[{"id":2304,"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，某学生用一根长度为20 cm的铁丝围成一个等腰三角形。已知底边长为6 cm，则这个等腰三角形的腰长是多少？","answer":"B","explanation":"等腰三角形有两条相等的腰和一条底边。已知铁丝总长为20 cm，即三角形的周长为20 cm，底边长为6 cm。设腰长为x cm，则根据周长公式可得：2x + 6 = 20。解这个方程：2x = 20 - 6 = 14，所以x = 7。因此，腰长为7 cm。选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:44:33","updated_at":"2026-01-10 10:44:33","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":"6 cm","is_correct":0},{"id":"B","content":"7 cm","is_correct":1},{"id":"C","content":"8 cm","is_correct":0},{"id":"D","content":"10 cm","is_correct":0}]},{"id":458,"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30名同学进行调查，记录了他们每周课外阅读的小时数。整理数据后发现，阅读时间在3小时及以下的有6人，4小时的有8人，5小时的有10人，6小时的有4人，7小时的有2人。请问这组数据的众数是多少？","answer":"C","explanation":"众数是一组数据中出现次数最多的数值。根据题目提供的数据：阅读3小时的有6人，4小时的有8人，5小时的有10人，6小时的有4人，7小时的有2人。其中，阅读5小时的人数最多，为10人，因此这组数据的众数是5小时。选项C正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:47:41","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小时","is_correct":0},{"id":"B","content":"4小时","is_correct":0},{"id":"C","content":"5小时","is_correct":1},{"id":"D","content":"6小时","is_correct":0}]},{"id":131,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个长方形的长比宽多5厘米，若其周长为30厘米，则这个长方形的宽是______厘米。","answer":"5","explanation":"设长方形的宽为x厘米，则长为(x + 5)厘米。根据长方形周长公式：周长 = 2 × (长 + 宽)，代入得：2 × (x + x + 5) = 30，即2 × (2x + 5) = 30，化简得4x + 10 = 30，解得4x = 20，x = 5。因此，宽为5厘米。本题结合代数设未知数与一元一次方程求解，符合初一学生对方程和几何基础的学习要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:59:12","updated_at":"2025-12-24 08:59:12","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":553,"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时间（单位：小时）时，记录了以下5个数据：2.5，3，3.5，4，4.5。如果他想用这组数据制作频数分布表，并将数据分为两组：3小时以下（不含3小时）和3小时及以上，那么这两组的频数分别是多少？","answer":"A","explanation":"首先明确分组标准：第一组是“3小时以下（不含3小时）”，即小于3；第二组是“3小时及以上”，即大于或等于3。原始数据为：2.5，3，3.5，4，4.5。其中，只有2.5小于3，属于第一组，频数为1；其余数据3、3.5、4、4.5均大于或等于3，属于第二组，共4个数据，频数为4。因此，两组的频数分别是1和4，正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:11:16","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":"1和4","is_correct":1},{"id":"B","content":"2和3","is_correct":0},{"id":"C","content":"3和2","is_correct":0},{"id":"D","content":"4和1","is_correct":0}]},{"id":2486,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在观察一个圆柱形水杯的正投影时，发现当水杯直立放置在水平桌面上，且光线从正前方水平照射时，其投影为一个矩形。若将水杯绕其底面圆心顺时针旋转30°，则此时水杯的正投影最可能是什么形状？","answer":"D","explanation":"圆柱形水杯直立时，其正投影为矩形，因为圆柱的侧面投影为矩形，底面和顶面投影为线段。当水杯绕底面圆心旋转30°后，圆柱的轴线不再垂直于投影面，而是倾斜了30°。此时，圆柱的侧面投影会因倾斜而变为平行四边形（上下底边仍平行且等长，但侧边倾斜），而底面和顶面的圆形投影变为椭圆弧，但在正投影中通常不可见或退化为线段。因此整体投影呈现为平行四边形。选项D正确。选项A错误，因为旋转后不再垂直；选项B仅描述局部；选项C不符合旋转后的几何特征。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:11:24","updated_at":"2026-01-10 15:11:24","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":"一个矩形","is_correct":0},{"id":"B","content":"一个椭圆","is_correct":0},{"id":"C","content":"一个矩形上方叠加一个半圆","is_correct":0},{"id":"D","content":"一个平行四边形","is_correct":1}]},{"id":473,"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完成数学作业所用的时间（单位：分钟），并将数据整理如下：30, 35, 40, 40, 45, 50, 55, 60, 60, 60。如果他想用一个统计量来代表大多数同学完成作业的时间，最合适的统计量是：","answer":"C","explanation":"题目中给出的数据是：30, 35, 40, 40, 45, 50, 55, 60, 60, 60。观察数据发现，60分钟出现了3次，是出现次数最多的数据，因此众数是60。题目要求用一个统计量来代表‘大多数’同学的时间，而‘众数’正是反映数据集中趋势、体现出现频率最高的值，最适合描述‘大多数’的情况。虽然平均数和中位数也能反映集中趋势，但它们不一定对应实际出现最多的数值；极差只反映数据范围，不能代表典型情况。因此最合适的统计量是众数。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:55:53","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":"平均数","is_correct":0},{"id":"B","content":"中位数","is_correct":0},{"id":"C","content":"众数","is_correct":1},{"id":"D","content":"极差","is_correct":0}]},{"id":2337,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一个几何问题时，发现一个等腰三角形ABC，其中AB = AC，且底边BC的长度为8。若从顶点A向底边BC作高AD，垂足为D，且高AD的长度为√15。现以BC所在直线为x轴，点D为原点建立平面直角坐标系，则顶点A的坐标可能是下列哪一项？","answer":"A","explanation":"由于△ABC是等腰三角形，AB = AC，底边为BC，因此从顶点A向底边BC所作的高AD必垂直于BC，并且平分底边BC。已知BC = 8，所以BD = DC = 4。题目中以BC所在直线为x轴，点D为原点建立坐标系，因此点D的坐标为(0, 0)。又因为AD是高，长度为√15，且A点在BC的上方（通常默认向上为正方向），所以点A位于y轴正方向上，坐标为(0, √15)。若A在下方则为(0, -√15)，但题目未说明方向时一般取正方向。结合坐标系设定和等腰三角形性质，正确答案为A选项(0, √15)。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 10:57:22","updated_at":"2026-01-10 10:57:22","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, √15)","is_correct":1},{"id":"B","content":"(4, √15)","is_correct":0},{"id":"C","content":"(0, -√15)","is_correct":0},{"id":"D","content":"(8, √15)","is_correct":0}]},{"id":1324,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善交通状况，计划在一条主干道旁修建一个矩形绿化带。绿化带的一边紧贴道路（不需要围栏），其余三边用总长为60米的环保材料围栏围成。为了提升生态效益，绿化带被划分为两个区域：一个正方形种植区用于种植灌木，另一个矩形区域用于种植草本植物。正方形种植区的一边与道路平行，且其边长比草本植物区域的宽度多2米。已知草本植物区域的长度与正方形种植区的边长相等。设草本植物区域的宽度为x米。\n\n（1）用含x的整式表示绿化带的总长度和总宽度；\n（2）根据围栏总长为60米，列出关于x的一元一次方程，并求出x的值；\n（3）若每平方米灌木种植成本为80元，草本植物为50元，求整个绿化带的总种植成本；\n（4）若城市规划要求绿化带面积不得小于200平方米，请验证该设计方案是否满足要求，并说明理由。","answer":"（1）设草本植物区域的宽度为x米，则正方形种植区的边长为(x + 2)米。\n由于草本植物区域的长度与正方形边长相等，也为(x + 2)米。\n\n绿化带的总长度（与道路平行的方向）为：正方形边长 + 草本植物区域长度 = (x + 2) + (x + 2) = 2x + 4（米）。\n\n绿化带的总宽度（垂直于道路的方向）为：草本植物区域的宽度 = x 米。\n\n答：绿化带总长度为(2x + 4)米，总宽度为x米。\n\n（2）围栏用于三边：两条宽（左右两侧）和一条长（远离道路的一侧）。\n围栏总长 = 2 × 宽度 + 长度 = 2x + (2x + 4) = 4x + 4（米）。\n\n根据题意，围栏总长为60米：\n4x + 4 = 60\n4x = 56\nx = 14\n\n答：x的值为14。\n\n（3）当x = 14时：\n正方形种植区边长 = 14 + 2 = 16（米），面积 = 16 × 16 = 256（平方米）。\n草本植物区域面积 = 长度 × 宽度 = 16 × 14 = 224（平方米）。\n\n总种植成本 = 256 × 80 + 224 × 50 = 20480 + 11200 = 31680（元）。\n\n答：总种植成本为31680元。\n\n（4）绿化带总面积 = 正方形面积 + 草本植物面积 = 256 + 224 = 480（平方米）。\n\n因为480 > 200，所以该设计方案满足绿化带面积不得小于200平方米的要求。\n\n答：满足要求，因为总面积为480平方米，大于200平方米。","explanation":"本题综合考查了整式的加减、一元一次方程、几何图形初步及实际问题的建模能力。第（1）问要求学生根据文字描述建立代数表达式，理解图形结构；第（2）问通过围栏总长建立方程，体现方程建模思想；第（3）问结合有理数运算与面积计算，考查多步运算能力；第（4）问引入不等式思想（虽未直接使用不等式符号，但需比较大小），检验方案合理性。题目情境贴近生活，结构层层递进，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 10:55:37","updated_at":"2026-01-06 10:55:37","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":2761,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在参观博物馆时看到一件刻有文字的青铜器，讲解员介绍说这种文字是商周时期刻在龟甲和兽骨上的，主要用于占卜记事。这件文物上的文字最可能是：","answer":"A","explanation":"题干中提到文字刻在‘龟甲和兽骨上’，并用于‘占卜记事’，这正是甲骨文的典型特征。甲骨文是商朝时期王室用于占卜记事而在龟甲或兽骨上契刻的文字，是中国已发现的古代文字中年代最早、体系较为完整的文字。虽然金文也出现在商周时期，但它主要铸刻在青铜器上；小篆和隶书则是秦朝统一后及之后流行的字体，时间较晚。因此，正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:39:54","updated_at":"2026-01-12 10:39:54","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":"甲骨文","is_correct":1},{"id":"B","content":"金文","is_correct":0},{"id":"C","content":"小篆","is_correct":0},{"id":"D","content":"隶书","is_correct":0}]},{"id":1969,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某次校园义卖活动中不同商品的销售情况时，记录了五种商品的销售额（单位：元）：125.6, 98.4, 142.3, 110.8, 135.7。为了分析这组数据的集中趋势，该学生计算了这组数据的中位数和平均数，并发现两者存在一定差异。若将这组数据按从小到大的顺序排列后，位于中间位置的数据与所有数据之和除以数据个数的结果之差最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中中位数与平均数的计算及比较。首先将五种商品的销售额从小到大排序：98.4, 110.8, 125.6, 135.7, 142.3。由于数据个数为5（奇数），中位数是第3个数，即125.6。接着计算平均数：(125.6 + 98.4 + 142.3 + 110.8 + 135.7) ÷ 5 = 612.8 ÷ 5 = 122.56。然后计算中位数与平均数之差：125.6 - 122.56 = 3.04。该值最接近选项B（2.8）。因此，正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:48:51","updated_at":"2026-01-07 14:48:51","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":"1.2","is_correct":0},{"id":"B","content":"2.8","is_correct":1},{"id":"C","content":"3.6","is_correct":0},{"id":"D","content":"4.4","is_correct":0}]}]