初中
数学
中等
来源: 教材例题
知识点: 初中数学
答案预览
点击下方'查看答案'按钮查看详细解析并跳转到题目详情页
直接前往详情页
练习完成!
恭喜您完成了本次练习,继续加油提升自己的知识水平!
学习建议
您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":443,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中,某学生记录了连续5天每天收集的废纸重量(单位:千克),数据如下:2.5,3.0,2.8,3.2,2.7。为了分析数据变化趋势,该学生计算了这组数据的平均数,并发现如果将每天的重量都增加0.3千克,则新的平均数比原来多多少?","answer":"C","explanation":"首先计算原始数据的平均数:(2.5 + 3.0 + 2.8 + 3.2 + 2.7) ÷ 5 = 14.2 ÷ 5 = 2.84(千克)。如果每天的数据都增加0.3千克,则新的数据为:2.8,3.3,3.1,3.5,3.0。新的平均数为:(2.8 + 3.3 + 3.1 + 3.5 + 3.0) ÷ 5 = 15.7 ÷ 5 = 3.14(千克)。新旧平均数之差为:3.14 - 2.84 = 0.3(千克)。也可以直接理解:当一组数据中每个数都增加同一个值时,其平均数也增加相同的值。因此,平均数增加了0.3千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:42:45","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"0.1千克","is_correct":0},{"id":"B","content":"0.2千克","is_correct":0},{"id":"C","content":"0.3千克","is_correct":1},{"id":"D","content":"0.5千克","is_correct":0}]},{"id":2422,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个菱形花坛,设计师提供了以下四个方案。已知菱形的两条对角线长度分别为 d₁ 和 d₂,且满足 d₁ = 2√3 米,d₂ = 6 米。为了确保花坛结构稳定,施工方需要验证该菱形是否可以被分割成两个全等的等边三角形。以下说法正确的是:","answer":"C","explanation":"首先,根据菱形性质,对角线互相垂直且平分。已知 d₁ = 2√3 米,d₂ = 6 米,则每条对角线的一半分别为 √3 米和 3 米。利用勾股定理可求出菱形边长:边长 = √[(√3)² + 3²] = √(3 + 9) = √12 = 2√3 米。若该菱形能分割成两个等边三角形,则每个三角形的三边都应相等,即边长应等于 2√3 米,且每个内角为60°。但通过计算一个内角:tan(θ\/2) = (√3)\/3 = 1\/√3,得 θ\/2 = 30°,所以 θ = 60°,看似符合。然而,菱形被一条对角线分成的两个三角形是全等等腰三角形,只有当边长等于对角线一半构成的直角三角形斜边,且所有边相等时才为等边。此处虽然一个角为60°,但其余弦定理验证:若为等边三角形,三边均为 2√3,但由对角线分割出的三角形两边为 2√3,底边为 d₁ = 2√3,看似可能,但实际另一条对角线为6米,意味着另一方向的跨度不满足等边条件。更关键的是,若两个等边三角形组成菱形,则对角线比应为 √3 : 1,而本题中 d₁:d₂ = 2√3 : 6 = √3 : 3 ≠ √3 : 1,矛盾。因此,尽管部分角度为60°,整体无法构成两个全等等边三角形。正确判断应基于边长与结构一致性,故选C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:35:01","updated_at":"2026-01-10 12:35:01","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"可以分割成两个全等的等边三角形,因为对角线互相垂直且平分","is_correct":0},{"id":"B","content":"可以分割成两个全等的等边三角形,因为每条边长都等于 √3 米","is_correct":0},{"id":"C","content":"不能分割成两个全等的等边三角形,因为计算出的边长与等边三角形要求不符","is_correct":1},{"id":"D","content":"不能分割成两个全等的等边三角形,因为菱形的内角不是60°","is_correct":0}]},{"id":19,"subject":"地理","grade":"初二","stage":"初中","type":"填空题","content":"我国最大的河流是______,最长的内流河是______。","answer":"长江, 塔里木河","explanation":"长江是我国最长的河流,塔里木河是我国最长的内流河。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":2,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1838,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一个直角三角形的两条直角边,分别为√12 cm和√27 cm。若该三角形的斜边长度为c cm,则c²的值是多少?","answer":"C","explanation":"根据勾股定理,直角三角形中斜边的平方等于两条直角边的平方和。已知两条直角边分别为√12 cm和√27 cm,因此:c² = (√12)² + (√27)² = 12 + 27 = 39。选项C正确。本题考查了二次根式的平方运算与勾股定理的综合应用,难度适中,符合八年级学生的认知水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:50:23","updated_at":"2026-01-06 16:50:23","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"13","is_correct":0},{"id":"B","content":"25","is_correct":0},{"id":"C","content":"39","is_correct":1},{"id":"D","content":"51","is_correct":0}]},{"id":609,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"14","answer":"待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:34:08","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1084,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在整理班级同学最喜欢的运动项目调查数据时,共收集了60份有效问卷。其中喜欢篮球的人数占总人数的$\\frac{1}{3}$,喜欢足球的人数是喜欢篮球人数的$\\frac{1}{2}$,其余同学喜欢羽毛球。那么喜欢羽毛球的同学有___人。","answer":"30","explanation":"总人数为60人。喜欢篮球的人数为60 × $\\frac{1}{3}$ = 20人。喜欢足球的人数是篮球人数的$\\frac{1}{2}$,即20 × $\\frac{1}{2}$ = 10人。因此,喜欢羽毛球的人数为60 - 20 - 10 = 30人。本题考查了数据的收集与整理,以及有理数的乘法与加减运算,符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:54:27","updated_at":"2026-01-06 08:54:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":187,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"下列各数中,最小的数是( )。","answer":"A","explanation":"本题考查有理数的大小比较。在数轴上,负数位于0的左侧,正数位于0的右侧,因此负数小于0,0小于正数。给出的四个数中,-3是唯一的负数,其余都是非负数(0和正数),所以-3是最小的数。也可以通过比较数值大小直接判断:-3 < 0 < 1 < 2。因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:01:29","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"-3","is_correct":1},{"id":"B","content":"0","is_correct":0},{"id":"C","content":"1","is_correct":0},{"id":"D","content":"2","is_correct":0}]},{"id":814,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在调查班级同学最喜欢的课外活动时,收集了以下数据:阅读、运动、绘画、音乐。他将这些数据整理成扇形统计图,其中表示‘运动’的扇形圆心角为108度。如果全班共有40名学生,那么喜欢‘运动’的学生人数是___人。","answer":"12","explanation":"扇形统计图中,每个部分的圆心角占整个圆(360度)的比例等于该部分数据占总数据的比例。‘运动’对应的圆心角是108度,因此喜欢运动的学生所占比例为108 ÷ 360 = 0.3。全班共有40名学生,所以喜欢运动的学生人数为40 × 0.3 = 12人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:30:51","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":502,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛,参赛学生共收集了120张答题卡。老师将这些答题卡按正确率分为A、B、C三个等级,其中A级占总数的一半,B级比C级多20张。请问C级答题卡有多少张?","answer":"A","explanation":"设C级答题卡有x张,则B级有(x + 20)张。已知A级占总数的一半,总数为120张,所以A级有120 ÷ 2 = 60张。根据总数量关系列方程:60 + (x + 20) + x = 120。化简得:2x + 80 = 120,解得2x = 40,x = 20。因此C级答题卡有20张,正确答案是A。本题考查一元一次方程的实际应用,结合数据的整理与描述,符合七年级数学知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:10:21","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"20张","is_correct":1},{"id":"B","content":"30张","is_correct":0},{"id":"C","content":"40张","is_correct":0},{"id":"D","content":"50张","is_correct":0}]},{"id":2392,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次数学实践活动中,某学生测量了一块四边形土地的四个顶点坐标分别为 A(0, 0)、B(4, 0)、C(5, 2) 和 D(1, 2)。他通过计算发现该四边形的一组对边平行且相等,另一组对边也平行且相等。若他想进一步验证这个四边形是否为平行四边形,并计算其面积,以下哪种方法最合理?","answer":"B","explanation":"本题考查平行四边形的判定与面积计算,融合了坐标几何、一次函数斜率、向量思想和数据分析能力。选项 B 是最科学合理的方法:首先,通过一次函数斜率判断 AB 与 CD 是否平行(k_AB = (0-0)\/(4-0) = 0,k_CD = (2-2)\/(1-5) = 0,故平行),同理 AD 与 BC 的斜率均为 2\/1 = 2,说明两组对边分别平行,符合平行四边形定义;其次,可进一步用距离公式验证对边长度相等,增强结论可靠性;最后,面积可通过向量 AB = (4,0) 与 AD = (1,2) 的叉积 |4×2 - 0×1| = 8 得到,或使用分割法、坐标法(如鞋带公式)计算,方法严谨且符合八年级知识范围。选项 A 虽部分正确,但未利用坐标优势,效率较低;选项 C 错误,因角度并非直角;选项 D 混淆了轴对称与平行四边形的关系,平行四边形不一定是轴对称图形。因此,B 为最佳方法。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:52:06","updated_at":"2026-01-10 11:52:06","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"利用勾股定理分别计算四条边的长度,若对边相等,则该四边形是平行四边形,再用底乘高计算面积。","is_correct":0},{"id":"B","content":"利用一次函数的斜率判断 AB 与 CD、AD 与 BC 是否分别平行,再通过向量法或距离公式验证对边相等,最后用向量叉积或分割法求面积。","is_correct":1},{"id":"C","content":"直接假设该四边形是矩形,用长乘宽计算面积,因为所有角看起来都是直角。","is_correct":0},{"id":"D","content":"将该四边形沿 y 轴对折,若两部分完全重合,则说明是轴对称图形,因此是平行四边形,面积可用对称性估算。","is_correct":0}]}]