习题练习

[{"id":1317,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动，要求测量并绘制校园内一个不规则多边形花坛的平面图。已知该花坛的边界由五条线段首尾相连组成，形成一个凸五边形。测量小组在平面直角坐标系中确定了五个顶点的坐标分别为 A(2, 3)、B(5, 7)、C(9, 6)、D(8, 2)、E(4, 1)。为了计算花坛的面积，一名学生采用‘分割法’，将五边形 ABCDE 分割为一个三角形和一个梯形。他首先连接对角线 AC，将原五边形分为四边形 ABCE 和三角形 ACD，但发现计算复杂。后来他改用另一种方法：利用坐标几何中的‘鞋带公式’（Shoelace Formula）直接计算多边形面积。请根据该学生的方法，使用鞋带公式计算该五边形花坛的面积，并验证结果是否合理。此外，若每平方米种植 4 株花，且预算允许最多种植 120 株，问该花坛是否适合按标准种植？请说明理由。","answer":"解题步骤如下：\n\n第一步：列出五边形顶点坐标，并按顺时针或逆时针顺序排列（此处按 A→B→C→D→E→A 顺序）：\nA(2, 3)\nB(5, 7)\nC(9, 6)\nD(8, 2)\nE(4, 1)\n回到 A(2, 3)\n\n第二步：应用鞋带公式计算面积。\n鞋带公式为：\n面积 = 1\/2 |Σ(x_i * y_{i+1}) - Σ(y_i * x_{i+1})|\n\n计算第一组乘积和（x_i * y_{i+1}）：\n2×7 = 14\n5×6 = 30\n9×2 = 18\n8×1 = 8\n4×3 = 12\n总和 = 14 + 30 + 18 + 8 + 12 = 82\n\n计算第二组乘积和（y_i * x_{i+1}）：\n3×5 = 15\n7×9 = 63\n6×8 = 48\n2×4 = 8\n1×2 = 2\n总和 = 15 + 63 + 48 + 8 + 2 = 136\n\n第三步：代入公式求面积：\n面积 = 1\/2 × |82 - 136| = 1\/2 × |-54| = 1\/2 × 54 = 27\n\n因此，五边形花坛的面积为 27 平方米。\n\n第四步：计算可种植的花株数量。\n每平方米种植 4 株，则总株数 = 27 × 4 = 108 株。\n\n第五步：判断是否适合种植。\n预算允许最多种植 120 株，而实际需要 108 株，108 < 120，因此在预算范围内。\n\n答：该花坛的面积为 27 平方米，最多可种植 108 株花，未超过预算上限，适合按标准种植。","explanation":"本题综合考查了平面直角坐标系、多边形面积计算（鞋带公式）、有理数运算及实际应用能力。鞋带公式是七年级学生在学习坐标系后可以拓展掌握的一种高效计算任意多边形面积的方法，尤其适用于顶点坐标已知的情况。题目通过真实情境引入，要求学生正确排序顶点、准确进行有理数乘法和加减运算，并最终结合不等式思想（108 ≤ 120）做出合理判断。解题关键在于理解公式的结构、避免符号错误，并能将数学结果应用于实际问题决策中，体现了数学建模的核心素养。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:53:04","updated_at":"2026-01-06 10:53:04","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":1757,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动，要求将学生分成若干小组，每组人数相同。已知若每组安排6人，则最后一组只有4人；若每组安排8人，则最后一组只有6人；若每组安排9人，则最后一组只有7人。问：该校七年级参加活动的学生至少有多少人？请通过建立方程或不等式模型，并结合整除性质进行分析求解。","answer":"设参加活动的学生总人数为x人。\n\n根据题意，可列出以下同余关系：\n\nx ≡ 4 (mod 6) ——（1）\n\nx ≡ 6 (mod 8) ——（2）\n\nx ≡ 7 (mod 9) ——（3）\n\n观察发现，每个余数都比除数少2：\n\n即：x + 2 ≡ 0 (mod 6)\n\nx + 2 ≡ 0 (mod 8)\n\nx + 2 ≡ 0 (mod 9)\n\n说明 x + 2 是 6、8、9 的公倍数。\n\n先求6、8、9的最小公倍数：\n\n分解质因数：\n\n6 = 2 × 3\n\n8 = 2³\n\n9 = 3²\n\n取各质因数最高次幂：2³ × 3² = 8 × 9 = 72\n\n所以 x + 2 是72的倍数，即 x + 2 = 72k（k为正整数）\n\n因此 x = 72k - 2\n\n当k = 1时，x = 72 - 2 = 70\n\n验证：\n\n70 ÷ 6 = 11组余4人 → 符合（1）\n\n70 ÷ 8 = 8组余6人 → 符合（2）\n\n70 ÷ 9 = 7组余7人 → 符合（3）\n\n当k = 2时，x = 144 - 2 = 142，也满足，但题目要求“至少”有多少人。\n\n所以最小满足条件的x为70。\n\n答：该校七年级参加活动的学生至少有70人。","explanation":"本题考查学生对同余概念的理解与转化能力，结合整除性质和一元一次方程建模思想。关键在于发现三个条件中余数与除数的关系：余数均为除数减2，从而转化为x + 2是6、8、9的公倍数。通过求最小公倍数得到最小解。题目融合了整数的整除性、最小公倍数、方程建模与逻辑推理，属于典型的困难级别应用题，要求学生具备较强的观察力与抽象思维能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:33:35","updated_at":"2026-01-06 14:33:35","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":271,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"6人","answer":"答案待完善","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 15:30:11","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":2476,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图，在平面直角坐标系中，点A(0, 4)，点B(6, 0)，点C在x轴正半轴上，且△ABC是以AB为斜边的等腰直角三角形。点D是线段AC的中点，点E在y轴上，使得△BDE是以BD为底边的等腰三角形，且DE = BE。直线l经过点D和点E，与x轴交于点F。已知某学生测量了五组实验数据，记录了F点的横坐标x与对应线段DF的长度d，如下表所示：\\n\\n| x | d |\\n|-----|--------|\\n| 2.8 | 3.16 |\\n| 3.0 | 3.00 |\\n| 3.2 | 2.83 |\\n| 3.4 | 2.65 |\\n| 3.6 | 2.45 |\\n\\n(1) 求点C的坐标；\\n(2) 求直线l的解析式；\\n(3) 利用勾股定理和一次函数性质，验证当x = 3时，d = 3是否成立；\\n(4) 根据表中数据，用最小二乘法思想估算当d = 2.00时，x的近似值（保留两位小数）。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:57:40","updated_at":"2026-01-10 14:57:40","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":2527,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在操场上观察旗杆的投影。已知旗杆高6米，太阳光线与地面形成的仰角为30°，则此时旗杆在地面的投影长度为多少米？","answer":"A","explanation":"本题考查锐角三角函数的应用。旗杆、投影和太阳光线构成一个直角三角形，其中旗杆为对边，投影为邻边，太阳光线与地面的夹角为30°。根据正切函数定义：tan(30°) = 对边 \/ 邻边 = 6 \/ x。因为 tan(30°) = √3 \/ 3，所以有 √3 \/ 3 = 6 \/ x，解得 x = 6 \/ (√3 \/ 3) = 6 × 3 \/ √3 = 18 \/ √3。将分母有理化：18 \/ √3 = (18√3) \/ 3 = 6√3。因此，旗杆的投影长度为6√3米，正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:11:59","updated_at":"2026-01-10 16:11:59","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":"A","content":"6√3","is_correct":1},{"id":"B","content":"3√3","is_correct":0},{"id":"C","content":"12","is_correct":0},{"id":"D","content":"2√3","is_correct":0}]},{"id":557,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保活动中，某学校七年级学生收集了可回收垃圾的重量数据如下：塑料瓶 2.5 千克，废纸 3.8 千克，金属罐 1.2 千克，玻璃瓶 4.1 千克。请问这些可回收垃圾的总重量是多少千克？","answer":"B","explanation":"本题考查的是有理数的加法运算，属于数据的收集与整理范畴。题目给出了四种可回收垃圾的重量：塑料瓶 2.5 千克，废纸 3.8 千克，金属罐 1.2 千克，玻璃瓶 4.1 千克。要求总重量，只需将这些小数相加：2.5 + 3.8 = 6.3；6.3 + 1.2 = 7.5；7.5 + 4.1 = 11.6。因此，总重量为 11.6 千克，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 19:21:34","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":"10.6 千克","is_correct":0},{"id":"B","content":"11.6 千克","is_correct":1},{"id":"C","content":"12.6 千克","is_correct":0},{"id":"D","content":"13.6 千克","is_correct":0}]},{"id":655,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保主题活动中，某学生记录了连续5天每天节约用水的升数，分别为：3.5升、4.2升、3.8升、4.0升、3.6升。这5天平均每天节约用水______升。","answer":"3.82","explanation":"要计算平均每天节约用水的升数，需将5天的用水量相加后除以天数。计算过程为：(3.5 + 4.2 + 3.8 + 4.0 + 3.6) ÷ 5 = 19.1 ÷ 5 = 3.82（升）。本题考查的是数据的收集、整理与描述中的平均数计算，属于七年级数学中数据处理的基础知识，难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:13:04","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":1826,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一块直角三角形纸片的三边长度，分别为5 cm、12 cm和13 cm。他将其沿一条直线折叠，使得直角顶点恰好落在斜边的中点上。折叠后，原直角三角形被分成了两个部分。若其中一个部分的周长为15 cm，则另一个部分的周长是多少？","answer":"B","explanation":"首先，根据勾股定理验证：5² + 12² = 25 + 144 = 169 = 13²，因此这是一个直角三角形，直角位于5 cm和12 cm两边之间，斜边为13 cm。斜边中点将斜边分为两段，每段长6.5 cm。折叠时，直角顶点（设为点C）被折到斜边AB的中点M上，折痕是对称轴，即CM的垂直平分线。折叠后，点C与点M重合，形成轴对称图形。折叠线将三角形分成两个部分，其中一个部分的周长已知为15 cm。由于折叠是轴对称操作，折痕上的点不动，而点C移动到M，因此其中一个部分包含原三角形的一部分边和折痕，另一个部分也类似。通过分析可知，折叠后形成的两个部分共享折痕，且其中一个部分的边界包括原三角形的两条直角边的一部分和折痕，另一个部分包括斜边的一半、折痕和另一段路径。利用几何对称性和周长守恒思想，整个原三角形周长为5 + 12 + 13 = 30 cm。折叠不改变总边长分布，但折痕被重复计算。设折痕长为x，则两个部分的周长之和为30 + 2x（因为折痕在两个部分中各出现一次）。已知一个部分周长为15，设另一个为y，则15 + y = 30 + 2x → y = 15 + 2x。通过几何分析或构造辅助线可求得折痕长度约为2.5 cm（具体可通过坐标法或相似三角形得出），代入得y ≈ 15 + 5 = 20 cm。因此另一个部分的周长为20 cm。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:30:04","updated_at":"2026-01-06 16:30:04","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":"18 cm","is_correct":0},{"id":"B","content":"20 cm","is_correct":1},{"id":"C","content":"22 cm","is_correct":0},{"id":"D","content":"24 cm","is_correct":0}]},{"id":471,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中，某班级共收集了120份有效问卷。统计结果显示，喜欢垃圾分类宣传活动的学生人数是喜欢节水宣传活动的2倍，而喜欢节水宣传活动的学生比喜欢低碳出行宣传活动的多10人。设喜欢低碳出行宣传活动的学生有x人，则根据题意可列出一元一次方程为：","answer":"A","explanation":"设喜欢低碳出行宣传活动的学生有x人。根据题意，喜欢节水宣传活动的学生比喜欢低碳出行的多10人，因此为(x + 10)人；喜欢垃圾分类宣传活动的学生是喜欢节水宣传的2倍，即为2(x + 10)人。三类人数之和等于总有效问卷数120，因此方程为：x + (x + 10) + 2(x + 10) = 120。选项A正确列出了该方程。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:54:03","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":"A","content":"x + (x + 10) + 2(x + 10) = 120","is_correct":1},{"id":"B","content":"x + (x - 10) + 2x = 120","is_correct":0},{"id":"C","content":"x + 2x + (x + 10) = 120","is_correct":0},{"id":"D","content":"x + (x + 10) + 2x = 120","is_correct":0}]},{"id":391,"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情况时，随机抽取了50名学生进行调查，发现其中喜欢阅读小说的有28人，喜欢阅读科普书的有15人，两种都不喜欢的有10人。那么既喜欢阅读小说又喜欢阅读科普书的学生至少有多少人？","answer":"A","explanation":"总人数为50人，两种都不喜欢的有10人，因此至少喜欢一种书的学生有50 - 10 = 40人。设既喜欢小说又喜欢科普书的学生人数为x。根据容斥原理，喜欢小说或科普书的人数 = 喜欢小说的人数 + 喜欢科普书的人数 - 两者都喜欢的人数。即：28 + 15 - x = 40。解得：43 - x = 40，所以x = 3。因此，既喜欢阅读小说又喜欢阅读科普书的学生至少有3人。选项A正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:13:28","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":"3人","is_correct":1},{"id":"B","content":"5人","is_correct":0},{"id":"C","content":"8人","is_correct":0},{"id":"D","content":"13人","is_correct":0}]}]