习题练习

[{"id":1922,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间时，收集了以下数据（单位：小时）：2.5，3，1.5，4，2.5，3.5，2。若将这组数据按从小到大的顺序排列，则位于中间位置的数是：","answer":"A","explanation":"首先将数据从小到大排列：1.5，2，2.5，2.5，3，3.5，4。共有7个数据，为奇数个，因此中位数是正中间的那个数，即第4个数。第4个数是2.5，所以答案是A。本题考查数据的整理与描述中的中位数概念，属于简单难度，符合七年级学生认知水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:15:15","updated_at":"2026-01-07 13:15:15","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":"2.5","is_correct":1},{"id":"B","content":"3","is_correct":0},{"id":"C","content":"2.75","is_correct":0},{"id":"D","content":"3.5","is_correct":0}]},{"id":1957,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生参加学校组织的‘健康生活’主题调查，记录了连续7天每天步行的步数（单位：千步），数据如下：6.2, 5.8, 7.1, 6.5, 6.9, 5.5, 7.3。若该学生希望估算自己一个月（按30天计算）的总步行步数，并假设每日步数服从这组数据的平均水平，则估算结果最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中利用样本平均数估计总体的应用。首先计算7天步行步数的平均数：(6.2 + 5.8 + 7.1 + 6.5 + 6.9 + 5.5 + 7.3) ÷ 7 = 45.3 ÷ 7 ≈ 6.471（千步\/天）。然后估算30天的总步数：6.471 × 30 ≈ 194.13（千步），最接近195千步。因此选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:47:02","updated_at":"2026-01-07 14:47:02","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":"180千步","is_correct":0},{"id":"B","content":"195千步","is_correct":1},{"id":"C","content":"200千步","is_correct":0},{"id":"D","content":"210千步","is_correct":0}]},{"id":2014,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园艺术节中，某学生设计了一个轴对称图案，图案由两个全等的直角三角形拼接而成，形成一个等腰三角形。已知其中一个直角三角形的两条直角边分别为5 cm和12 cm，则这个等腰三角形的周长是多少？","answer":"C","explanation":"首先，根据勾股定理计算直角三角形的斜边：斜边 = √(5² + 12²) = √(25 + 144) = √169 = 13 cm。由于两个全等的直角三角形沿斜边拼接，形成的等腰三角形的两条腰分别为5 cm和12 cm中较长的一条边（即12 cm）作为底边？不，实际上，当两个全等直角三角形沿斜边拼接时，形成的是以两条直角边为腰的等腰三角形？不对。正确理解是：若沿直角边拼接，则可能形成等腰三角形。但题意是‘拼接成一个等腰三角形’，最合理的方式是将两个直角三角形沿长度为12 cm的直角边重合，这样两个5 cm的直角边成为等腰三角形的两腰，底边为13 cm + 13 cm？不成立。正确拼接方式应为：将两个直角三角形沿斜边以外的边拼接，使非直角边对应相等。实际上，标准做法是将两个全等直角三角形沿直角边12 cm拼接，使两个5 cm边成为等腰三角形的两腰，此时底边为两个斜边之和？不，这样不形成三角形。正确方式：将两个直角三角形沿长度为5 cm的直角边拼接，使两个12 cm边成为等腰三角形的两腰，底边为两个斜边？也不对。重新分析：要形成等腰三角形，应将两个全等直角三角形沿一条直角边拼接，使得另外两条相等的边成为等腰三角形的两腰。若沿5 cm边拼接，则两腰为12 cm，底边为两个斜边？不，底边应为两个直角顶点的连线，即两个直角三角形的另一条直角边（12 cm）平行，底边为斜边？混乱。正确理解：将两个全等直角三角形沿斜边以外的边拼接，使形成的三角形有两条边相等。最合理的是：将两个直角三角形沿12 cm边拼接，使两个5 cm边在同一直线上，形成底边为10 cm，两腰为13 cm的等腰三角形？但这样不是由两个直角三角形直接拼接成一个大三角形。正确拼接方式：将两个直角三角形沿直角边12 cm重合，使两个5 cm边成为等腰三角形的两腰，此时两个直角顶点重合，两个斜边成为等腰三角形的两条边？不成立。实际上，正确方式是：将两个全等直角三角形沿直角边5 cm拼接，使两个12 cm边在同一直线上，形成底边为24 cm，两腰为13 cm的等腰三角形？也不对。重新思考：若两个全等直角三角形沿一条直角边拼接，且该边不是斜边，则形成的大三角形有两条边为原斜边，一条边为两倍直角边。但要使大三角形为等腰三角形，必须使两条边相等。因此，只有当两个直角三角形沿直角边拼接后，两条斜边作为等腰三角形的两腰，底边为两倍","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:29:49","updated_at":"2026-01-09 10:29:49","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":"30 cm","is_correct":0},{"id":"B","content":"34 cm","is_correct":0},{"id":"C","content":"36 cm","is_correct":1},{"id":"D","content":"40 cm","is_correct":0}]},{"id":2507,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，一个圆锥的底面半径为3 cm，高为4 cm。若将该圆锥沿高旋转180°，则旋转后的几何体与原圆锥组合成一个新的立体图形。求这个新立体图形的主视图（从正前方正视）的形状。","answer":"A","explanation":"原圆锥底面半径为3 cm，高为4 cm。将其沿高旋转180°后，相当于将另一个相同的圆锥倒置拼接在原圆锥上方，两个圆锥的底面重合，顶点朝相反方向。组合后的立体图形是一个上下对称的“双圆锥”，总高度为4 + 4 = 8 cm，底面直径仍为6 cm。从正前方正视（主视图）时，看到的轮廓是两个等腰三角形拼接而成的等腰三角形，底边为原底面直径6 cm，总高为8 cm。因此主视图是一个底边长为6 cm、高为8 cm的等腰三角形。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:30:19","updated_at":"2026-01-10 15:30:19","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":"一个底边长为6 cm，高为8 cm的等腰三角形","is_correct":1},{"id":"B","content":"一个底边长为6 cm，高为4 cm的等腰三角形","is_correct":0},{"id":"C","content":"一个直径为6 cm的圆","is_correct":0},{"id":"D","content":"一个底边长为6 cm，高为4 cm的矩形","is_correct":0}]},{"id":1773,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市计划在一条东西走向的主干道旁建设一个矩形公园，公园的四个顶点分别位于平面直角坐标系中的A(2, 3)、B(x, 3)、C(x, y)、D(2, y)，其中x > 2，y > 3。已知公园的周长为28个单位长度，面积为48平方单位。现需在公园内铺设一条从点A到点C的对角线路径，并在路径两侧各安装一排路灯，每排路灯间距为1个单位长度（包括起点和终点）。若每盏路灯的安装成本为50元，求铺设该路径所需安装路灯的总成本。","answer":"1. 由题意，矩形公园的四个顶点为A(2,3)、B(x,3)、C(x,y)、D(2,y)，其中x > 2，y > 3。\n2. 矩形的长为|x - 2| = x - 2，宽为|y - 3| = y - 3。\n3. 周长公式：2[(x - 2) + (y - 3)] = 28\n 化简得：(x - 2) + (y - 3) = 14 → x + y = 19 ①\n4. 面积公式：(x - 2)(y - 3) = 48 ②\n5. 设a = x - 2，b = y - 3，则a > 0，b > 0，且：\n a + b = 14\n ab = 48\n6. 解这个方程组：由a + b = 14得b = 14 - a，代入ab = 48：\n a(14 - a) = 48 → 14a - a² = 48 → a² - 14a + 48 = 0\n 解得：a = [14 ± √(196 - 192)] \/ 2 = [14 ± √4] \/ 2 = [14 ± 2]\/2\n 所以a = 8 或 a = 6\n 对应b = 6 或 b = 8\n7. 因此有两种可能：\n (a,b) = (8,6) → x = 10, y = 9\n 或 (a,b) = (6,8) → x = 8, y = 11\n8. 计算对角线AC的长度：\n 情况一：A(2,3), C(10,9) → AC = √[(10-2)² + (9-3)²] = √(64 + 36) = √100 = 10\n 情况二：A(2,3), C(8,11) → AC = √[(8-2)² + (11-3)²] = √(36 + 64) = √100 = 10\n 两种情况下AC长度均为10单位。\n9. 路径AC上每1单位长度安装一盏路灯，包括起点和终点，因此路灯数量为：10 ÷ 1 + 1 = 11盏（每排）\n10. 两侧各一排，共2排，总灯数：11 × 2 = 22盏\n11. 每盏成本50元，总成本：22 × 50 = 1100元\n答案：1100元","explanation":"本题综合考查平面直角坐标系中点的坐标、矩形周长与面积、二元一次方程组的建立与求解、勾股定理求距离以及实际应用中的计数问题。关键在于通过设辅助变量简化方程，并利用对称性发现两种情况下的对角线长度相同，从而避免重复计算。最后注意路灯安装包含端点，需用‘距离÷间距+1’计算数量。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 15:13:26","updated_at":"2026-01-06 15:13:26","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":2131,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 2(x - 3) = 4 时，第一步将方程两边同时除以2，得到 x - 3 = 2。接下来他应该进行的正确步骤是：","answer":"B","explanation":"方程 x - 3 = 2 中，为了求出 x，需要将 -3 消去。根据等式性质，应在等式两边同时加上3，得到 x = 5。这是七年级一元一次方程求解中的基本步骤，符合课程标准要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 12:56:39","updated_at":"2026-01-09 12:56:39","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":"两边同时减去3，得到 x = -1","is_correct":0},{"id":"B","content":"两边同时加上3，得到 x = 5","is_correct":1},{"id":"C","content":"两边同时乘以3，得到 x = 6","is_correct":0},{"id":"D","content":"两边同时除以3，得到 x = 2\/3","is_correct":0}]},{"id":2217,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在记录一周内每天的温度变化时，发现某天的气温比前一天上升了5℃，记作+5℃；第二天又下降了3℃，应记作____℃。","answer":"-3","explanation":"根据正负数表示相反意义的量的知识点，气温上升用正数表示，下降则用负数表示。下降了3℃，应记作-3℃，符合七年级学生对正负数在实际生活中应用的学习要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","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":1060,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了废旧纸张和塑料瓶共12件，其中废旧纸张比塑料瓶多4件。设塑料瓶的数量为x件，则根据题意可列出一元一次方程：_x + (x + 4) = 12_，解得x = _4_，因此塑料瓶有_4_件，废旧纸张有_8_件。","answer":"x + (x + 4) = 12；4；4；8","explanation":"设塑料瓶数量为x件，则废旧纸张数量为x + 4件。根据总数量为12件，可列方程x + (x + 4) = 12。解这个方程：2x + 4 = 12 → 2x = 8 → x = 4。因此塑料瓶有4件，废旧纸张有4 + 4 = 8件。本题考查一元一次方程的建立与求解，符合七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:51:55","updated_at":"2026-01-06 08:51:55","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":2456,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在△ABC中，∠C=90°，AC=5，BC=12。若点D在斜边AB上，且CD⊥AB，则CD的长度为______。","answer":"60\/13","explanation":"先由勾股定理得AB=13，再利用等面积法：S△ABC=½×5×12=½×13×CD，解得CD=60\/13。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:00:56","updated_at":"2026-01-10 14:00:56","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":413,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学每天使用手机的时间（单位：分钟），并将数据整理成如下频数分布表：\n\n| 使用时间区间 | 频数（人数） |\n|---------------|--------------|\n| 0–30 | 8 |\n| 31–60 | 12 |\n| 61–90 | 15 |\n| 91–120 | 10 |\n| 121以上 | 5 |\n\n请问这组数据的中位数最可能落在哪个区间？","answer":"C","explanation":"首先计算总人数：8 + 12 + 15 + 10 + 5 = 50人。中位数是第25和第26个数据的平均值。累计频数：0–30分钟有8人，31–60分钟累计为8+12=20人，61–90分钟累计为20+15=35人。由于第25和第26个数据都落在累计频数超过25的区间，即61–90分钟区间内，因此中位数最可能落在61–90分钟。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:30:07","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":"0–30分钟","is_correct":0},{"id":"B","content":"31–60分钟","is_correct":0},{"id":"C","content":"61–90分钟","is_correct":1},{"id":"D","content":"91–120分钟","is_correct":0}]}]