习题练习

[{"id":1216,"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，要求学生测量校园内一个不规则花坛的边界，并用数学方法估算其面积。花坛的边界由五条线段组成，形成一个凸五边形ABCDE。学生们在平面直角坐标系中建立了模型，测得五个顶点的坐标分别为：A(0, 0)，B(4, 0)，C(6, 3)，D(3, 6)，E(0, 4)。为了估算面积，一名学生提出将五边形分割为三个三角形：△ABC、△ACD和△ADE。请根据该学生的分割方法，利用坐标几何知识，计算该五边形的面积。（提示：可使用向量叉积法或坐标法中的‘鞋带公式’，但需通过三角形面积公式逐步计算）","answer":"解：\n\n我们将五边形ABCDE分割为三个三角形：△ABC、△ACD和△ADE。利用平面直角坐标系中三角形面积的坐标公式：\n\n对于顶点为 (x₁, y₁)，(x₂, y₂)，(x₃, y₃) 的三角形，其面积为：\n\n面积 = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|\n\n第一步：计算△ABC的面积\nA(0, 0)，B(4, 0)，C(6, 3)\n\nS₁ = ½ |0×(0 - 3) + 4×(3 - 0) + 6×(0 - 0)|\n  = ½ |0 + 4×3 + 0| = ½ × 12 = 6\n\n第二步：计算△ACD的面积\nA(0, 0)，C(6, 3)，D(3, 6)\n\nS₂ = ½ |0×(3 - 6) + 6×(6 - 0) + 3×(0 - 3)|\n  = ½ |0 + 6×6 + 3×(-3)| = ½ |36 - 9| = ½ × 27 = 13.5\n\n第三步：计算△ADE的面积\nA(0, 0)，D(3, 6)，E(0, 4)\n\nS₃ = ½ |0×(6 - 4) + 3×(4 - 0) + 0×(0 - 6)|\n  = ½ |0 + 3×4 + 0| = ½ × 12 = 6\n\n第四步：求总面积\nS = S₁ + S₂ + S₃ = 6 + 13.5 + 6 = 25.5\n\n答：该五边形的面积为25.5平方单位。","explanation":"本题考查平面直角坐标系中多边形面积的坐标计算方法，属于几何与代数综合应用题。解题关键在于将不规则多边形合理分割为若干三角形，并运用坐标法中的三角形面积公式进行逐项计算。题目要求不使用直接套用鞋带公式，而是通过三角形分割的方式，训练学生的图形分析能力和坐标运算能力。该方法不仅巩固了平面直角坐标系的知识，还融合了整式运算（含绝对值与代数式化简），体现了数形结合的思想。难度较高，因涉及多个坐标点的代入、符号处理及多步运算，适合能力较强的七年级学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:23:18","updated_at":"2026-01-06 10:23:18","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":194,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明买了3支铅笔和2本笔记本，共花费18元。已知每本笔记本比每支铅笔贵3元。设每支铅笔的价格为x元，则下列方程正确的是（ ）","answer":"A","explanation":"题目中设每支铅笔的价格为x元，因为每本笔记本比每支铅笔贵3元，所以每本笔记本的价格为(x + 3)元。小明买了3支铅笔，总价为3x元；买了2本笔记本，总价为2(x + 3)元。根据总花费为18元，可列出方程：3x + 2(x + 3) = 18。因此，正确选项是A。其他选项错误地将笔记本价格设为比铅笔便宜，或混淆了数量与单价的关系。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:03:39","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":"3x + 2(x + 3) = 18","is_correct":1},{"id":"B","content":"3x + 2(x - 3) = 18","is_correct":0},{"id":"C","content":"3(x + 3) + 2x = 18","is_correct":0},{"id":"D","content":"3(x - 3) + 2x = 18","is_correct":0}]},{"id":243,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数的相反数时，误将该数加上了5，结果得到了8。那么这个数的正确相反数应该是____。","answer":"-3","explanation":"设这个数为x。根据题意，某学生误将x加上5得到8，即x + 5 = 8，解得x = 3。这个数的相反数是-3。因此，正确答案是-3。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:42:03","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":460,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"144度","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 17:48:43","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":1955,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学校七年级组织学生参加植树活动，计划在一条笔直的小路一侧每隔一定距离种一棵树。已知小路全长120米，起点和终点都种树，共种了13棵树。若每两棵相邻树之间的距离相等，且设这个距离为x米，则根据题意可列方程为：","answer":"A","explanation":"本题考查一元一次方程在实际问题中的应用，涉及植树问题中的间隔数与总长度的关系。已知小路全长120米，起点和终点都种树，共种了13棵树。在直线段上两端都种树的情况下，间隔数 = 树的数量 - 1。因此，有13 - 1 = 12个间隔。每个间隔距离为x米，总长度等于间隔数乘以每个间隔的距离，即12x = 120。选项A正确。其他选项错误地将树的数量或间隔数计算错误。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:46:45","updated_at":"2026-01-07 14:46:45","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":"12x = 120","is_correct":1},{"id":"B","content":"13x = 120","is_correct":0},{"id":"C","content":"11x = 120","is_correct":0},{"id":"D","content":"14x = 120","is_correct":0}]},{"id":1859,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划图中，两条平行轨道AB和CD被一条斜向联络线EF所截，形成多个角。已知∠1与∠2是同旁内角，且∠1的度数是∠2的2倍少30°。同时，在平面直角坐标系中，点A的坐标为(2, 3)，点B在x轴正方向上，且AB的长度为5个单位。若将线段AB向右平移3个单位，再向下平移2个单位得到线段A'B'，求：(1) ∠1和∠2的度数；(2) 点A'的坐标；(3) 若点C是线段A'B'的中点，求点C的坐标。","answer":"(1) 设∠2的度数为x°，则∠1 = (2x - 30)°。\n因为AB∥CD，EF为截线，∠1与∠2是同旁内角，所以∠1 + ∠2 = 180°。\n列方程：(2x - 30) + x = 180\n3x - 30 = 180\n3x = 210\nx = 70\n所以∠2 = 70°，∠1 = 2×70 - 30 = 110°。\n\n(2) 点A(2, 3)向右平移3个单位，横坐标加3，得(5, 3)；再向下平移2个单位，纵坐标减2，得(5, 1)。\n所以点A'的坐标为(5, 1)。\n\n(3) 点B在x轴正方向上，且AB = 5，A(2, 3)，设B(x, 0)，由距离公式：\n√[(x - 2)² + (0 - 3)²] = 5\n(x - 2)² + 9 = 25\n(x - 2)² = 16\nx - 2 = ±4\nx = 6 或 x = -2\n因为B在x轴正方向上，且从A向右延伸更合理（结合平移方向），取x = 6，即B(6, 0)。\n将B(6, 0)同样平移：向右3单位得(9, 0)，向下2单位得(9, -2)，即B'(9, -2)。\n点C是A'B'的中点，A'(5, 1)，B'(9,...","explanation":"本题综合考查平行线性质、一元一次方程、平面直角坐标系中的平移与坐标计算、中点公式。第(1)问利用同旁内角互补建立方程求解角度；第(2)问考查图形平移对坐标的影响；第(3)问需先通过距离公式确定点B坐标，再经平移得B'，最后用中点公式求C。关键步骤是正确理解几何关系与坐标变换规则，并准确进行代数运算。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:39:21","updated_at":"2026-01-07 09:39:21","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":2541,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆形花坛的半径为6米，现计划在花坛中心安装一个自动旋转喷水器，喷水范围形成一个扇形，其圆心角为θ（0° < θ < 360°）。已知喷水覆盖区域的面积S（平方米）与圆心角θ（度）之间的关系为 S = (θ\/360) × π × 6²。若要求喷水覆盖面积恰好为花坛总面积的1\/3，则θ的值应为多少？","answer":"B","explanation":"首先计算整个花坛的面积：π × 6² = 36π 平方米。题目要求喷水覆盖面积为总面积的1\/3，即 (1\/3) × 36π = 12π 平方米。根据题中给出的公式 S = (θ\/360) × 36π，代入 S = 12π 得：12π = (θ\/360) × 36π。两边同时除以π，得到 12 = (θ\/360) × 36。两边同除以12，得 1 = (θ\/360) × 3，即 θ\/360 = 1\/3，解得 θ = 120°。因此正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 16:50:58","updated_at":"2026-01-10 16:50:58","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":"90°","is_correct":0},{"id":"B","content":"120°","is_correct":1},{"id":"C","content":"150°","is_correct":0},{"id":"D","content":"180°","is_correct":0}]},{"id":745,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级大扫除中，某学生负责统计各小组打扫教室所用时间。第一组用了 0.75 小时，第二组用了 45 分钟，第三组用了 3\/4 小时。将三个小组所用时间统一换算成分钟，并按从小到大的顺序排列，排在中间的时间是____分钟。","answer":"45","explanation":"首先将各组时间统一换算为分钟：第一组 0.75 小时 = 0.75 × 60 = 45 分钟；第二组已经是 45 分钟；第三组 3\/4 小时 = 3\/4 × 60 = 45 分钟。三组时间均为 45 分钟，按从小到大排列后，中间的值仍然是 45 分钟。本题考查有理数与时间单位换算，以及数据的整理与排序，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:16:04","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":310,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生记录了连续5天的气温变化情况，每天的气温分别为-2℃、0℃、3℃、-1℃、4℃。这5天气温的平均值是多少？","answer":"A","explanation":"要计算这5天气温的平均值，首先将所有气温相加：(-2) + 0 + 3 + (-1) + 4 = 4。然后将总和除以天数5，得到平均值：4 ÷ 5 = 0.8。因此，这5天气温的平均值是0.8℃。本题考查有理数的加减运算以及数据的整理与描述中的平均数计算，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:35:35","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":"0.8℃","is_correct":1},{"id":"B","content":"1.0℃","is_correct":0},{"id":"C","content":"1.2℃","is_correct":0},{"id":"D","content":"1.4℃","is_correct":0}]},{"id":801,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中，某学生收集废旧电池的数量比另一名学生的3倍少5节。如果两人一共收集了27节电池，那么收集较少的学生收集了___节电池。","answer":"8","explanation":"设收集较少的学生收集了x节电池，则另一名学生收集了(3x - 5)节。根据题意，两人共收集27节，列出方程：x + (3x - 5) = 27。化简得4x - 5 = 27，解得4x = 32，x = 8。因此，收集较少的学生收集了8节电池。本题考查一元一次方程的实际应用，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":"2025-12-30 00:16:45","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":[]}]