[{"id":2298,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一个等腰三角形的底边长为8 cm，腰长为5 cm。若该三角形的一条对称轴将其分成两个全等直角三角形，则每个直角三角形的斜边长为多少？","answer":"A","explanation":"等腰三角形的对称轴是从顶角垂直平分底边的高，它将原三角形分成两个全等的直角三角形。每个直角三角形的底边为原底边的一半，即8 ÷ 2 = 4 cm，一条直角边为高（未知），另一条直角边为4 cm，斜边即为原等腰三角形的腰长，为5 cm。因此，每个直角三角形的斜边长为5 cm。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:43:17","updated_at":"2026-01-10 10:43:17","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":"5 cm","is_correct":1},{"id":"B","content":"6 cm","is_correct":0},{"id":"C","content":"8 cm","is_correct":0},{"id":"D","content":"10 cm","is_correct":0}]},{"id":1013,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保活动中，某班级收集了可回收垃圾共120千克，其中纸张占总重量的五分之三，塑料占总重量的四分之一，其余为金属。那么金属的重量是___千克。","answer":"18","explanation":"首先计算纸张的重量：120 × 3\/5 = 72 千克；然后计算塑料的重量：120 × 1\/4 = 30 千克；纸张和塑料共重 72 + 30 = 102 千克；因此金属的重量为 120 - 102 = 18 千克。本题考查有理数中的分数乘法与加减运算在实际问题中的应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:24:02","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":2454,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在一次校园绿化项目中，工人师傅用一根长为10米的绳子围成一个直角三角形花坛，其中一条直角边比另一条短2米。设较短的直角边为x米，则可列方程为____，解得较短的直角边长为____米。","answer":"x² + (x + 2)² = 10²；6","explanation":"根据勾股定理，两直角边平方和等于斜边平方。较短边为x，较长边为x+2，斜边为10，列方程x² + (x+2)² = 100，解得x=6（舍负）。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:58:36","updated_at":"2026-01-10 13:58:36","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":1757,"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加数学实践活动，要求将学生分成若干小组，每组人数相同。已知若每组安排6人，则最后一组只有4人；若每组安排8人，则最后一组只有6人；若每组安排9人，则最后一组只有7人。问：该校七年级参加活动的学生至少有多少人？请通过建立方程或不等式模型，并结合整除性质进行分析求解。","answer":"设参加活动的学生总人数为x人。\n\n根据题意，可列出以下同余关系：\n\nx ≡ 4 (mod 6) ——（1）\n\nx ≡ 6 (mod 8) ——（2）\n\nx ≡ 7 (mod 9) ——（3）\n\n观察发现，每个余数都比除数少2：\n\n即：x + 2 ≡ 0 (mod 6)\n\nx + 2 ≡ 0 (mod 8)\n\nx + 2 ≡ 0 (mod 9)\n\n说明 x + 2 是 6、8、9 的公倍数。\n\n先求6、8、9的最小公倍数：\n\n分解质因数：\n\n6 = 2 × 3\n\n8 = 2³\n\n9 = 3²\n\n取各质因数最高次幂：2³ × 3² = 8 × 9 = 72\n\n所以 x + 2 是72的倍数，即 x + 2 = 72k（k为正整数）\n\n因此 x = 72k - 2\n\n当k = 1时，x = 72 - 2 = 70\n\n验证：\n\n70 ÷ 6 = 11组余4人 → 符合（1）\n\n70 ÷ 8 = 8组余6人 → 符合（2）\n\n70 ÷ 9 = 7组余7人 → 符合（3）\n\n当k = 2时，x = 144 - 2 = 142，也满足，但题目要求“至少”有多少人。\n\n所以最小满足条件的x为70。\n\n答：该校七年级参加活动的学生至少有70人。","explanation":"本题考查学生对同余概念的理解与转化能力，结合整除性质和一元一次方程建模思想。关键在于发现三个条件中余数与除数的关系：余数均为除数减2，从而转化为x + 2是6、8、9的公倍数。通过求最小公倍数得到最小解。题目融合了整数的整除性、最小公倍数、方程建模与逻辑推理，属于典型的困难级别应用题，要求学生具备较强的观察力与抽象思维能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:33:35","updated_at":"2026-01-06 14:33:35","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":2218,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生记录了一周内每天的温度变化，规定比0℃高为正，比0℃低为负。其中某天的温度记为-3℃，另一天的温度比这一天高5℃，则这一天的温度记为___℃。","answer":"2","explanation":"题目中已知某天温度为-3℃，另一天比它高5℃，即计算-3 + 5。根据正负数加减法则，-3 + 5 = 2，因此这一天的温度记为2℃。该题考查正负数在实际情境中的加减运算，符合七年级学生对正负数意义的理解和应用要求。","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":2190,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在一条东西方向的直线上做实验，规定向东为正方向。他从原点出发，先向东走了5米，记作+5米，然后向西走了8米。此时他所在的位置应记作多少米？","answer":"D","explanation":"该学生从原点出发，向东走5米到达+5米的位置，再向西走8米，相当于从+5米减去8米，即5 - 8 = -3米。因此，他最终位置应记作-3米。选项D正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","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米","is_correct":0},{"id":"B","content":"+3米","is_correct":0},{"id":"C","content":"-3米","is_correct":0},{"id":"D","content":"-13米","is_correct":1}]},{"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":280,"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名学生的阅读时间（单位：小时\/周），并将数据整理如下：5, 6, 7, 5, 8, 6, 7, 9, 5, 6, 7, 8, 6, 5, 7, 8, 9, 6, 7, 5, 8, 7, 6, 5, 7, 8, 6, 7, 5, 6。为了分析这组数据的集中趋势，该学生想求出这组数据的中位数。请问这组数据的中位数是多少？","answer":"B","explanation":"首先将30个数据按从小到大的顺序排列：5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9。由于数据个数为30（偶数），中位数是第15个和第16个数据的平均数。第15个数是7，第16个数也是7，因此中位数为(7 + 7) ÷ 2 = 7。但仔细核对排序后发现：实际排序中第15个是6，第16个是7。正确排序后前14个为5和6，第15个是6，第16个是7，因此中位数为(6 + 7) ÷ 2 = 6.5。正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:31:13","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":"6","is_correct":0},{"id":"B","content":"6.5","is_correct":1},{"id":"C","content":"7","is_correct":0},{"id":"D","content":"7.5","is_correct":0}]},{"id":189,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明去文具店买笔记本，每本笔记本的价格是5元。他买了3本，付给收银员20元，应找回多少钱？","answer":"A","explanation":"首先计算小明购买3本笔记本的总花费：每本5元，3本就是 5 × 3 = 15 元。他付了20元，所以应找回的钱是 20 - 15 = 5 元。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 14:01:39","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":"5元","is_correct":1},{"id":"B","content":"10元","is_correct":0},{"id":"C","content":"15元","is_correct":0},{"id":"D","content":"20元","is_correct":0}]},{"id":345,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在平面直角坐标系中，点 A 的坐标是 (3, 4)，点 B 的坐标是 (3, -2)。这两点之间的距离是多少？","answer":"A","explanation":"点 A 和点 B 的横坐标相同，都是 3，说明它们位于同一条竖直线上。两点之间的距离等于它们纵坐标之差的绝对值。计算：|4 - (-2)| = |4 + 2| = |6| = 6。因此，两点之间的距离是 6。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:41:02","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":"6","is_correct":1},{"id":"B","content":"5","is_correct":0},{"id":"C","content":"7","is_correct":0},{"id":"D","content":"8","is_correct":0}]}]