习题练习

[{"id":890,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了12.5千克废纸，另一名同学收集的废纸比这名学生多3.7千克，两人一共收集了___千克废纸。","answer":"28.7","explanation":"首先，第二名同学收集的废纸重量为12.5 + 3.7 = 16.2千克。然后将两人收集的废纸重量相加：12.5 + 16.2 = 28.7千克。因此，两人一共收集了28.7千克废纸。本题考查的是有理数的加法运算，属于简单难度，符合七年级有理数知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:08:03","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":2429,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在一张方格纸上画了一个四边形ABCD，其顶点坐标分别为A(0, 0)、B(4, 0)、C(5, 2)、D(1, 2)。该学生声称这个四边形是平行四边形，并尝试通过计算对边长度和斜率来验证。若只根据坐标信息判断，以下哪个结论最能支持该四边形是平行四边形？","answer":"D","explanation":"判断一个四边形是否为平行四边形，有多种方法。在坐标系中，最直接且可靠的方法之一是验证对角线是否互相平分，即两条对角线的中点是否重合。计算对角线AC的中点：A(0,0)、C(5,2)，中点为((0+5)\/2, (0+2)\/2) = (2.5, 1)；对角线BD的中点：B(4,0)、D(1,2)，中点为((4+1)\/2, (0+2)\/2) = (2.5, 1)。两者中点相同，说明对角线互相平分，因此四边形ABCD是平行四边形。选项D正确。其他选项虽部分正确（如A、B、C中提到的边长或斜率关系），但单独使用可能存在反例（如等腰梯形满足某些边等长或斜率相同但不是平行四边形），而中点重合是平行四边形的充要条件之一，更具说服力。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:52:54","updated_at":"2026-01-10 12:52:54","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":"AB与CD的长度相等，且AD与BC的斜率相同","is_correct":0},{"id":"B","content":"AB与CD的斜率相同，且AD与BC的长度相等","is_correct":0},{"id":"C","content":"AB与CD的斜率相同，且AD与BC的斜率也相同","is_correct":0},{"id":"D","content":"对角线AC和BD的中点坐标相同","is_correct":1}]},{"id":273,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级调查中，某学生记录了10名同学的身高（单位：厘米）：150，152，155，155，158，160，162，165，168，170。这组数据的中位数是多少？","answer":"C","explanation":"中位数是将一组数据从小到大排列后，处于中间位置的数。本题共有10个数据，是偶数个，因此中位数是第5个和第6个数据的平均数。数据已按顺序排列：150，152，155，155，158，160，162，165，168，170。第5个数是158，第6个数是160。中位数为（158 + 160）÷ 2 = 318 ÷ 2 = 159。因此正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:30:20","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":"155","is_correct":0},{"id":"B","content":"158","is_correct":0},{"id":"C","content":"159","is_correct":1},{"id":"D","content":"160","is_correct":0}]},{"id":1732,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参与校园绿化规划活动，计划在校园内的一块矩形空地上种植花草。已知该矩形空地的周长为40米，且长比宽的3倍少2米。为了合理布置灌溉系统，需要在矩形空地的对角线交点处安装一个喷头，喷头覆盖范围为以交点为圆心、半径为√13米的圆形区域。现需判断该喷头是否能完全覆盖整个矩形空地。若不能完全覆盖，求喷头未覆盖区域的面积（精确到0.01平方米）。请通过建立数学模型并求解，回答上述问题。","answer":"设矩形空地的宽为x米，则长为(3x - 2)米。\n根据矩形周长公式：周长 = 2 × (长 + 宽)\n代入已知条件：\n2 × [x + (3x - 2)] = 40\n2 × (4x - 2) = 40\n8x - 4 = 40\n8x = 44\nx = 5.5\n因此，宽为5.5米，长为3 × 5.5 - 2 = 16.5 - 2 = 14.5米。\n\n矩形对角线长度由勾股定理得：\n对角线 = √(长² + 宽²) = √(14.5² + 5.5²) = √(210.25 + 30.25) = √240.5 ≈ 15.506米\n对角线的一半（即从中心到任一顶点的距离）为：15.506 ÷ 2 ≈ 7.753米\n\n喷头覆盖半径为√13 ≈ 3.606米\n由于7.753 > 3.606，说明喷头无法覆盖到矩形的四个顶点，因此不能完全覆盖整个矩形。\n\n喷头覆盖面积为：π × (√13)² = 13π ≈ 40.84平方米\n矩形总面积为：14.5 × 5.5 = 79.75平方米\n未覆盖区域面积为：79.75 - 40.84 = 38.91平方米\n\n答：喷头不能完全覆盖整个矩形空地，未覆盖区域的面积约为38.91平方米。","explanation":"本题综合考查了一元一次方程、实数运算、平面直角坐标系中的距离概念（隐含于勾股定理）、几何图形初步（矩形性质与圆覆盖）以及数据的计算与比较。解题关键在于：首先通过设未知数列方程求出矩形的长和宽；然后利用勾股定理计算对角线长度，进而判断喷头覆盖范围是否足够；最后通过面积差计算未覆盖部分。题目情境新颖，融合了实际生活问题，要求学生具备较强的建模能力和多知识点综合运用能力，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:18:29","updated_at":"2026-01-06 14:18:29","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":477,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"45分","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 17:57:43","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":2440,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一个等腰三角形ABC时，测得底边BC的长度为8 cm，腰AB与AC的长度均为5 cm。他尝试通过作底边BC上的高AD来分割该三角形，并利用勾股定理计算高AD的长度。随后，他将原三角形沿高AD对折，形成一个轴对称图形。若他将折叠后的图形放置在平面直角坐标系中，使点D与原点重合，点B位于x轴正半轴上，则点A的坐标可能为下列哪一项？","answer":"A","explanation":"首先，在等腰三角形ABC中，AB = AC = 5 cm，底边BC = 8 cm。作底边BC上的高AD，由等腰三角形性质可知，D为BC中点，因此BD = DC = 4 cm。在直角三角形ABD中，应用勾股定理：AD² = AB² - BD² = 5² - 4² = 25 - 16 = 9，故AD = 3 cm。由于三角形沿AD对折后具有轴对称性，且题目设定D与原点重合，B在x轴正半轴上，则B坐标为(4, 0)，C为(-4, 0)。高AD垂直于BC并位于y轴上，因此点A应在y轴正方向上，距离D为3个单位，即A点坐标为(0, 3)。选项A正确。选项C和D中的√39不符合计算结果，选项B的横坐标不为0，违背了对称轴为y轴的设定。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:18:26","updated_at":"2026-01-10 13:18:26","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":"A","content":"(0, 3)","is_correct":1},{"id":"B","content":"(4, 3)","is_correct":0},{"id":"C","content":"(0, √39)","is_correct":0},{"id":"D","content":"(4, √39)","is_correct":0}]},{"id":939,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保知识竞赛中，某班级学生共收集了120条有效答题记录。经统计，其中答对一题得3分，答错或不答扣1分。若该班级总得分为280分，则他们答对了____道题。","answer":"100","explanation":"设答对的题数为x，则答错或不答的题数为(120 - x)。根据得分规则，总得分为3x - 1×(120 - x) = 280。化简方程得：3x - 120 + x = 280，即4x = 400，解得x = 100。因此，他们答对了100道题。本题考查一元一次方程的实际应用，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:12:36","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":215,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"一个长方形的长是8厘米，宽是5厘米，它的面积是____平方厘米。","answer":"40","explanation":"长方形的面积计算公式是：面积 = 长 × 宽。题目中给出的长是8厘米，宽是5厘米，因此面积为 8 × 5 = 40 平方厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:08","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":410,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保活动中，某班学生收集了可回收垃圾和不可回收垃圾共120千克。已知可回收垃圾比不可回收垃圾多40千克，那么不可回收垃圾有多少千克？","answer":"A","explanation":"设不可回收垃圾为x千克，则可回收垃圾为(x + 40)千克。根据题意，两者之和为120千克，列出方程：x + (x + 40) = 120。化简得：2x + 40 = 120，移项得：2x = 80，解得：x = 40。因此，不可回收垃圾有40千克。本题考查一元一次方程的实际应用，属于简单难度，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":"2025-12-29 17:28:32","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":"40千克","is_correct":1},{"id":"B","content":"50千克","is_correct":0},{"id":"C","content":"60千克","is_correct":0},{"id":"D","content":"80千克","is_correct":0}]},{"id":2139,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 3(x - 2) = 2x + 1 时，第一步去括号得到 3x - 6 = 2x + 1，第二步移项得到 3x - 2x = 1 + 6，第三步合并同类项得到 x = 7。该学生解题过程中哪一步开始出错？","answer":"D","explanation":"该学生解题过程完全正确：第一步去括号正确，3(x - 2) 展开为 3x - 6；第二步移项正确，将 2x 移到左边变为 -2x，将 -6 移到右边变为 +6；第三步合并同类项，3x - 2x = x，1 + 6 = 7，得到 x = 7，符合解一元一次方程的步骤和规则，因此整个过程没有出错。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"第一步","is_correct":0},{"id":"B","content":"第二步","is_correct":0},{"id":"C","content":"第三步","is_correct":0},{"id":"D","content":"没有出错","is_correct":1}]}]