习题练习

[{"id":440,"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：20，25，30，35，40，45，50。如果将这组数据按从小到大的顺序排列后，位于正中间的那个数称为中位数。那么这组数据的中位数是多少？","answer":"B","explanation":"题目给出了一组7个数据：20，25，30，35，40，45，50。由于数据个数是奇数（7个），中位数就是排序后位于正中间的那个数，即第(7+1)\/2 = 4个数。将数据从小到大排列后，第4个数是35。因此，这组数据的中位数是35。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:41:12","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":"A","content":"30","is_correct":0},{"id":"B","content":"35","is_correct":1},{"id":"C","content":"40","is_correct":0},{"id":"D","content":"45","is_correct":0}]},{"id":2382,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化活动中，学校计划在一块直角三角形的空地上铺设草皮。已知该直角三角形的两条直角边长度分别为√12米和√27米。为了计算所需草皮的面积，一名学生需要先化简边长并应用勾股定理求出斜边长度，再计算面积。请问该直角三角形的面积是多少平方米？","answer":"A","explanation":"首先化简两条直角边：√12 = √(4×3) = 2√3，√27 = √(9×3) = 3√3。直角三角形的面积公式为(1\/2)×直角边1×直角边2，因此面积为(1\/2)×2√3×3√3 = (1\/2)×6×3 = (1\/2)×18 = 9（平方米）。注意题目仅要求面积，无需计算斜边。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:39:53","updated_at":"2026-01-10 11:39:53","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":"9","is_correct":1},{"id":"B","content":"6√3","is_correct":0},{"id":"C","content":"18","is_correct":0},{"id":"D","content":"9√3","is_correct":0}]},{"id":2530,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生投掷一枚均匀的六面骰子，连续投掷两次。两次点数之和为偶数的概率是多少？","answer":"C","explanation":"一枚均匀的六面骰子，每次投掷结果为1至6中的任意一个整数，且每个点数出现的概率相等。连续投掷两次，总共有6×6=36种等可能的结果。两次点数之和为偶数的情况有两种：两次都是奇数，或两次都是偶数。骰子上的奇数有1、3、5，共3个；偶数有2、4、6，也是3个。两次都是奇数的情况有3×3=9种，两次都是偶数的情况也有3×3=9种，因此和为偶数的总情况数为9+9=18种。所以概率为18\/36=1\/2。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:16:42","updated_at":"2026-01-10 16:16:42","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\/4","is_correct":0},{"id":"B","content":"1\/3","is_correct":0},{"id":"C","content":"1\/2","is_correct":1},{"id":"D","content":"2\/3","is_correct":0}]},{"id":1910,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织植树活动，计划将一批树苗平均分给若干小组。如果每组分配5棵树苗，则剩余3棵；如果每组分配6棵树苗，则最后一组不足3棵但至少有1棵。已知小组数量为整数，且树苗总数不超过50棵，则该班级最多可能有多少个小组？","answer":"B","explanation":"设小组数量为x（x为正整数），树苗总数为y。根据题意：\n\n1. 每组5棵，剩3棵：y = 5x + 3；\n2. 每组6棵时，最后一组不足3棵但至少有1棵，说明前(x−1)组每组6棵，最后一组有1、2棵，即：\n 6(x−1) + 1 ≤ y < 6(x−1) + 3\n 化简得：6x − 5 ≤ y < 6x − 3\n\n将y = 5x + 3代入不等式：\n6x − 5 ≤ 5x + 3 < 6x − 3\n\n解左边：6x − 5 ≤ 5x + 3 → x ≤ 8\n解右边：5x + 3 < 6x − 3 → 3 + 3 < x → x > 6\n\n所以x的取值范围是：6 < x ≤ 8，即x = 7 或 8\n\n又因为树苗总数不超过50棵：y = 5x + 3 ≤ 50 → 5x ≤ 47 → x ≤ 9.4，满足x=7和x=8\n\n当x=8时，y = 5×8 + 3 = 43\n验证第二种分法：前7组每组6棵，共42棵，最后一组43−42=1棵，符合“不足3棵但至少有1棵”\n\n当x=9时，y=48，但6×8 + 3 = 51 > 48，不满足y < 6x−3（即48 < 51成立），但检查分配：前8组48棵，最后一组0棵，不符合“至少有1棵”，故x=9不成立\n\n因此，满足所有条件的最大x为8。\n\n故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:11:51","updated_at":"2026-01-07 13:11:51","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":"7个","is_correct":0},{"id":"B","content":"8个","is_correct":1},{"id":"C","content":"9个","is_correct":0},{"id":"D","content":"10个","is_correct":0}]},{"id":1534,"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开展‘城市绿地规划’数学实践活动。活动要求学生在平面直角坐标系中设计一个矩形绿化区域，其四个顶点坐标均为整数，且满足以下条件：\n\n1. 矩形的一组对边平行于x轴，另一组对边平行于y轴；\n2. 矩形的周长为20个单位长度；\n3. 矩形的面积不小于24个单位面积；\n4. 矩形完全位于第一象限，且其左下角顶点位于原点(0, 0)；\n5. 设矩形的右上角顶点坐标为(x, y)，其中x和y均为正整数。\n\n现从所有满足上述条件的矩形中随机选取一个，求该矩形的面积恰好为24的概率。","answer":"解：\n\n由题意，矩形左下角顶点为(0, 0)，右上角顶点为(x, y)，其中x > 0，y > 0，且x、y均为正整数。\n\n因为矩形对边分别平行于坐标轴，所以其长为x，宽为y。\n\n根据条件2：周长为20，\n即：2(x + y) = 20 \n⇒ x + y = 10 \n（方程①）\n\n根据条件3：面积不小于24，\n即：xy ≥ 24 \n（不等式②）\n\n又x、y为正整数，且x + y = 10，我们可以列出所有满足方程①的正整数解：\n\n(x, y) 的可能组合为：\n(1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)\n\n计算每种组合的面积xy：\n1×9 = 9 < 24 → 不满足\n2×8 = 16 < 24 → 不满足\n3×7 = 21 < 24 → 不满足\n4×6 = 24 ≥ 24 → 满足\n5×5 = 25 ≥ 24 → 满足\n6×4 = 24 ≥ 24 → 满足\n7×3 = 21 < 24 → 不满足\n8×2 = 16 < 24 → 不满足\n9×1 = 9 < 24 → 不满足\n\n因此，满足所有条件的(x, y)组合有：\n(4,6), (5,5), (6,4)\n共3种。\n\n其中，面积恰好为24的有：(4,6) 和 (6,4)，共2种。\n\n注意：虽然(4,6)和(6,4)表示不同的矩形（长宽不同），但在坐标系中它们是不同的图形，应视为两个不同的矩形。\n\n因此，所求概率为：\n满足条件的矩形总数：3\n面积恰好为24的矩形数：2\n\n概率 = 2 \/ 3\n\n答：该矩形的面积恰好为24的概率是 2\/3。","explanation":"本题综合考查了平面直角坐标系、二元一次方程组、不等式与不等式组以及数据的整理与描述等知识点。解题关键在于：\n\n1. 利用矩形顶点坐标与边长的关系，将几何问题转化为代数问题；\n2. 由周长条件建立方程 x + y = 10；\n3. 由面积条件建立不等式 xy ≥ 24；\n4. 枚举所有满足方程的正整数解，并结合不等式筛选出符合条件的解；\n5. 在满足所有条件的样本空间中，计算目标事件（面积为24）发生的概率。\n\n本题难度较高，体现在需要综合运用多个知识点，并进行分类讨论与逻辑推理。同时，题目情境新颖，避免了传统应用题的套路，强调数学建模与数据分析能力，符合七年级数学课程的综合应用要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:17:55","updated_at":"2026-01-06 12:17:55","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":1950,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生测量一个不规则四边形花坛的四条边长，分别为3.5米、4.2米、5.1米和6.0米。若该花坛被一条对角线分成两个三角形，其中一个三角形的周长为12.8米，则另一个三角形的周长为______米。","answer":"6.0","explanation":"四边形总周长为3.5+4.2+5.1+6.0=18.8米。一个三角形周长为12.8米，包含两条边和对角线。另一三角形周长=总边长和−已知三角形中两条边+对角线=18.8−12.8=6.0米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:14:40","updated_at":"2026-01-07 14:14:40","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":2271,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上，点A表示的数是-4，点B表示的数是6。某学生在数轴上标出了点C，使得点C到点A的距离是点C到点B的距离的2倍。那么点C表示的数可能是多少？","answer":"D","explanation":"设点C表示的数为x。根据题意，点C到点A的距离为|x + 4|，点C到点B的距离为|x - 6|。由条件得：|x + 4| = 2|x - 6|。分情况讨论：当x ≥ 6时，x + 4 = 2(x - 6)，解得x = 16；当-4 ≤ x < 6时，x + 4 = 2(6 - x)，解得x = 16\/3；当x < -4时，-(x + 4) = 2(6 - x)，解得x = -16。经检验，x = -16和x = 16\/3均满足原方程，因此点C表示的数可能是-16或16\/3。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 16:09:15","updated_at":"2026-01-09 16:09:15","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","is_correct":0},{"id":"B","content":"8\/3","is_correct":0},{"id":"C","content":"16","is_correct":0},{"id":"D","content":"-16或16\/3","is_correct":1}]},{"id":1211,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校组织七年级学生参加数学实践活动，要求测量校园内一个不规则四边形花坛ABCD的面积。学生在平面直角坐标系中测得四个顶点的坐标分别为：A(0, 0)，B(4, 0)，C(5, 3)，D(1, 4)。为了验证测量数据的合理性，他们决定通过计算该四边形的面积来进行检验。已知在测量过程中，可能存在坐标误差，因此要求计算结果保留两位小数。请你根据所学知识，计算该四边形花坛的面积，并判断该四边形是否为凸四边形。","answer":"解：\n\n第一步：利用坐标计算四边形面积的常用方法是“分割法”或“坐标公式法”（鞋带公式）。这里采用坐标公式法（Shoelace Formula）。\n\n设四边形顶点按顺序为 A(x₁, y₁), B(x₂, y₂), C(x₃, y₃), D(x₄, y₄)，则面积为：\n\n面积 = ½ |x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁ - (y₁x₂ + y₂x₃ + y₃x₄ + y₄x₁)|\n\n代入坐标：\nA(0, 0), B(4, 0), C(5, 3), D(1, 4)\n\n计算第一部分：x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁\n= 0×0 + 4×3 + 5×4 + 1×0\n= 0 + 12 + 20 + 0 = 32\n\n计算第二部分：y₁x₂ + y₂x₃ + y₃x₄ + y₄x₁\n= 0×4 + 0×5 + 3×1 + 4×0\n= 0 + 0 + 3 + 0 = 3\n\n面积 = ½ |32 - 3| = ½ × 29 = 14.50\n\n所以，四边形花坛的面积为 14.50 平方单位。\n\n第二步：判断是否为凸四边形。\n\n判断方法：若四边形的所有内角都小于180度，或任意一条对角线都在四边形内部，则为凸四边形。\n\n我们可以通过向量叉积判断每条边的转向是否一致（即是否同向旋转）。\n\n计算各边向量：\n向量 AB = (4 - 0, 0 - 0) = (4, 0)\n向量 BC = (5 - 4, 3 - 0) = (1, 3)\n向量 CD = (1 - 5, 4 - 3) = (-4, 1)\n向量 DA = (0 - 1, 0 - 4) = (-1, -4)\n\n计算连续边的叉积（z分量）：\nAB × BC = 4×3 - 0×1 = 12 > 0\nBC × CD = 1×1 - 3×(-4) = 1 + 12 = 13 > 0\nCD × DA = (-4)×(-4) - 1×(-1) = 16 + 1 = 17 > 0\nDA × AB = (-1)×0 - (-4)×4 = 0 + 16 = 16 > 0\n\n所有叉积均为正，说明四边形顶点按逆时针顺序排列，且转向一致，因此是凸四边形。\n\n答：该四边形花坛的面积为 14.50 平方单位，且为凸四边形。","explanation":"本题综合考查了平面直角坐标系、几何图形初步和整式运算的知识。解题关键在于掌握利用坐标计算多边形面积的鞋带公式，并能通过向量叉积判断四边形的凹凸性。学生需要理解坐标与几何图形的关系，具备一定的代数运算能力和逻辑推理能力。题目设置了真实情境（测量花坛），要求精确计算并做出几何判断，体现了数学在实际问题中的应用，难度较高，适合学有余力的学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:21:53","updated_at":"2026-01-06 10:21:53","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":227,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生计算一个长方形花坛的面积，已知长为8米，宽为5米，那么这个花坛的面积是_平方米。","answer":"40","explanation":"长方形的面积计算公式是：面积 = 长 × 宽。题目中给出的长是8米，宽是5米，因此面积为 8 × 5 = 40 平方米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:49","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":2434,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"如图，在平面直角坐标系中，一次函数 y = -x + 4 的图像与 x 轴、y 轴分别交于点 A 和点 B。点 P 是线段 AB 上的一个动点，过点 P 作 x 轴的垂线，垂足为点 C，作 y 轴的垂线，垂足为点 D。当矩形 PCOD 的面积最大时，点 P 的坐标为（  ）。","answer":"B","explanation":"首先，求出一次函数 y = -x + 4 与坐标轴的交点。当 x = 0 时，y = 4，所以点 B 坐标为 (0, 4)；当 y = 0 时，x = 4，所以点 A 坐标为 (4, 0)。因此，线段 AB 上的任意点 P 可表示为 (x, -x + 4)，其中 0 ≤ x ≤ 4。\n\n点 P 向 x 轴作垂线，垂足 C 的坐标为 (x, 0)；向 y 轴作垂线，垂足 D 的坐标为 (0, -x + 4)。则矩形 PCOD 的顶点为 P(x, -x+4)、C(x,0)、O(0,0)、D(0,-x+4)，其长为 |x|，宽为 |-x+4|。由于在区间 [0,4] 上，x ≥ 0 且 -x+4 ≥ 0，故矩形面积为 S = x(4 - x) = -x² + 4x。\n\n这是一个关于 x 的二次函数，开口向下，最大值出现在顶点处。顶点横坐标为 x = -b\/(2a) = -4\/(2×(-1)) = 2。代入得 y = -2 + 4 = 2，所以点 P 坐标为 (2, 2)。\n\n因此，当矩形面积最大时，点 P 的坐标为 (2, 2)，正确答案为 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:02:02","updated_at":"2026-01-10 13:02:02","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, 3)","is_correct":0},{"id":"B","content":"(2, 2)","is_correct":1},{"id":"C","content":"(3, 1)","is_correct":0},{"id":"D","content":"(4, 0)","is_correct":0}]}]