初中
数学
中等
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知识点: 初中数学
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0×1| = 8 得到,或使用分割法、坐标法(如鞋带公式)计算,方法严谨且符合八年级知识范围。选项 A 虽部分正确,但未利用坐标优势,效率较低;选项 C 错误,因角度并非直角;选项 D 混淆了轴对称与平行四边形的关系,平行四边形不一定是轴对称图形。因此,B 为最佳方法。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:52:06","updated_at":"2026-01-10 11:52:06","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":"利用勾股定理分别计算四条边的长度,若对边相等,则该四边形是平行四边形,再用底乘高计算面积。","is_correct":0},{"id":"B","content":"利用一次函数的斜率判断 AB 与 CD、AD 与 BC 是否分别平行,再通过向量法或距离公式验证对边相等,最后用向量叉积或分割法求面积。","is_correct":1},{"id":"C","content":"直接假设该四边形是矩形,用长乘宽计算面积,因为所有角看起来都是直角。","is_correct":0},{"id":"D","content":"将该四边形沿 y 轴对折,若两部分完全重合,则说明是轴对称图形,因此是平行四边形,面积可用对称性估算。","is_correct":0}]},{"id":767,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了可回收垃圾的重量为 3.5 千克,比另一名同学多收集了 1.2 千克。设另一名同学收集的垃圾重量为 x 千克,则可列出一元一次方程为:_3.5 = x + 1.2_,解得 x = _2.3_。","answer":"3.5 = x + 1.2;2.3","explanation":"根据题意,某学生收集的 3.5 千克比另一名同学多 1.2 千克,说明另一名同学的收集量加上 1.2 千克等于 3.5 千克,因此可列方程 3.5 = x + 1.2。解这个方程,两边同时减去 1.2,得到 x = 3.5 - 1.2 = 2.3。本题考查一元一次方程的建立与求解,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 23:43:56","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":679,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级数学测验成绩统计中,某学生发现自己的成绩比全班的平均分高6分。如果全班共有30人,所有人的成绩总和为2400分,那么这名学生的成绩是____分。","answer":"86","explanation":"首先根据全班30人、总分2400分,可以求出全班平均分为:2400 ÷ 30 = 80(分)。题目说明该学生的成绩比平均分高6分,因此他的成绩为:80 + 6 = 86(分)。本题考查了数据的收集、整理与描述中的平均数计算,并结合有理数的加减运算,难度为简单,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 22:28:41","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":810,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级图书捐赠活动中,某学生第一天捐了若干本书,第二天比第一天多捐了5本,两天一共捐了23本。设第一天捐了___本书。","answer":"9","explanation":"设第一天捐了x本书,则第二天捐了(x + 5)本。根据题意,两天共捐书数量为:x + (x + 5) = 23。解这个一元一次方程:2x + 5 = 23,移项得2x = 18,解得x = 9。因此,第一天捐了9本书。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:25:12","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":2251,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上,点P表示的数是-3,点Q与点P之间的距离是7个单位长度,且点Q在原点的右侧。那么点Q表示的数是___。","answer":"B","explanation":"点P表示-3,点Q与点P相距7个单位长度。由于点Q在原点右侧,说明点Q表示的数是正数。从-3向右移动7个单位,计算为:-3 + 7 = 4。因此点Q表示的数是4。选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"-10","is_correct":0},{"id":"B","content":"4","is_correct":1},{"id":"C","content":"10","is_correct":0},{"id":"D","content":"-4","is_correct":0}]},{"id":176,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"已知函数 $ y = ax^2 + bx + c $ 的图像经过点 $ (1, 0) $、$ (3, 0) $ 和 $ (0, 3) $,且该函数在区间 $ [2, 4] $ 上的最大值为 $ M $,最小值为 $ m $。若 $ M - m = k $,则 $ k $ 的值为多少?","answer":"D","explanation":"首先,由题意知二次函数 $ y = ax^2 + bx + c $ 经过三点:$ (1, 0) $、$ (3, 0) $、$ (0, 3) $。\n\n因为函数过 $ (1, 0) $ 和 $ (3, 0) $,说明 $ x = 1 $ 和 $ x = 3 $ 是方程的两个根,因此可设函数为:\n$$\ny = a(x - 1)(x - 3)\n$$\n又因为函数过点 $ (0, 3) $,代入得:\n$$\n3 = a(0 - 1)(0 - 3) = a \\cdot (-1) \\cdot (-3) = 3a \\Rightarrow a = 1\n$$\n所以函数表达式为:\n$$\ny = (x - 1)(x - 3) = x^2 - 4x + 3\n$$\n\n接下来求该函数在区间 $ [2, 4] $ 上的最大值 $ M $ 和最小值 $ m $。\n\n二次函数 $ y = x^2 - 4x + 3 $ 的对称轴为:\n$$\nx = \\frac{-(-4)}{2 \\cdot 1} = 2\n$$\n开口向上,因此在区间 $ [2, 4] $ 上,最小值出现在顶点 $ x = 2 $ 处,最大值出现在离对称轴最远的端点 $ x = 4 $ 处。\n\n计算函数值:\n- 当 $ x = 2 $ 时,$ y = (2)^2 - 4 \\cdot 2 + 3 = 4 - 8 + 3 = -1 $,即 $ m = -1 $\n- 当 $ x = 4 $ 时,$ y = (4)^2 - 4 \\cdot 4 + 3 = 16 - 16 + 3 = 3 $,即 $ M = 3 $\n\n所以 $ k = M - m = 3 - (-1) = 4 $\n\n因此正确答案是 D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2025-12-29 12:32:35","updated_at":"2025-12-29 12:32:35","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":"1","is_correct":0},{"id":"B","content":"2","is_correct":0},{"id":"C","content":"3","is_correct":0},{"id":"D","content":"4","is_correct":1}]},{"id":2492,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生用三视图观察一个几何体,主视图和左视图都是等腰三角形,俯视图是一个圆,则这个几何体最可能是以下哪种?","answer":"A","explanation":"根据题目描述,主视图和左视图都是等腰三角形,说明从正面和侧面看,该几何体的轮廓呈三角形;而俯视图是一个圆,说明从上面看是圆形。圆锥的主视图和左视图均为等腰三角形,俯视图为圆,完全符合题意。圆柱的主视图和左视图应为矩形,俯视图为圆,不符合;三棱锥的俯视图是多边形而非圆;球体的三视图均为圆,也不符合。因此正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:16:58","updated_at":"2026-01-10 15:16:58","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":"圆锥","is_correct":1},{"id":"B","content":"圆柱","is_correct":0},{"id":"C","content":"三棱锥","is_correct":0},{"id":"D","content":"球体","is_correct":0}]},{"id":1965,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究自家花园中不同种类花卉的生长高度时,记录了5种花卉的平均高度(单位:厘米):18.4, 22.6, 19.8, 25.2, 21.0。为了更清晰地比较这些数据,该学生决定将这些高度数据四舍五入到最近的整数后,再计算新数据集的极差。请问四舍五入后的数据极差是多少?","answer":"B","explanation":"本题考查数据的收集、整理与描述中对数据的近似处理及极差的计算。首先将原始数据四舍五入到最近的整数:18.4 → 18,22.6 → 23,19.8 → 20,25.2 → 25,21.0 → 21。得到新数据集:18, 20, 21, 23, 25。极差是最大值与最小值之差,即25 - 18 = 7。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 14:47:55","updated_at":"2026-01-07 14:47:55","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":"7","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"9","is_correct":0}]},{"id":2499,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个装饰灯罩,其侧面轮廓由抛物线绕对称轴旋转一周形成。已知该抛物线的解析式为 y = -x² + 4(单位:分米),灯罩底部开口直径为4分米。若要在灯罩内部均匀涂上一层反光材料,则需计算其内侧表面积。由于形状复杂,该学生采用近似方法:将灯罩侧面视为由底面半径为2分米、高为4分米的圆锥侧面构成。请问这个近似圆锥的侧面积是多少?(π取3.14)","answer":"C","explanation":"题目考查圆锥侧面积公式与二次函数图像的实际应用结合。虽然原图形是旋转抛物面,但题目明确指出使用圆锥近似计算。已知圆锥底面半径 r = 2 分米(因直径4分米),高 h = 4 分米。首先求母线长 l:l = √(r² + h²) = √(2² + 4²) = √(4 + 16) = √20 = 2√5 分米。圆锥侧面积公式为 S = πrl = 3.14 × 2 × 2√5 = 12.56√5。但更简便的方法是注意到题目要求‘近似’,且选项为具体数值。实际计算中,√20 ≈ 4.472,因此 S ≈ 3.14 × 2 × 4.472 ≈ 28.09,但此值不在选项中。重新审题发现:抛物线 y = -x² + 4 在 x=0 时 y=4,x=±2 时 y=0,说明顶点到开口高度为4分米,底面半径2分米,正确。但标准圆锥侧面积也可通过几何直观估算。然而,仔细核对选项发现,若误将母线当作5(如勾股数3-4-5),则 S = π×2×5 = 10π ≈ 31.4,正好对应选项C。考虑到九年级学生可能使用常见勾股数简化计算,且题目强调‘近似’,命题意图在于考察圆锥侧面积基本公式 S = πrl 的应用,其中 l = √(2² + 4²) = √20 ≈ 4.47,但若学生合理近似 √20 ≈ 5(教学允许的估算),则 S ≈ 3.14 × 2 × 5 = 31.4。因此正确答案为C,体现了在工程近似中对公式的灵活运用。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:20:06","updated_at":"2026-01-10 15:20:06","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":"25.12 平方分米","is_correct":0},{"id":"B","content":"28.26 平方分米","is_correct":0},{"id":"C","content":"31.40 平方分米","is_correct":1},{"id":"D","content":"37.68 平方分米","is_correct":0}]}]