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
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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。
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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.3,4,5
3.若是最简二次根式,则a的值可能是( )
A.-3 B. C.5 D.8
4.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-7
C.-36的平方根是6 D.25的算术平方根是5
5.要使式子有意义,x的值可以是
A.2021 B.2022 C.2023 D.2024
6.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D
7.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)
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