初中
数学
中等
来源: 教材例题
知识点: 初中数学
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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":135,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"已知一个长方形的长比宽多3厘米,且它的周长是26厘米,那么这个长方形的宽是____厘米。","answer":"5","explanation":"本题考查初一学生对方程的应用能力,结合几何图形(长方形)的周长公式进行列方程求解。题目涉及有理数运算和一元一次方程的建立与求解,符合初一数学课程中‘代数初步’与‘简单几何应用’的学习内容。通过设未知数、列方程、解方程的过程,帮助学生理解实际问题向数学模型的转化。","solution_steps":"设长方形的宽为x厘米,则长为(x + 3)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入已知条件得:2 × (x + x + 3) = 26。化简得:2 × (2x + 3) = 26 → 4x + 6 = 26 → 4x = 20 → x = 5。因此,宽是5厘米。","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 09:40:59","updated_at":"2025-12-24 09:40:59","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":2376,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生用一张矩形纸片制作一个无盖长方体盒子,纸片的长为 24 cm,宽为 18 cm。从四个角各剪去一个边长为 x cm 的正方形,然后将四边折起形成盒子。若要求盒子的容积为 400 cm³,则 x 的值应满足的方程是:","answer":"A","explanation":"制作无盖长方体盒子时,从矩形纸片的四个角各剪去一个边长为 x 的正方形后,折起四边形成盒子。此时,盒子的高为 x cm,底面的长为 (24 - 2x) cm,宽为 (18 - 2x) cm。容积 = 长 × 宽 × 高,即 V = x(24 - 2x)(18 - 2x)。题目给出容积为 400 cm³,因此方程为 x(24 - 2x)(18 - 2x) = 400。选项 A 正确。选项 B 错误,因为未考虑两边都剪去 x;选项 C 缺少高度项 x;选项 D 错误地将 x 平方,不符合实际几何意义。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:27:47","updated_at":"2026-01-10 11:27:47","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":"x(24 - 2x)(18 - 2x) = 400","is_correct":1},{"id":"B","content":"x(24 - x)(18 - x) = 400","is_correct":0},{"id":"C","content":"(24 - x)(18 - x) = 400","is_correct":0},{"id":"D","content":"x²(24 - 2x)(18 - 2x) = 400","is_correct":0}]},{"id":714,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在某次班级数学测验中,某学生答对了全部题目的五分之三,共答对了12道题。那么这次测验一共有____道题。","answer":"20","explanation":"设这次测验一共有x道题。根据题意,某学生答对了全部题目的五分之三,即(3\/5)x = 12。解这个一元一次方程:两边同时乘以5,得3x = 60;再两边同时除以3,得x = 20。因此,这次测验一共有20道题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:49:59","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":1986,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在纸上画了一个边长为8 cm的正方形,并在正方形内部以其中一条对角线为对称轴,画了一个与该对角线重合的等腰直角三角形。若将该三角形绕正方形的中心顺时针旋转90°,则旋转前后两个三角形重叠部分的面积是多少?(π取3.14)","answer":"A","explanation":"本题考查旋转与几何图形的综合应用,重点在于理解旋转对称性和图形重叠关系。正方形边长为8 cm,其对角线长度为√(8² + 8²) = √128 = 8√2 cm。以其中一条对角线为对称轴画的等腰直角三角形,其两条直角边均为8 cm,面积为(1\/2) × 8 × 8 = 32 cm²。正方形中心是对角线的交点,也是旋转中心。当该三角形绕正方形中心顺时针旋转90°时,由于正方形具有90°旋转对称性,且原三角形关于中心对称,旋转后的三角形将与原三角形关于中心成轴对称。两个三角形重叠的部分是一个较小的等腰直角三角形,其直角边为原三角形直角边的一半,即4 cm。因此,重叠部分面积为(1\/2) × 4 × 4 = 8 cm²。但进一步分析发现,实际重叠区域是由两个45°-45°-90°三角形组成,每个面积为8 cm²,总重叠面积为16 cm²。故正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 15:05:54","updated_at":"2026-01-07 15:05:54","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":"16 cm²","is_correct":1},{"id":"B","content":"24 cm²","is_correct":0},{"id":"C","content":"32 cm²","is_correct":0},{"id":"D","content":"8 cm²","is_correct":0}]},{"id":1830,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一次函数与轴对称图形的综合问题时,发现函数 y = 2x + 4 的图像与坐标轴围成的三角形区域关于某条直线对称后,恰好与原图形重合。若将该三角形的三个顶点坐标分别代入表达式 |x| + |y|,则这三个值的平均数为多少?","answer":"B","explanation":"首先确定一次函数 y = 2x + 4 与坐标轴的交点。令 x = 0,得 y = 4,即与 y 轴交于点 A(0, 4);令 y = 0,得 0 = 2x + 4,解得 x = -2,即与 x 轴交于点 B(-2, 0)。原点 O(0, 0) 是坐标轴交点,因此所围成的三角形为 △AOB,顶点为 O(0,0)、A(0,4)、B(-2,0)。\n\n题目指出该三角形关于某条直线对称后与原图形重合。观察可知,该三角形不是轴对称图形本身,但若考虑其关于直线 x = -1 对称,则点 B(-2,0) 对称后为 (0,0),点 O(0,0) 对称后为 (-2,0),点 A(0,4) 对称后为 (-2,4),并不重合。进一步分析发现,实际上题目暗示的是:整个图形(包括位置)在某种对称变换下不变,但更合理的理解是考察三角形顶点坐标的绝对值表达式计算,对称性在此处主要用于确认图形结构合理性。\n\n接下来计算每个顶点代入 |x| + |y| 的值:\n- 对于 O(0,0):|0| + |0| = 0\n- 对于 A(0,4):|0| + |4| = 4\n- 对于 B(-2,0):|-2| + |0| = 2\n\n三个值分别为 0、4、2,其平均数为 (0 + 4 + 2) ÷ 3 = 6。\n\n因此正确答案为 B。本题综合考查了一次函数图像与坐标轴交点、三角形顶点坐标、绝对值运算以及数据的平均数计算,同时隐含轴对称思想的初步应用,符合八年级知识范围,难度适中且情境新颖。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:48:29","updated_at":"2026-01-06 16:48:29","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":"A","content":"4","is_correct":0},{"id":"B","content":"6","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"10","is_correct":0}]},{"id":2299,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一块三角形花坛的三边长度,分别为5米、12米和13米。他想知道这块花坛是否为直角三角形,以便合理规划灌溉系统。根据所学知识,可以判断该三角形是直角三角形吗?","answer":"A","explanation":"根据勾股定理,若一个三角形是直角三角形,则其两条较短边的平方和等于最长边(斜边)的平方。本题中,三边分别为5、12、13,其中13为最长边。计算得:5² + 12² = 25 + 144 = 169,而13² = 169,两者相等,满足勾股定理的逆定理,因此该三角形是直角三角形。选项A正确。选项B错误,因为三边不等并不影响是否为直角三角形;选项C错误,三边为整数只是勾股数的特征,不能单独作为判断依据;选项D错误,13确实是三边中最长的,符合斜边条件。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:43:35","updated_at":"2026-01-10 10:43:35","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":"A","content":"是,因为5² + 12² = 13²","is_correct":1},{"id":"B","content":"不是,因为三边长度不相等","is_correct":0},{"id":"C","content":"是,因为三边长度都是整数","is_correct":0},{"id":"D","content":"不是,因为13不是最长边","is_correct":0}]},{"id":595,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织一次环保知识竞赛,参赛学生分为若干小组。已知每两个小组之间都要进行一次答题对决,共进行了45场对决。问该班级共有多少个小组参赛?","answer":"C","explanation":"本题考查的是组合问题与一元二次方程的实际应用,属于七年级数学中‘一元一次方程’的拓展应用(虽涉及一元二次,但在七年级可通过枚举或简单推理解决)。每两个小组进行一场对决,属于从n个小组中任选2个的组合问题,总场数为C(n,2) = n(n-1)\/2。题目给出总场数为45,因此列出方程:n(n-1)\/2 = 45。两边同乘以2得:n(n-1) = 90。尝试代入选项验证:当n=10时,10×9=90,满足条件。因此共有10个小组。此题虽形式上为一元二次方程,但七年级学生可通过试值法轻松解决,符合简单难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 20:45:46","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":"8个","is_correct":0},{"id":"B","content":"9个","is_correct":0},{"id":"C","content":"10个","is_correct":1},{"id":"D","content":"11个","is_correct":0}]},{"id":205,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生计算一个数的相反数时,将 -5 写成了 5,那么他计算的是 ___ 的相反数。","answer":"-5","explanation":"相反数的定义是:一个数 a 的相反数是 -a。题目中说某学生将 -5 写成了 5,说明他实际上是把原数的相反数算成了 5。也就是说,-a = 5,那么 a = -5。因此,他计算的是 -5 的相反数。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:31","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":2210,"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气温变化时,发现某天的气温比前一天上升了5℃,记作+5℃;而另一天的气温比前一天下降了3℃,应记作____℃。","answer":"-3","explanation":"根据正数和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。题目中气温下降3℃,因此应记作-3℃。这符合七年级学生对正负数在实际生活中应用的理解要求。","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":[]}]