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[{"id":1803,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一块直角三角形纸片的两条直角边,长度分别为5厘米和12厘米。若他想用一根细线沿着纸片的边缘完整绕一圈,至少需要多长的细线?","answer":"B","explanation":"题目要求计算直角三角形的周长。已知两条直角边分别为5厘米和12厘米,首先利用勾股定理求斜边长度:斜边 = √(5² + 12²) = √(25 + 144) = √169 = 13厘米。然后将三边相加得到周长:5 + 12 + 13 = 30厘米。因此,至少需要30厘米的细线才能绕边缘一圈。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 16:17:08","updated_at":"2026-01-06 16:17:08","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":"17厘米","is_correct":0},{"id":"B","content":"30厘米","is_correct":1},{"id":"C","content":"25厘米","is_correct":0},{"id":"D","content":"34厘米","is_correct":0}]},{"id":1808,"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6厘米,两腰各为5厘米。若以该三角形的底边为轴进行轴对称变换,得到的新三角形与原三角形组成的图形是什么?","answer":"D","explanation":"原三角形是等腰三角形,底边为6厘米,两腰为5厘米。以底边为轴作轴对称变换后,会得到一个与原三角形完全对称的新三角形,两个三角形共用底边,顶点分别在底边两侧。这样形成的四边形有两组对边分别相等(每条腰5厘米,底边6厘米被对称复制),且由于对称性,对边平行,因此构成一个平行四边形。由于边长不等(5≠6),不是菱形;角度不是直角,也不是矩形或正方形。故正确答案为D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 16:18:06","updated_at":"2026-01-06 16:18:06","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":"菱形","is_correct":0},{"id":"B","content":"矩形","is_correct":0},{"id":"C","content":"正方形","is_correct":0},{"id":"D","content":"平行四边形","is_correct":1}]},{"id":489,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"17个","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 18:03:23","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":1091,"subject":"数学","grade":"七年级","stage":"小学","type":"填空题","content":"某学生在整理班级同学的身高数据时,将数据按从小到大的顺序排列,发现最矮的同学身高为148厘米,最高的同学身高为165厘米。如果将所有同学的身高都增加3厘米,则新的数据中,最高身高与最矮身高的差是___厘米。","answer":"17","explanation":"原数据中最高身高为165厘米,最矮为148厘米,两者相差165 - 148 = 17厘米。当所有数据都增加相同的数值(3厘米)时,数据的分布形状不变,极差(最大值与最小值之差)保持不变。因此,新的最高身高为165 + 3 = 168厘米,新的最矮身高为148 + 3 = 151厘米,差值为168 - 151 = 17厘米。所以答案是17。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:35","updated_at":"2026-01-06 08:55:35","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":2140,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时,将方程 2(x - 3) = 4 的两边同时除以2,得到 x - 3 = 2,然后解得 x = 5。这一解法的依据是等式的哪一条性质?","answer":"D","explanation":"该学生在解方程时,将方程两边同时除以2,这是运用了等式的基本性质:等式两边同时除以同一个不为零的数,等式仍然成立。这一步骤是解一元一次方程的常用方法,符合七年级数学课程中关于等式性质的教学内容。","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":"等式两边同时加上同一个数,等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数,等式仍然成立","is_correct":0},{"id":"C","content":"等式两边同时乘以同一个数,等式仍然成立","is_correct":0},{"id":"D","content":"等式两边同时除以同一个不为零的数,等式仍然成立","is_correct":1}]},{"id":1815,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在计算一个二次根式时,将√(12) + √(27) 化简为最简形式。以下哪个选项是正确的结果?","answer":"A","explanation":"首先将每个二次根式化为最简形式:√12 = √(4×3) = 2√3,√27 = √(9×3) = 3√3。然后将它们相加:2√3 + 3√3 = 5√3。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":"2026-01-06 16:20:11","updated_at":"2026-01-06 16:20:11","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√3","is_correct":1},{"id":"B","content":"7√3","is_correct":0},{"id":"C","content":"13√3","is_correct":0},{"id":"D","content":"3√5","is_correct":0}]},{"id":1099,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级图书捐赠活动中,某学生捐出的图书数量比班级平均每人捐书数量的2倍还多3本。如果班级共有30名学生,总共捐了150本书,那么这名学生捐了___本书。","answer":"13","explanation":"首先根据题意,班级共有30名学生,总共捐了150本书,因此平均每人捐书数量为150 ÷ 30 = 5本。题目中说某学生捐出的图书数量比平均每人捐书数量的2倍还多3本,即2 × 5 + 3 = 10 + 3 = 13本。因此,这名学生捐了13本书。本题考查了有理数的四则运算和一元一次方程的基本思想,通过平均数建立数量关系,适合七年级学生水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:57:21","updated_at":"2026-01-06 08:57:21","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":2527,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在操场上观察旗杆的投影。已知旗杆高6米,太阳光线与地面形成的仰角为30°,则此时旗杆在地面的投影长度为多少米?","answer":"A","explanation":"本题考查锐角三角函数的应用。旗杆、投影和太阳光线构成一个直角三角形,其中旗杆为对边,投影为邻边,太阳光线与地面的夹角为30°。根据正切函数定义:tan(30°) = 对边 \/ 邻边 = 6 \/ x。因为 tan(30°) = √3 \/ 3,所以有 √3 \/ 3 = 6 \/ x,解得 x = 6 \/ (√3 \/ 3) = 6 × 3 \/ √3 = 18 \/ √3。将分母有理化:18 \/ √3 = (18√3) \/ 3 = 6√3。因此,旗杆的投影长度为6√3米,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 16:11:59","updated_at":"2026-01-10 16:11:59","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√3","is_correct":1},{"id":"B","content":"3√3","is_correct":0},{"id":"C","content":"12","is_correct":0},{"id":"D","content":"2√3","is_correct":0}]},{"id":335,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的运动项目数据时,制作了如下频数分布表。已知喜欢篮球的人数占总人数的30%,总人数为40人,则喜欢篮球的人数是多少?","answer":"B","explanation":"题目要求计算喜欢篮球的人数。已知总人数为40人,喜欢篮球的人数占总人数的30%。计算方法是:40 × 30% = 40 × 0.3 = 12。因此,喜欢篮球的人数是12人。本题考查的是数据的收集、整理与描述中的百分比计算,属于七年级数学中数据处理的基础知识,难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:39:57","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":"10","is_correct":0},{"id":"B","content":"12","is_correct":1},{"id":"C","content":"15","is_correct":0},{"id":"D","content":"18","is_correct":0}]},{"id":658,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保主题活动中,某学生收集了不同种类的可回收物品,其中废纸的重量为3.5千克,塑料瓶的重量比废纸多1.2千克,金属罐的重量是塑料瓶的一半。那么金属罐的重量是______千克。","answer":"2.35","explanation":"首先根据题意,塑料瓶的重量比废纸多1.2千克,废纸为3.5千克,因此塑料瓶重量为3.5 + 1.2 = 4.7千克。金属罐的重量是塑料瓶的一半,即4.7 ÷ 2 = 2.35千克。本题考查有理数的加减与乘除运算在实际问题中的应用,属于简单难度的综合计算题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:14:38","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":[]}]