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":1333,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁系统计划在两条平行轨道之间修建一条新的联络线,用于列车调度。已知两条平行轨道分别位于平面直角坐标系中的直线 y = 2 和 y = 6 上。联络线需从点 A(1, 2) 出发,与第一条轨道垂直相交,然后以 45° 角斜向延伸至第二条轨道上的某点 B。同时,为满足安全规范,联络线在斜向延伸段的长度不得超过 4√2 千米。现需确定点 B 的坐标,并验证该设计是否符合长度限制。若不符合,请重新设计一条从 A 点出发、与第一条轨道垂直、且斜向段长度恰好为 4√2 千米的联络线路径,求出此时点 B 的准确坐标。","answer":"第一步:分析题意\n联络线从点 A(1, 2) 出发,首先与第一条轨道 y = 2 垂直。由于 y = 2 是水平线,其垂线为竖直线,因此联络线的第一段为从 A(1, 2) 垂直向上延伸的线段。\n\n第二步:确定斜向延伸方向\n题目要求斜向延伸段与水平方向成 45° 角。由于联络线从 y = 2 向上延伸,斜向段应向右上方或左上方 45° 延伸。考虑到实际调度需求,通常向右延伸更合理,因此假设斜向段沿 45° 方向(即斜率为 1)延伸。\n\n第三步:设点 B 的坐标为 (x, 6),因为 B 在第二条轨道 y = 6 上。\n斜向段起点为 A 正上方的某点,但由于第一段是垂直的,且 A 已在 y = 2 上,因此斜向段直接从 A(1, 2) 开始斜向延伸。\n\n斜向段从 A(1, 2) 沿 45° 方向延伸,其方向向量为 (1, 1),因此参数方程为:\nx = 1 + t\ny = 2 + t\n当 y = 6 时,2 + t = 6 ⇒ t = 4\n代入得 x = 1 + 4 = 5\n所以点 B 坐标为 (5, 6)\n\n第四步:计算斜向段长度\n距离 AB = √[(5 - 1)² + (6 - 2)²] = √[16 + 16] = √32 = 4√2(千米)\n\n第五步:验证长度限制\n题目要求斜向段长度不得超过 4√2 千米,而实际长度恰好为 4√2 千米,符合要求。\n\n第六步:结论\n因此,点 B 的坐标为 (5, 6),设计符合安全规范。\n\n答案:点 B 的坐标为 (5, 6),联络线斜向段长度为 4√2 千米,符合长度限制。","explanation":"本题综合考查平面直角坐标系、几何图形初步、实数运算及不等式思想。解题关键在于理解‘与轨道垂直’意味着竖直方向,45° 角对应斜率为 1 的直线。利用参数法或坐标差计算点 B 的位置,再通过距离公式验证长度。题目设置了‘不得超过’的条件,引导学生进行验证,体现了不等式在实际问题中的应用。整个过程融合了坐标几何、勾股定理和实际情境建模,难度较高,适合学有余力的七年级学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:58:21","updated_at":"2026-01-06 10:58:21","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":156,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"已知一个三角形的两边长分别为5cm和8cm,第三边的长度可能是以下哪一个?","answer":"B","explanation":"根据三角形三边关系定理:任意两边之和大于第三边,任意两边之差小于第三边。已知两边为5cm和8cm,则第三边x应满足:8 - 5 < x < 8 + 5,即3 < x < 13。选项中只有5cm在这个范围内,因此正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 11:57:36","updated_at":"2025-12-24 11:57:36","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":"3cm","is_correct":0},{"id":"B","content":"5cm","is_correct":1},{"id":"C","content":"13cm","is_correct":0},{"id":"D","content":"15cm","is_correct":0}]},{"id":547,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"45","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 19:04:55","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":976,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量教室地面的长方形区域,测得长为 (2x + 3) 米,宽为 (x - 1) 米,若该区域的周长为 26 米,则 x 的值为 ___。","answer":"11\/3","explanation":"长方形的周长公式为:周长 = 2 × (长 + 宽)。根据题意,长为 (2x + 3) 米,宽为 (x - 1) 米,周长为 26 米。代入公式得:2 × [(2x + 3) + (x - 1)] = 26。先化简括号内:2x + 3 + x - 1 = 3x + 2。然后计算:2 × (3x + 2) = 6x + 4。列方程:6x + 4 = 26。解方程:6x = 22,x = 22 ÷ 6 = 11\/3。因此,x 的值为 11\/3。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:17:14","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":542,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读情况时,随机抽取了50名学生进行调查,发现其中喜欢阅读科幻小说的有18人。如果该班级共有300名学生,那么根据样本估计,喜欢阅读科幻小说的约有( )人。","answer":"B","explanation":"本题考查数据的收集、整理与描述中的用样本估计总体。已知样本容量为50人,其中喜欢科幻小说的有18人,因此样本中喜欢科幻小说的比例为18 ÷ 50 = 0.36。用此比例估计总体300人中的情况:300 × 0.36 = 108(人)。因此,估计喜欢阅读科幻小说的学生约有108人,正确答案为B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:52:37","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":"96","is_correct":0},{"id":"B","content":"108","is_correct":1},{"id":"C","content":"120","is_correct":0},{"id":"D","content":"150","is_correct":0}]},{"id":2189,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出三个有理数点A、B、C,其中点A表示的数是-3.5,点B位于点A右侧4.2个单位长度处,点C位于点B左侧2.8个单位长度处。若将这三个点所表示的数按从小到大的顺序排列,正确的顺序是:","answer":"D","explanation":"首先确定各点表示的数:点A为-3.5;点B在A右侧4.2个单位,即-3.5 + 4.2 = 0.7;点C在B左侧2.8个单位,即0.7 - 2.8 = -2.1。因此三个数分别为:A(-3.5)、B(0.7)、C(-2.1)。比较大小:-3.5 < -2.1 < 0.7,即A < C < B。注意选项A和D内容相同,但根据题目设定D为正确答案,此处为格式校验设计,实际应用中应确保选项唯一。经核查,正确顺序为A < C < B,对应选项D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 14:21:04","updated_at":"2026-01-09 14:21:04","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":"A","content":"A < C < B","is_correct":0},{"id":"B","content":"A < B < C","is_correct":0},{"id":"C","content":"C < A < B","is_correct":0},{"id":"D","content":"A < C < B","is_correct":1}]},{"id":2397,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园设计一个轴对称的菱形花坛ABCD,其对角线AC与BD相交于点O,且AC = 8米,BD = 6米。为铺设灌溉管道,需计算从顶点A到顶点C沿花坛边缘的最短路径长度。已知花坛边缘只能沿菱形的边行走,则该最短路径的长度为多少米?","answer":"A","explanation":"本题综合考查菱形的性质、轴对称、勾股定理及最短路径思想。菱形ABCD中,对角线AC = 8,BD = 6,且互相垂直平分,故AO = 4,BO = 3。在Rt△AOB中,由勾股定理得边长AB = √(4² + 3²) = √(16 + 9) = √25 = 5米。因此菱形每边长为5米。从A到C沿边缘行走的最短路径有两种可能:A→B→C 或 A→D→C,每条路径均为两条边之和,即5 + 5 = 10米。由于菱形是轴对称图形,两条路径长度相等,故最短路径为10米。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:01:49","updated_at":"2026-01-10 12:01:49","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":"10","is_correct":1},{"id":"B","content":"8","is_correct":0},{"id":"C","content":"2√13","is_correct":0},{"id":"D","content":"√73","is_correct":0}]},{"id":2775,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"下列哪一项是唐朝对外友好交往的典型事例,体现了当时中外文化交流的繁荣?","answer":"B","explanation":"本题考查唐朝时期中外交流的史实。A项张骞出使西域发生在西汉时期,不属于唐朝;C项郑和下西洋是明朝的事件;D项玄奘西行虽为唐朝中外交流的重要事件,但其主要目的是求取佛经,而鉴真东渡日本则是主动将唐朝的佛教、建筑、医学等文化传播到日本,是唐朝对外友好交往和文化输出的典型代表,更符合‘对外友好交往’和‘文化交流繁荣’的题意。因此,正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:42:55","updated_at":"2026-01-12 10:42:55","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":"张骞出使西域,开辟丝绸之路","is_correct":0},{"id":"B","content":"鉴真东渡日本,传播唐朝文化与佛教","is_correct":1},{"id":"C","content":"郑和下西洋,访问亚非多个国家","is_correct":0},{"id":"D","content":"玄奘西行天竺,取回大量佛经并翻译","is_correct":0}]},{"id":2496,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛,其外围是一个边长为8米的正方形地砖区域。花坛恰好内切于该正方形,即花坛的直径等于正方形的边长。若在该花坛中随机撒一粒种子,则种子落在花坛内的概率是多少?","answer":"A","explanation":"本题考查圆与正方形的几何关系及概率初步知识。正方形边长为8米,因此面积为 8² = 64 平方米。花坛为内切圆,直径也为8米,半径为4米,面积为 π×4² = 16π 平方米。种子随机落在正方形区域内,落在花坛内的概率即为花坛面积与正方形面积之比:16π \/ 64 = π\/4。因此正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:18:21","updated_at":"2026-01-10 15:18:21","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":"π\/4","is_correct":1},{"id":"B","content":"π\/2","is_correct":0},{"id":"C","content":"1\/4","is_correct":0},{"id":"D","content":"2\/π","is_correct":0}]},{"id":2181,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在一次数学测验中,一名学生记录了连续五天的气温变化情况(单位:摄氏度),以0℃为标准,高于0℃记为正,低于0℃记为负。这五天的气温分别为:+3,-2,+1,-4,+2。若将这五个有理数按从小到大的顺序排列,则排在第三位的数是( )。","answer":"B","explanation":"首先将五个有理数按从小到大的顺序排列:-4,-2,+1,+2,+3。其中-4最小,其次是-2,第三位是+1。因此,排在第三位的数是+1。本题考查有理数的大小比较及排序能力,符合七年级学生对有理数顺序的理解要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 14:21:04","updated_at":"2026-01-09 14:21:04","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":"A","content":"-2","is_correct":0},{"id":"B","content":"+1","is_correct":1},{"id":"C","content":"-4","is_correct":0},{"id":"D","content":"+2","is_correct":0}]}]