陕西省咸阳市实验中学2021-2022学年八年级上学期第一次月考数学试题

这是一份陕西省咸阳市实验中学2021-2022学年八年级上学期第一次月考数学试题，共9页。试卷主要包含了本试卷分为第一部分，领到试卷和答题卡后，请用0，有理数-8的立方根是，比较大小等内容，欢迎下载使用。
试卷类型：A咸阳市实验中学2021~2022学年度第一学期第一次月考八年级数学试题注意事项：1.本试卷分为第一部分（选择题）和第二部分（非选择题）。全卷共4页，总分120分。考试时间120分钟。2.领到试卷和答题卡后，请用0.5毫米黑色墨水签字笔，分别在试卷和答题卡上填写姓名和准考证号，同时用2B铅笔在答题卡上填涂对应的试卷类型信息点（A或B）。3.请在答题卡上各题的指定区域内作答，否则作答无效。4.作图时，先用铅笔作图，再用规定签字笔描黑。5.考试结束，本试卷和答题卡一并交回。第一部分（选择题  共24分）一、选择题（共8小题，每小题3分，计24分.每小题只有一个选项是符合题意的）1.下列各数是无理数的是（    ）A. B. C. D.2.下列三个数中，能组成一组勾股数的是（    ）A.，， B.，， C.，， D.3，4，53.若是最简二次根式，则a的值可能是（    ）A.-3 B. C.5 D.84.下列说法正确的是（    ）A.64的算术平方根是±8 B.49的平方根是-7C.-36的平方根是6  D.25的算术平方根是55.要使式子有意义，x的值可以是A.2021 B.2022 C.2023 D.20246.如图，数轴上的点可以近似的表示的值的是（    ）A.点A B.点B C.点C D.点D7.在《算法统宗》中有一道“荡秋千”的问题：“平地秋千未起，踏板一尺离地.送行二步与人齐，五尺人高曾记.仕女佳人争蹴，终朝笑语欢嬉.良工高士素好奇，算出索长有几.”此问题可理解为：如图，有一架秋千，当它静止时，踏板离地距离长度为1尺.将它往前水平推送10尺（即尺）时，秋千的踏板离地距离就和身高5尺的人一样高.若运动过程中秋千的绳索始终拉得很直，则绳索长为（    ）A.15尺 B.14.5尺 C.14尺 D.13.5尺8.勾股定理是人类最伟大的科学发现之一，在我国古代《周髀算经》中早有记载.如图①，以直角三角形的各边为边分别向外作正方形，再把较小的两张正方形纸片按图②的方式放置在最大正方形内.若图②中阴影部分图形的面积为3，则较小两个正方形重叠部分图形的面积为（    ）A.6 B.5 C.3 D.2第二部分（非选择题  共96分）二、填空题（共5小题，每小题3分，计15分）9.有理数-8的立方根是.10.比较大小：____________.（填“>”、“<”或“=”）11.如图，在四边形中，点E为的中点，于点E，，，，，则四边形的面积为_____________.12.如图，圆柱形容器外壁距离下底面3的A处有一只蚂蚁，它想吃到正对面外壁距离上底面3的B处的米粒，若圆柱的高为12，底面周长为24.则蚂蚁爬行的最短距离为.13.图中的螺旋形由一系列直角三角形组成，则以第n个三角形的斜边长为边长的正方形的面积为___________.（用含n的代数式表示）三、解答题（共13小题，计81分.解答应写出过程）14.（5分）计算：.15.（5分）在如图所示的数轴上作出表示的点.（不写作法，保留作图痕迹）16.（5分）在一个长、宽、高分别为8，4，2的长方体容器中装满水，将容器中的水全部倒入一个正方体容器中，恰好倒满（两容器的厚度忽略不计），求此正方体容器的棱长.17.（5分）已知，，求的值.18.（5分）如图，每个小正方形的边长都为1，A，B，C是小正方形的顶点.（1）求和的长；（2）求的度数.19.（5分）已知某三角形的面积等于长、宽分别为、的长方形的面积，若该三角形的一条边长为，求这条边上的高.20.（5分）每年的十月一日是一年一度的国庆节，是我们伟大祖国母亲的生日，大街小巷挂满了彩旗，成为一道靓丽的风景线，格外引人瞩目，催人奋进.长方形彩旗完全展平时的尺寸图如图1所示（单位：），其中长方形是由双层白布缝制的穿旗杆用的旗裤，长方形为绸缎旗面，将穿好彩旗的旗杆垂直插在地面上.从旗顶M到地面的高度为220，在无风的天气里，彩旗自然下垂时的长度与相等（如图2所示）.求彩旗自然下垂时最低处离地面的高度h.21.（6分）已知，.（1）若x的算术平方根为3，求a的值；（2）如果一个正数的平方根分别为x，y，求这个正数.22.（7分）如图，大长方形内有两个相邻的正方形，面积分别为9和6.（1）小正方形边长的值在哪两个连续的整数之间？（2）求图中阴影部分的面积.23.（7分）今年9月，第十四届全国运动会在我市隆重举行.这是我市人民期待已久的一次盛会，也是宣传西安发展、推介西安之美、展示西安形象的绝好机遇.为美化城市，加大绿化力度，某公园有一块如图所示的四边形空地，现计划在空地上种植花草，经测量，米，米，米，米.求四边形空地的面积.24.（8分）【阅读材料】把分母中的根号化去，使分母转化为有理数的过程，叫做分母有理化.通常把分子、分母乘以同一个不等于0的式子，以达到化去分母中根号的目的.例如：化简.解：.【理解应用】（1）化简：①；②.（2）计算：.25.（8分）我们新定义一种三角形：两边平方和等于第三边平方的4倍的三角形叫做常态三角形.例如：某三角形三边长分别是5，6和8，因为，所以这个三角形是常态三角形.（1）若三边长分别是2，和4，试判断此三角形是否为常态三角形；（2）如图，在中，点D在边上，连接，，，，若是常态三角形，求的长.26.（10分）如图，已知在中，，，，D是上的一点，，点P从B点出发沿射线方向以每秒2个单位的速度向右运动.设点P的运动时间为t，连接.（1）当秒时，求的长度（结果保留根号）；（2）当为等腰三角形时，求t的值；（3）过点D作于点E，连接.在点P的运动过程中，当t为何值时，平分？  试卷类型：A咸阳市实验中学2021~2022学年度第一学期第一次月考八年级数学试题参考答案及评分标准一、选择题（共8小题，每小题3分，计24分.每小题只有一个选项是符合题意的）1.A  2.D  3.C  4.D  5.A  6.B  7.B  8.C二、填空题（共5小题，每小题3分，计15分）9.-2  10.>  11.  12.  13.三、解答题（共13小题，计81分.解答应写出过程）14.解：原式········································································（3分）.·················································································（5分）15.解：如图所示，点A是表示的点.·················································································（5分）16.解：由于装满水的长方体容器中的水，全部倒入正方体容器中，恰好倒满，所以它们的体积相等，而长方体容器的体积，································································（2分）所以正方体容器的体积为64，···························································（3分）所以此正方体容器的棱长为.····························································（5分）17.解：因为，，所以，，··········································································（4分）所以.·············································································（5分）18.解：（1）根据勾股定理可以得到：，，所以，.············································································（2分）（2）连接，如图.因为，，而，即，所以是等腰直角三角形，且，···························································（4分）所以.·············································································（5分）19.解：长方形的面积，································································（1分）所以该三角形这条边上的高····························································（3分）.·················································································（5分）20.解：在中，，，所以，············································································（3分）所以彩旗自然下垂时最低处离地面的高度.··················································（5分）21.解：（1）因为x的算术平方根为3，所以，即，··············································································（1分）所以.·············································································（2分）（2）根据题意得：，即：，············································································（3分）所以，············································································（4分）所以，············································································（5分）所以这个正数为.·····································································（6分）22.解：（1）因为小正方形的面积为6，所以小正方形的边长为，······························································（2分）因为，所以小正方形边长的值在2和3之间.·······················································（4分）（2）因为阴影部分的面积的和为一个长为，宽为的长方形面积，所以阴影部分的面积.·································································（7分）23.解：如图，连接.···································································（1分）在中，，米，米，所以（米）.········································································（3分）在中，米，米，米，所以.所以是直角三角形，且.································································（5分）所以（平方米）.所以四边形空地的面积为234平方米.······················································（7分）24.解：（1）①原式.··································································（2分）②原式.············································································（4分）（2）原式··········································································（6分）.·················································································（8分）25.解：（1）因为，所以此三角形是常态三角形.····························································（2分）（2）在中，，所以，，而，即，所以，故.所以是直角三角形.···································································（3分）已知是常态三角形，分和两种情况进行讨论：①当时，由，可得时，解得：，··········································································（4分）则，在中，.············································································（6分）②当时，由，可得，解得：，则，在中，，，不符合题意，舍去.故的长为.··········································································（8分）26.解：（1）根据题意，得，，，在中，根据勾股定理，得.······························································（2分）（2）在中，，，根据勾股定理，得.若，则，解得；·····································································（4分）若，则，即，解得；··································································（5分）若，则，解得.答：当为等腰三角形时，t的值为或16或5.··················································（6分）（3）①当点P在线段上时，如图1所示：因为，所以，因为平分，所以，又因为，所以，所以，，因为，所以，············································································（7分）所以，在中，由勾股定理得：，解得：；··········································································（8分）②当点P在线段的延长线上时，如图2所示：同①得：，所以，，因为，所以，所以，············································································（9分）在中，由勾股定理得：，解得：.综上所述，在点P的运动过程中，当t的值为5或11时，平分.·····································（10分）