习题练习

[{"id":1413,"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，要求学生在平面直角坐标系中设计一个由直线段构成的封闭图形。已知该图形由以下四条线段围成：线段AB、线段BC、线段CD和线段DA。其中，点A的坐标为(0, 0)，点B的坐标为(4, 0)，点C位于第一象限且满足直线BC与x轴正方向的夹角为45°，点D位于y轴上，且线段CD与线段AB平行。若该封闭图形的面积为10平方单位，求点C和点D的坐标。","answer":"解：\n\n已知点A(0, 0)，点B(4, 0)，线段AB在x轴上，长度为4。\n\n由于线段CD与线段AB平行，而AB在x轴上（水平），所以CD也是水平线段，即点C和点D的纵坐标相同。\n\n又因为点D在y轴上，设点D的坐标为(0, y)，则点C的纵坐标也为y。\n\n点C在第一象限，且直线BC与x轴正方向夹角为45°，说明直线BC的斜率为tan(45°) = 1。\n\n点B坐标为(4, 0)，设点C坐标为(x, y)，则由斜率公式：\n(y - 0)\/(x - 4) = 1\n即 y = x - 4 ①\n\n又因点C纵坐标为y，且点D为(0, y)，CD为水平线段，长度为|x - 0| = |x|。由于C在第一象限，x > 0，所以CD长度为x。\n\n现在考虑图形ABCD：\n- A(0,0), B(4,0), C(x,y), D(0,y)\n\n这是一个梯形，上底为CD = x，下底为AB = 4，高为y（因为上下底平行于x轴，垂直距离为y）。\n\n梯形面积公式：S = (上底 + 下底) × 高 ÷ 2\n代入得：\n10 = (x + 4) × y ÷ 2\n即 (x + 4)y = 20 ②\n\n将①式 y = x - 4 代入②式：\n(x + 4)(x - 4) = 20\nx² - 16 = 20\nx² = 36\nx = 6 或 x = -6\n\n由于点C在第一象限，x > 0，故x = 6\n代入①得：y = 6 - 4 = 2\n\n因此，点C坐标为(6, 2)，点D坐标为(0, 2)\n\n验证：\n- CD长度为6，AB长度为4，高为2\n- 面积 = (6 + 4) × 2 ÷ 2 = 10，符合条件\n- BC斜率 = (2 - 0)\/(6 - 4) = 2\/2 = 1，对应45°角，正确\n- D在y轴上，C在第一象限，均满足\n\n答：点C的坐标为(6, 2)，点D的坐标为(0, 2)。","explanation":"本题综合考查平面直角坐标系、一次函数斜率、几何图形面积计算以及方程组的建立与求解。解题关键在于识别图形为梯形，并利用几何条件（平行、角度、坐标位置）建立代数关系。首先由角度确定直线BC的斜率为1，建立点C坐标与点B的关系；再由CD与AB平行且D在y轴上，得出C与D纵坐标相同；最后利用梯形面积公式建立方程，联立求解。整个过程涉及坐标系、直线斜率、方程求解和几何面积，综合性强，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:29:18","updated_at":"2026-01-06 11:29:18","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":862,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生在整理班级同学的课外阅读情况时，发现喜欢阅读科幻小说的人数占总人数的30%，喜欢阅读历史书籍的人数比科幻小说的少10%，其余12人喜欢阅读其他类型书籍。那么该班级共有___名学生。","answer":"30","explanation":"设该班级共有x名学生。根据题意，喜欢科幻小说的人数为30%x = 0.3x，喜欢历史书籍的人数比科幻小说少10%，即少0.1x，因此喜欢历史书籍的人数为0.3x - 0.1x = 0.2x。其余12人喜欢其他类型书籍。根据总人数关系可得方程：0.3x + 0.2x + 12 = x，即0.5x + 12 = x。解这个一元一次方程：x - 0.5x = 12，0.5x = 12，x = 24 ÷ 0.5 = 30。因此，该班级共有30名学生。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:16:35","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":1941,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A(2, 3)和点B(8, 7)是某矩形的两个对角顶点，且该矩形的边分别平行于坐标轴。若该矩形的周长是面积的$\\frac{1}{2}$，则这个矩形的另外两个顶点坐标的横坐标之和为____。","answer":"10","explanation":"由A、B为对角顶点且边平行坐标轴，可得另两点为(2,7)和(8,3)。设长为6，宽为4，周长=20，面积=24。验证20 = 24×½成立。另两点横坐标为2和8，和为10。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:11:56","updated_at":"2026-01-07 14:11:56","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":1971,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某次学校科技节中各参赛小组完成项目所用时间时，记录了八个小组的数据（单位：分钟）：28.5, 32.1, 26.8, 30.4, 29.7, 33.6, 27.9, 31.2。为了分析这组数据的集中趋势和离散程度，该学生先计算了平均数，再计算了各数据与平均数之差的绝对值，并求出这些绝对值的平均数（即平均绝对偏差，MAD）。请问这组数据的平均绝对偏差最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中平均绝对偏差（MAD）的概念与计算。首先计算八个小组所用时间的平均数：(28.5 + 32.1 + 26.8 + 30.4 + 29.7 + 33.6 + 27.9 + 31.2) ÷ 8 = 240.2 ÷ 8 = 30.025。然后计算每个数据与平均数之差的绝对值：|28.5−30.025|=1.525，|32.1−30.025|=2.075，|26.8−30.025|=3.225，|30.4−30.025|=0.375，|29.7−30.025|=0.325，|33.6−30.025|=3.575，|27.9−30.025|=2.125，|31.2−30.025|=1.175。将这些绝对值相加：1.525 + 2.075 + 3.225 + 0.375 + 0.325 + 3.575 + 2.125 + 1.175 = 14.4。最后求平均绝对偏差：14.4 ÷ 8 = 1.8。1.8 最接近选项 B 的 1.7，因此答案为 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:49:19","updated_at":"2026-01-07 14:49:19","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":"1.5","is_correct":0},{"id":"B","content":"1.7","is_correct":1},{"id":"C","content":"1.9","is_correct":0},{"id":"D","content":"2.1","is_correct":0}]},{"id":1940,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD，其中A(0, 0)，B(4, 0)，C(4, 3)，D(0, 5)。若将该四边形绕原点逆时针旋转90°，得到新四边形A'B'C'D'，则点C'的坐标为___。","answer":"(-3, 4)","explanation":"绕原点逆时针旋转90°，坐标变换公式为(x, y) → (-y, x)。C(4, 3)变换后为(-3, 4)。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:11:30","updated_at":"2026-01-07 14:11:30","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":290,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的运动项目调查数据时，制作了如下统计表。已知喜欢篮球的人数比喜欢足球的多6人，且喜欢篮球和足球的总人数为30人。那么喜欢足球的人数是多少？","answer":"B","explanation":"设喜欢足球的人数为x人，则喜欢篮球的人数为(x + 6)人。根据题意，两者总人数为30人，可列出一元一次方程：x + (x + 6) = 30。解这个方程：2x + 6 = 30，2x = 24，x = 12。因此，喜欢足球的人数是12人，对应选项B。本题考查了一元一次方程在数据整理中的简单应用，符合七年级数学课程标准要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:32:12","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":"10人","is_correct":0},{"id":"B","content":"12人","is_correct":1},{"id":"C","content":"18人","is_correct":0},{"id":"D","content":"24人","is_correct":0}]},{"id":749,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级大扫除中，某学生负责统计清洁工具的分配情况。已知每2名学生共用1把扫帚，每3名学生共用1个拖把，每4名学生共用1个水桶。如果总共使用了26件清洁工具，那么参加大扫除的学生人数是___人。","answer":"24","explanation":"设参加大扫除的学生人数为x。根据题意，扫帚的数量为x\/2，拖把的数量为x\/3，水桶的数量为x\/4。总工具数为26件，因此可列方程：x\/2 + x\/3 + x\/4 = 26。通分后得(6x + 4x + 3x)\/12 = 26，即13x\/12 = 26。两边同乘以12，得13x = 312，解得x = 24。因此，参加大扫除的学生人数是24人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:22:54","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":2239,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上从原点出发，先向右移动5个单位长度，再向左移动8个单位长度，接着向右移动3个单位长度，最后向左移动6个单位长度。此时该学生所在位置对应的数是___。","answer":"-6","explanation":"该问题考查正负数在数轴上的实际应用与连续运算能力。向右移动表示正方向，用正数表示；向左移动表示负方向，用负数表示。因此，整个移动过程可表示为：+5 + (-8) + 3 + (-6)。逐步计算：5 - 8 = -3；-3 + 3 = 0；0 - 6 = -6。最终位置对应的数是-6。此题融合了正负数的加减运算与数轴直观理解，符合七年级课程标准中对有理数运算和数形结合的要求，且避免了常见题型结构，具有一定的综合性和思维难度。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:39:22","updated_at":"2026-01-09 14:39:22","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":584,"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时，随机抽取了30名学生进行调查，发现每天阅读时间在0.5小时到1.5小时之间。他将这些数据分为5组，并制作了频数分布表。若每组组距相同，则每组的组距是多少小时？","answer":"B","explanation":"题目中给出的数据范围是从0.5小时到1.5小时，因此全距为1.5 - 0.5 = 1.0小时。将数据分为5组，且每组组距相同，则组距 = 全距 ÷ 组数 = 1.0 ÷ 5 = 0.2小时。因此正确答案是B选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 20:12:03","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":"0.1","is_correct":0},{"id":"B","content":"0.2","is_correct":1},{"id":"C","content":"0.3","is_correct":0},{"id":"D","content":"0.4","is_correct":0}]},{"id":549,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验，成绩分布如下表所示。已知成绩在80分及以上的学生占总人数的40%，成绩在60分到79分之间的学生比成绩在60分以下的学生多10人，且全班共有50名学生。那么，成绩在60分以下的学生有多少人？","answer":"A","explanation":"设成绩在60分以下的学生有x人，则成绩在60分到79分之间的学生有(x + 10)人。根据题意，成绩在80分及以上的学生占总人数的40%，即50 × 40% = 20人。全班总人数为50人，因此可以列出方程：x + (x + 10) + 20 = 50。化简得：2x + 30 = 50，解得2x = 20，x = 10。所以，成绩在60分以下的学生有10人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:08:26","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人","is_correct":1},{"id":"B","content":"15人","is_correct":0},{"id":"C","content":"20人","is_correct":0},{"id":"D","content":"25人","is_correct":0}]}]