB=60°.【解答】解:∵△ABC是等边三角形∴∠A=∠ACB=60°,AC=BC,∵AD=CE,∴△ADC≌△CEB,∴∠ACD=∠CB,∴∠BCD+∠CBE=∠BCD+∠ACD=∠ACB=60°.故答案为60.19.【答案】20°或80°【解析】解:①若100°的外角是此等腰三角形的顶角的邻角,则此顶角为:180°−100°=80°;②若100°的外角是此等腰三角形的底角的邻角,此底角为:180°−100°=80°;∴顶角为:180°−80°−80°=20°;故这个等腰三角形的顶角为:20°或80°.故答案为:20°或80°.由等腰三角形的一个外角是100°,可分别从①若100°的外角是此等腰三角形的顶角的邻角;②若100°的外角是此等腰三角形的底角的邻角去分析求解,即可求得答案.此题考查了等腰三角形的性质.此题比较简单,解题的关键是注意分类讨论思想的应用,小心别漏解.20.【答案】(a,−b)【解析】解:∵点A第一次关于x轴对称后在第四象限,点A第二次关于y轴对称后在第三象限,点A第三次关于x轴对称后在第二象限,点A第四次关于y轴对称后在第一象限,即点A回到原始位置,∴每四次对称为一个循环组依次循环,∵2021÷4=505…1,∴经过第2013次变换后所得的A 点与第一次变换的位置相同,在第四象限,坐标为(a,−b), 故答案为(a,−b).观察图形可知每四次对称为一个循环组依次循环,用2021除以4,然后根据商和余数的情况确定出变换后的点A 所在的象限,然后解答即可.本题考查了轴对称的性质,点的坐标变换规律,读懂题目信息,观察出每四次对称为一个循环组依次循环是解题的关键,也是本题的难点.21.【答案】解:如图所示,点P 就是所求的点.【解析】使P 到点M 、N 的距离相等,即画MN 的垂直平分线,且到∠AOB 的两边的距离相等,即画它的角平分线,两线的交点就是点P 的位置.此题主要考查了线段的垂直平分线和角平分线的作法.这些基本作图要熟练掌握,注意保留作图痕迹.22.【答案】解:如图所示,设AD =DC =x ,BC =y ,由题意得{x +2x =12y +x =21,或{x +2x =21y +x =12, 解得{x =4y =17或{x =7y =5, 当{x =4y =17,等腰三角形的三边为8,8,17,显然不符合三角形的三边关系;当{x =7y =5时,等腰三角形的三边为14,14,5, 所以,这个等腰三角形的底边长是5,综上所述,这个等腰三角形的底边长5.【解析】作出图形,设AD =DC =x ,BC =y ,然后分两种情况列出方程组求解,再根据三角形的三边关系判断即可得解. 本题考查了等腰三角形的性质性质,三角形的三边关系,难点在于分情况讨论,作出图形更形象直观.23.【答案】解:如图所示.【解析】根据关于坐标轴对称的点的坐标特点画出图形即可.本题考查的是作图−轴对称变换,熟知关于坐标轴对称的点的坐标特点是解答此题的关键.24.【答案】证明:∵OC=OD,∴△ODC是等腰三角形,∴∠C=∠D,又∵AB//DC,∴∠A=∠C,∠B=∠D,∴∠A=∠B,∴△AOB是等腰三角形,∴OA=OB.【解析】根据OC=OD得,△ODC是等腰三角形;根据AB//DC,得出对应角相等,求得△AOB是等腰三角形,证明最后结果.本题主要考查了等腰三角形的判定和平行线的性质:两直线平行,内错角相等.25.【答案】证明:(1)∵三角形ABC是等边△ABC,∴∠ACB=∠ABC=60°,又∵CE=CD,∴∠E=∠CDE,又∵∠ACB=∠E+∠CDE,∠ACB=30°,∴∠E=12∵DM⊥BC,∴DE=2DM;(2)连接BD,∵等边△ABC中,D是AC的中点,∴∠DBC=12∠ABC=12×60°=30°由(1)知∠E=30°∴∠DBC=∠E=30°∴DB=DE又∵DM⊥BC∴M是BE的中点.【解析】(1)由等边△ABC的性质可得:∠ACB=∠ABC=60°,然后根据等边对等角可得:∠E=∠CDE,最后根据外角的性质可求∠E=30°,根据含30°角的直角三角形的性质即可求证;(2)连接BD,由等边三角形的三线合一的性质可得:∠DBC=12∠ABC=12×60°=30°,由(1)可得:∠DBC=∠E=30°,然后根据等角对等边,可得:DB=DE,最后根据等腰三角形的三线合一的性质可得M是BE的中点.此题考查了等边三角形的有关性质,重点考查了等边三角形的三线合一的性质.26.【答案】(1)证明:∵∠MAN=120°,AC平分∠MAN,∴∠DAC=∠BAC=60°∵∠ABC=∠ADC=90°,∴∠DCA=∠BCA=30°,在Rt△ACD中,∠DCA=30°,Rt△ACB中,∠BCA=30°∴AC=2AD,AC=2AB,∴AD+AB=AC;(2)解:结论AD+AB=AC成立.理由如下:在AN上截取AE=AC,连接CE,∵∠BAC=60°,∴△CAE为等边三角形,∴AC=CE,∠AEC=60°,∵∠DAC=60°,∴∠DAC=∠AEC,∵∠ABC+∠ADC=180°,∠ABC+∠EBC=180°,∴∠ADC=∠EBC,∴△ADC≌△EBC,∴DC=BC,DA=BE,∴AD+AB=AB+BE=AE,∴AD+AB=AC.【解析】本题主要考查了30°的直角三角形的边角关系以及全等三角形的判定和性质问题,能够利用其性质求解一些简单的计算、证明问题.(1)由题中条件可得,∠DCA=∠BCA=30°,在直角三角形中可得AC=2AD,AC=2AB,所以AD+AB=AC.(2)在AN上截取AE=AC,连接CE,可得△CAE为等边三角形,进而可得△ADC≌△EBC,即DC=BC,DA=BE,进而结论得证.27.【答案】解:(1)∵Rt△AOB面积是2,且OB=2,∴OA=2,即A(0,2);(2)如图1所示,由P的速度为1个单位/秒,得到OP=t,设直线AB 解析式为y =kx +b ,把A(0,2)和B(2,0)代入得:{b =22k +b =0, 解得:k =−1,b =2,即AB 解析式为y =−x +2,把x =t 代入直线AB 解析式y =−x +2中得:y =−t +2,即MP =−t +2, ∴S △MBQ =12BQ ⋅MP ,即y =12×(4−2)×(−t +2)=−t +2(0≤t <2); (3)当点P 在线段OB 延长线上运动时,存在某一时刻t =3或4(秒)时,使△MBQ 是以QM 为腰的等腰三角形,如图2所示:∵BM =QM ,MP ⊥BQ ,∴BP =QP =12BQ =1,∴OP =OB +BP =2+1=3,若P 与Q 重合,即t =4秒时,△MBQ 为等腰三角形,则当点P 在线段OB 延长线上运动时,存在某一时刻t =3或4秒时,使△MBQ 是以QM 为腰的等腰三角形.【解析】(1)由直角三角形AOB 面积,以及B 的坐标,求出OB 的长,进而求出OA 的长,确定出A 的坐标即可;(2)如图1所示,作出相应的图形,表示出OP 的长,利用待定系数法求出直线AB 解析式,表示出M 纵坐标,即为MP 的长,由BQ 为底,MP 为高表示出三角形MBQ 面积,即可确定出y 与t 的函数解析式;(3)当点P在线段OB延长线上运动时,存在某一时刻t(秒),使△MBQ是以QM为腰的等腰三角形,如图2所示,求出此时OP的长,即可确定出此时的时间.此题属于一次函数综合题,涉及的知识有:待定系数法求一次函数解析式,三角形面积求法,坐标与图形性质,以及等腰三角形的性质,熟练掌握待定系数法是解本题的关键.。
萧红中学2021届八年级4月份线上教学质量自我监测英语学科出题教师:王宏一、单项选择(本题共20分,每小题1分)( ) 1. Which of the following words has the same sound as the underlined letters of the word “inst ea d”?A.dealB. deafC. realize( ) 2. Which pair of the words with the underlined letters has different sounds?A.stress lendB. waste snackC. since kid( ) 3. Which word of the following doesn’t have the same stress as the others?A.provideB. allowC. proper( ) 4. Playing computer games and copying others’ homework are both bad habits for you. If you have them, you should ____ them ____.A.cut upB. cut downC. cut out( ) 5. In school, students need to share the cleaning work ______ the classroom clean and tidy.A. for keepB. with keepingC. to keep( ) 6. ---Could I get ______ to eat? I’m hungry. --- Of course, you ______.A.something, couldB. something, canC. anything, could( ) 7. ---I’m worried about making friends in the new class.---Don’t worry. I’m sure you will ____ your classmates if you are nice and friendly to them.A.catch up withB. compare withC. get on with( ) 8. --- I went to see you at 4: 00 yesterday afternoon, but you weren't at home.--- Sorry, I _____ with my friends at that time.A.. was swimingB. swamC. was swimming( ) 9. I hope you can go shopping with me tomorrow. If you don’t, ______.A. so will IB. neither I willC. neither will I( ) 10. Mary is my close friend. And she often offers ______ me with my English.A.to helpB. helpingC. help( )11.---Alex, you can sing as ______ as a professional(专业的) singer. I think you will be a singer in the future.---Thank you for saying so.A. beautifulB. beautifullyC. more beautifully( ) 12. ---What’s in today’s newspaper?--- According to the newspaper, the charity(慈善机构) provides homeless people ______ food and clothes.A.for B. with C. to( ) 13. Studying is hard for me. But whenever I want to give up, my teacher always tells me to ______.A.continueB. competeC. compare( ) 14. My father was reading a newspaper while my mother ________ dinner.A. is cookingB. cookedC. was cooking( ) 15. --- How long will you stay here?---As soon as I ______ my homework, I ______ back home.A. finish, will goB. will finish, goC. will finish, will go( ) 16. Besides studying at school, we also need to make some friends ______ we can help each other when being in trouble.A.in order toB. so thatC. such that( ) 17. --- ______ Mr. White was tired, ______ he kept working in his office.--- What a hard-working teacher he is!A.Though, butB. Although, /C. /, so( ) 18. It is a holiday in western countries. It comes on the fourth Thursday in November. On this day, families come together and have delicious food, like turkey. When it comes, people often say ______.A. Merry ChristmasB. Happy Thanksgiving DayC. Happy Halloween( ) 19. Safety always comes the first. When an earthquake (地震) happens, you should ______.①hide yourself under the solid(结实的) things②run out of the room and go to another place at once③stay in the safe place until the shaking stops④look for your family and help them⑤stay away from glass and windowsA.①②③B. ③④⑤C.①③⑤( ) 20. Many teenagers help with the chores. They are glad to do so. Here is a survey of 200 students. According to theA.20B.180C.60二、完型填空(本题共10分,每小题1分)Are your parents often angry with you? Or do you disagree with your parents? Maybe the relations between you and your parents are 21 . It's difficult to make your home a happy and healthy place. 22 , you can change the situations by the following advice.Respect(尊敬)It's important to show your respect and care about your parents' feelings. Though you have disagreements(分歧), at least try to understand 23 they’re trying to tell you. You shouldn’t keep your own thoughts only.Communication You can communicate with your parents 24 a polite way. You may not get what you want, but explaining your thoughts calmly(冷静地) and rationally(理智地) may show your parents that you are no longer a kid, not just a kid.Trust How can your parents build trust in you? It 25 what you do. For example, if you get home late without 26 first, next time your parents may not believe you. To get their trust, you must pay attention to the ways you do things __27 fighting with them. It will make your life much easier.Taking time together Some teenagers find that sharing time together can build trust and communication. Your parents may 28 going shopping, hanging out and playing cards with you. Spending time together will drive away the“ black cloud” __29 over your home.Your parents give all their love to you. Why 30 try these ways actively to own a happy family as you want?( ) 21. A. badly B. bad C. better( ) 22. A. In fact B. In the end C. As a result( ) 23. A. whatever B. whenever C. wherever( ) 24. A. on B. with C. in( ) 25. A. depends on B. takes after C. makes a difference( ) 26. A. to call B. calling C. calls( ) 27. A. of B. instead C. instead of( ) 28. A. be interested in B. be good at C. be used to( ) 29. A. to hang B. hangs C. hanging( ) 30.A. not B. not you C. don’t三、阅读理解(本题共20分,每小题1分)(A)It is expected(被期望) that students who come to university should have the ability(能力) to take a good set of notes.( ) 31. During listening to a lesson, the students ____.A. may lose interest after 10 minutesB. should keep up with the teacherC. should take down all the words the teacher says( ) 32. To look for main points, you can do many things EXCEPT ____.A. paying attention to teachers’ body languageB. showing your notes to the teachersC. discussing with classmates during the last minutes of a class( ) 33.If you don’t go over new notes in time, you may only remember ____of it.A. not more than 50%B. around 60%C. not more than 80%( ) 34. Which of the following is NOT true according to the passage?A. Reading the text before the class can help you find out questions ahead.B. Taking notes in class can help you get the main idea better.C. One should go over his lesson within 10 minutes after class.( ) 35. The article is probably for ____.A. school pupilsB. high school studentsC. university students(B)On a Friday night, a poor young violinist(小提琴手) was playing his violin at the entrance(人口) of the subway station. He played such great music that people slowed down to listen and put some money into his hat.The next day, the violinist came to the entrance again. He put down his hat as usual. Beside the hat was a piece of paper with some words on it. It said: A George Black has put an important thing into my hat by mistake. Welcome to claim (认领) it. Soon the people were all interested in the words and wondered what it could be. After a while, a man rushed into the crowd, took the violinist’s shoulders and said, "I knew you would certainly come. The violinist asked “Did you lose anything?”“ Lottery(彩票)” The violinist took out a lottery ticket with M r. Black’s name on it. Mr. Bla ck took it and danced happily.Mr. Black was an office worker. He bought the lottery and won a prize of $500,000. But when he gave $50 to the violinist for his wonderful music, the lottery ticket was thrown in the hat, too. The violinist was a college student. That morning, he was going to fly to Vienna for higher education. However, when he found the ticket, he put off the flight and went back to the entrance. Later someone asked the violinist why he didn’t take the lottery ticket to pay for his tuition(学费). He said, “Although I don’t have much money, I live a happy life. If I lose honesty, I won’t be happy forever.”It is true that being honest can get respect from others. People can’t lose honesty.( ) 36. The violinist sang songs so well that people stopped to listen and put money into his hat.( ) 37.The violinist came to the entrance of the subway station again on Saturday.( ) 38.Mr. Black was the owner of the lottery and he won a prize of Y500,000.( ) 39.The violinist put off his flight to Vienna because he wanted to find the person who lost the lottery.( ) 40. From the passage we know that the violinist was a person who was poor but being honest.( C )Which hand do you use when writing? About 8 to 15 percent of people are left handed. They often have to use tools that are designed (设计) for right-handed people. So it is difficult for left-handers to use most tools. In the past, people thought it was strange to use the left hand. Young students looked down upon their left-handed classmates. Some children were even punished (惩罚) for using their left hands to write. But these days parents and teachers have accepted that. In almost every school in the world, left-handed students can use their left hands to write.What causes people to be left-handed or right-handed? Experts have searched long and hard on this. They conclude (得出结论) that left-handed people are left handed for the same reason that right-handed people are right handed. One out of every ten people just is a left-hander. It's simply like the color of our eyes- some people have brown eyes, while some othershave black eyes.However, many researchers think that left-handers and right-handers are different in some places. Right-handers are more outgoing than left-handers. Many right-handed people understand spoken words better. Research shows that left-handers are creative and artistic. Many famous performers are left handed. If you prefer one hand, but you are still good at writing with the other, you are mixed. Research shows mixed-handers can remember everyday things better than other people. What did you eat for lunch two weeks ago? If you're mixed handed, you can probably remember.( ) 41. The underlined word "artistic” in the passage means ______ in Chinese.A.骄傲的B. 蠢笨的C.艺术的( ) 42. From the passage, we can learn that left-handers ______.A. are ordinary(普通的) people like right- handersB. may be more outgoingC. might remember everyday things better( ) 43. In the past, people thought left-handers are ______.A. very cleverB. very strangeC. creative and artistic( ) 44. Which of the following is TRUE according to the passage?A. In almost every school in the world in the past, left-handed students can use their left hands to write.B. Experts conclude that the reason why left-handed people are left handed is different from that right-handedpeople are right hand.C. Research shows mixed-handed can remember everyday things better than other people.( ) 45. Which is the main idea of this passage?A. Left-handers are looked down upon by others.B. How do we write with left hands?C. It tells us something about left -hander and the difference between left-handers and right-handers.(D)Summer is here. One of the best ways to cool off on a hot summer day is to jump into a swimming pool. Swimming is great exercise, and a trip to a pool or beach is a good way to have fun with friends. __46 . So it's important to stay safe as you do it. Here are a few things you can do to swim safely.Swim only in places set for swimming.Don't swim unless there is a lifeguard(救生员). 47 . In some places, water runs fast, which can sweep swimmers away without warning. So if a sign says "No swimming, please take care.”Be careful when diving(跳水) into the water headfirst(头向前地)Most pools have deep ends where diving is safe. In natural water, don't dive unless you know that the water is deep enough.48 . Make sure you have someone with you. Nobody knows what will happen while you are swimming. If something goes wrong, you can get help from that person as soon as possible. 49 .Pay close attention to children s safetyKids who don’t know how to swim should wear life jackets. 50 . If you have a pool at home, put a fence(围栏) around it in case children fall in. And if a child is missing, check(检查) the pool first.根据短文内容。
萧红中学2024-2025学年度上学期八学年上·语文学科·阶梯作业命题教师:王宁审题教师:王文明一、积累与运用(32分)南京大屠杀,早已是所有正义力量的集体记忆,唯有日本右翼.分子仍在梦中yì语。
国家公祭日之长鸣警钟zhèn聋发kuì,那些装睡梦游的罪恶灵魂无处dùn形。
80年,cāng海桑田。
1937年12月18日,《纽约时报》在一则报道中写道:“大规.模抢劫、侵.犯妇女、杀害平民……日军将南京变成了一座恐怖之城。
”2017年9月,国际和平城市协会宣布,南京成为“国际和平城市”。
国际和平城市协会项目执行会长弗雷德·阿门特指出,南京这座城市是第二次世界大战中饱受战火摧残.的一个典型,如今成为国际和平城市后,方便全世界的人们更多地了解中华民族追求、热爱和平的悠久历史。
1.(3分)选段中加点字注音正确的一项是()A.右翼.(jì)B.规模.(mú)C.侵.(qīn)犯D.摧残.(cǎn)2.(5分)根据拼音写出相应的汉字A、yì()语B、zhèn()聋发kuì()C、dùn()形D、cāng()海桑田3.(3分)对于选段中画线的病句,修改正确的一项是()A.南京这座城市是第二次世界大战中饱受战火摧残的一个典范,如今成为国际和平城市后,方便全世界的人们更多地了解中华民族追求、热爱和平的悠久历史。
B.南京这座城市是第二次世界大战中饱受战火摧残的一个典型,如今成为国际和平城市后,方便全世界的人们更多地了解中华民族追求、热爱和平的历史悠久。
C.南京这座城市是第二次世界大战中饱受战火摧残的一个典型,如今成为国际和平城市后,方便全世界的人们更多地了解中华民族热爱、追求和平的悠久历史。
D.南京这座城市是第二次世界大战中充分饱受战火摧残的一个典型,如今成为国际和平城市后,方便全世界的人们更多地了解中华民族追求、热爱和平的悠久历史。