习题练习

[{"id":1426,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生参加数学实践活动，要求学生利用平面直角坐标系设计一个‘校园寻宝’路线。已知校园平面图上以正门为原点O(0,0)，向东为x轴正方向，向北为y轴正方向。第一个藏宝点A位于(3,4)，第二个藏宝点B位于(-2,6)，第三个藏宝点C位于(5,-3)。一名学生从正门出发，依次经过A、B、C三个点后返回正门。若该学生每走1个单位长度需要消耗2分钟，且在每个藏宝点停留整理数据的时间为5分钟。已知该学生总共用时不超过150分钟，问：该学生是否能在规定时间内完成整个寻宝任务？如果不能，最多可以跳过几个藏宝点（只能跳过B或C，不能跳过A），才能确保总时间不超过150分钟？请通过计算说明。","answer":"首先计算从原点O(0,0)到A(3,4)的距离：\n距离OA = √[(3-0)² + (4-0)²] = √(9+16) = √25 = 5\n\n从A(3,4)到B(-2,6)的距离：\n距离AB = √[(-2-3)² + (6-4)²] = √[(-5)² + 2²] = √(25+4) = √29 ≈ 5.385\n\n从B(-2,6)到C(5,-3)的距离：\n距离BC = √[(5+2)² + (-3-6)²] = √[7² + (-9)²] = √(49+81) = √130 ≈ 11.402\n\n从C(5,-3)返回原点O(0,0)的距离：\n距离CO = √[(5-0)² + (-3-0)²] = √(25+9) = √34 ≈ 5.831\n\n总行走距离 = OA + AB + BC + CO ≈ 5 + 5.385 + 11.402 + 5.831 = 27.618（单位长度）\n\n行走时间 = 27.618 × 2 ≈ 55.236（分钟）\n\n停留时间：共3个藏宝点，每个停留5分钟，总停留时间 = 3 × 5 = 15（分钟）\n\n总用时 ≈ 55.236 + 15 = 70.236（分钟）\n\n由于70.236 < 150，因此该学生能在规定时间内完成整个寻宝任务。\n\n但题目要求判断“是否能在规定时间内完成”，并进一步问“如果不能，最多可以跳过几个点”。然而根据计算，实际用时远小于150分钟，因此无需跳过任何点。\n\n但为严谨起见，我们验证是否存在理解偏差：题目中“总共用时不超过150分钟”是上限，而实际仅需约70分钟，远低于限制。\n\n因此结论是：该学生能在规定时间内完成整个寻宝任务，不需要跳过任何藏宝点。\n\n答案：能完成，不需要跳过任何点。","explanation":"本题综合考查了平面直角坐标系中两点间距离公式、实数的运算、近似计算以及实际问题的建模能力。解题关键在于正确运用距离公式√[(x₂−x₁)²+(y₂−y₁)²]计算各段路径长度，再结合时间与距离的关系（每单位2分钟）和停留时间进行总时间估算。虽然题目设置了‘是否超时’和‘跳过点’的复杂情境，但通过精确计算发现实际耗时远低于限制，体现了数学建模中数据验证的重要性。本题难度较高，因其融合了多个知识点并要求学生进行多步推理和实际判断，符合困难级别要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:34:57","updated_at":"2026-01-06 11:34:57","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":747,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级图书角统计中，某学生发现科普类书籍占总数的30%，文学类书籍比科普类多20本，其余40本是历史类书籍。那么图书角共有____本书。","answer":"100","explanation":"设图书角总共有x本书。根据题意，科普类书籍占30%，即0.3x本；文学类比科普类多20本，即(0.3x + 20)本；历史类有40本。三类书籍总和等于总数，因此可列方程：0.3x + (0.3x + 20) + 40 = x。化简得：0.6x + 60 = x，移项得：60 = 0.4x，解得x = 150 ÷ 1.5 = 100。所以图书角共有100本书。本题考查一元一次方程的实际应用，结合百分数与数据整理背景，符合七年级知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:21:52","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":470,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学每天用于课外阅读的时间（单位：分钟），并将数据整理如下：15, 20, 25, 30, 35, 40, 45。如果再加入一个数据后，这组数据的中位数变为30，那么加入的这个数据可能是多少？","answer":"B","explanation":"原数据有7个数，按从小到大排列为：15, 20, 25, 30, 35, 40, 45。中位数是第4个数，即30。加入一个数据后，总共有8个数，中位数是第4个和第5个数的平均数。要使中位数为30，则第4个和第5个数的平均数必须为30。若加入30，则新数据为：15, 20, 25, 30, 30, 35, 40, 45，此时第4个数是30，第5个数也是30，中位数为(30+30)÷2=30，符合条件。其他选项加入后，中位数均不等于30。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:53:54","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":"25","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"40","is_correct":0}]},{"id":210,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生用一根长为20厘米的铁丝围成一个长方形，若长方形的长为6厘米，则宽为_空白处_厘米。","answer":"4","explanation":"长方形的周长公式为：周长 = 2 × (长 + 宽)。已知周长为20厘米，长为6厘米，代入公式得：20 = 2 × (6 + 宽)。两边同时除以2，得10 = 6 + 宽，因此宽 = 10 - 6 = 4厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:48","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":2419,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个轴对称的三角形花坛，设计图显示该花坛为等腰三角形，底边长为8米，两腰相等。施工过程中，测量员在底边中点处垂直向上挖掘一条深沟，用于铺设灌溉管道，测得沟深为3米，恰好到达顶点。若花坛的对称轴即为这条垂直线，则该花坛的面积为多少平方米？","answer":"C","explanation":"本题综合考查轴对称、等腰三角形性质和三角形面积计算。花坛为等腰三角形，底边为8米，对称轴为底边的垂直平分线，且从底边中点垂直向上3米到达顶点，说明高为3米。等腰三角形的高将底边平分，因此底边一半为4米，高为3米，符合勾股定理中直角三角形的两直角边（3和4），斜边为5米，即腰长为5米，但本题不需求腰长。三角形面积公式为：面积 = (底 × 高) ÷ 2 = (8 × 3) ÷ 2 = 24 平方米。因此正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:30:12","updated_at":"2026-01-10 12:30:12","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":"A","content":"12","is_correct":0},{"id":"B","content":"18","is_correct":0},{"id":"C","content":"24","is_correct":1},{"id":"D","content":"36","is_correct":0}]},{"id":1868,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级学生参加数学实践活动，需在平面直角坐标系中设计一个轴对称图形。已知图形由三个点 A、B、C 构成，其中点 A 的坐标为 (2, 3)，点 B 在 x 轴上，点 C 在 y 轴上。若该图形关于直线 y = x 对称，且点 B 与点 C 到原点的距离之和为 10，求点 B 和点 C 的坐标。","answer":"设点 B 的坐标为 (a, 0)，点 C 的坐标为 (0, b)，其中 a 和 b 为实数。\n\n由于图形关于直线 y = x 对称，点 A(2, 3) 关于 y = x 的对称点为 A'(3, 2)，该点也应在图形上。\n\n因为图形由 A、B、C 三点构成，且整体关于 y = x 对称，所以点 B 和点 C 必须互为关于直线 y = x 的对称点。即：若 B 为 (a, 0)，则其对称点为 (0, a)，因此点 C 的坐标应为 (0, a)，即 b = a。\n\n同理，若 C 为 (0, b)，其对称点为 (b, 0)，则点 B 应为 (b, 0)，即 a = b。\n\n综上，可得 a = b。\n\n根据题意，点 B 到原点的距离为 |a|，点 C 到原点的距离为 |b| = |a|，因此距离之和为：\n|a| + |a| = 2|a| = 10\n解得：|a| = 5 ⇒ a = 5 或 a = -5\n\n因此，点 B 和点 C 的坐标有两种可能：\n情况一：a = 5 ⇒ B(5, 0)，C(0, 5)\n情况二：a = -5 ⇒ B(-5, 0)，C(0, -5)\n\n验证对称性：\n- 点 B...","explanation":"本题结合平面直角坐标系与轴对称性质，考查对称点坐标关系及绝对值的实际应用。关键突破口是理解图形关于 y = x 对称意味着任意一点的对称点也应在图形上，从而推出 B 与 C 必须互为对称点，进而得到它们的坐标关系。再利用距离公式建立方程求解。难点在于将几何对称性转化为代数关系，并正确处理绝对值方程。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:40:43","updated_at":"2026-01-07 09:40:43","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":1800,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织一次数学知识竞赛，参赛学生的成绩被整理成频数分布表如下：\n\n| 成绩区间（分） | 频数（人） |\n|----------------|------------|\n| 60 ≤ x < 70 | 5 |\n| 70 ≤ x < 80 | 12 |\n| 80 ≤ x < 90 | 18 |\n| 90 ≤ x ≤ 100 | 10 |\n\n已知该班参赛学生总人数为45人，且所有成绩均为整数。若将成绩按从高到低排列，则第23名学生的成绩最可能落在哪个区间？","answer":"C","explanation":"本题考查数据的整理与描述中的频数分布及中位数思想的应用。总人数为45人，将成绩从高到低排列，第23名是正中间的位置，即中位数所在位置。\n\n首先计算累计频数（从高分段开始累加）：\n- 90 ≤ x ≤ 100：10人（第1~10名）\n- 80 ≤ x < 90：18人 → 累计10 + 18 = 28人（第11~28名）\n\n因此，第23名落在第11到第28名之间，即属于“80 ≤ x < 90”这一组。\n\n虽然不能确定具体分数，但根据分组数据的中位数估计方法，第23名最可能落在80到90分区间内。\n\n故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:13:28","updated_at":"2026-01-06 16:13:28","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":"60 ≤ x < 70","is_correct":0},{"id":"B","content":"70 ≤ x < 80","is_correct":0},{"id":"C","content":"80 ≤ x < 90","is_correct":1},{"id":"D","content":"90 ≤ x ≤ 100","is_correct":0}]},{"id":403,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD，已知点A的坐标为(1, 2)，点B的坐标为(4, 2)，点C的坐标为(4, 5)，点D的坐标为(1, 5)。该学生想判断这个四边形的形状，以下说法正确的是：","answer":"B","explanation":"首先根据坐标确定四边形各边的位置：AB从(1,2)到(4,2)，是水平线段，长度为3；BC从(4,2)到(4,5)，是垂直线段，长度为3；CD从(4,5)到(1,5)，是水平线段，长度为3；DA从(1,5)到(1,2)，是垂直线段，长度为3。因此四条边长度均为3，且相邻边互相垂直，说明四个角都是直角。虽然四条边相等且角为直角，看似是正方形，但进一步观察发现，正方形是特殊的矩形，而题目中并未强调‘邻边相等’这一正方形的关键特征是否被学生验证。然而，根据坐标可直接看出：对边平行（AB∥CD，AD∥BC），且四个角均为90度，符合矩形的定义。同时，由于所有边长也相等，它实际上是一个正方形，但选项中D的描述虽然正确，但‘正方形’属于更特殊的分类，而题目要求选择‘正确’的说法，B和D都看似合理。但考虑到七年级学生对图形的初步认识，通常先掌握矩形定义（直角+对边相等），且题目中坐标明确显示水平与垂直边构成直角，最直接、稳妥的判断是矩形。此外，选项D虽数学上正确，但‘正方形’需额外验证邻边相等，而题目未突出这一点。综合教学重点和选项表述，B为最符合七年级认知水平的正确答案。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:17:13","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":"这是一个平行四边形，因为有两组对边分别平行","is_correct":0},{"id":"B","content":"这是一个矩形，因为四个角都是直角且对边相等","is_correct":1},{"id":"C","content":"这是一个菱形，因为四条边长度都相等","is_correct":0},{"id":"D","content":"这是一个正方形，因为四条边相等且四个角都是直角","is_correct":0}]},{"id":422,"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数据时，发现一周内每天阅读时间（单位：分钟）分别为：25，30，28，35，32，27，33。为了分析阅读时间的分布情况，该学生计算了这组数据的平均数。请问这组数据的平均数是多少？","answer":"C","explanation":"要计算这组数据的平均数，需要将所有数据相加，然后除以数据的个数。具体步骤如下：首先，将每天的阅读时间相加：25 + 30 + 28 + 35 + 32 + 27 + 33 = 210（分钟）。然后，用总和除以天数（7天）：210 ÷ 7 = 30（分钟）。因此，这组数据的平均数是30分钟，正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:32:46","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":"28分钟","is_correct":0},{"id":"B","content":"29分钟","is_correct":0},{"id":"C","content":"30分钟","is_correct":1},{"id":"D","content":"31分钟","is_correct":0}]},{"id":2520,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛，其边缘由一段圆弧和两条半径围成，形成一个扇形区域。已知该扇形的圆心角为60°，面积为6π平方米。若在该扇形区域内接一个最大的等边三角形（三个顶点均在扇形边界上），则这个等边三角形的边长是多少？","answer":"A","explanation":"首先，根据扇形面积公式 S = (θ\/360°) × πr²，其中θ = 60°，S = 6π，代入得：6π = (60\/360) × πr² → 6π = (1\/6)πr² → r² = 36 → r = 6米。因此扇形半径为6米。由于圆心角为60°，若将扇形的两条半径端点与圆心连接，可构成一个边长为6米的等边三角形（因为两边为半径，夹角60°，三边相等）。此三角形完全位于扇形内，且是内接于该扇形中最大的等边三角形（任何其他构造都会导致边长更短或超出边界）。故该等边三角形边长为6米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:56:03","updated_at":"2026-01-10 15:56:03","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米","is_correct":1},{"id":"B","content":"3√3米","is_correct":0},{"id":"C","content":"4√3米","is_correct":0},{"id":"D","content":"2√6米","is_correct":0}]}]