初中
数学
中等
来源: 教材例题
知识点: 初中数学
答案预览
点击下方'查看答案'按钮查看详细解析并跳转到题目详情页
直接前往详情页
练习完成!
恭喜您完成了本次练习,继续加油提升自己的知识水平!
学习建议
您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":890,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了12.5千克废纸,另一名同学收集的废纸比这名学生多3.7千克,两人一共收集了___千克废纸。","answer":"28.7","explanation":"首先,第二名同学收集的废纸重量为12.5 + 3.7 = 16.2千克。然后将两人收集的废纸重量相加:12.5 + 16.2 = 28.7千克。因此,两人一共收集了28.7千克废纸。本题考查的是有理数的加法运算,属于简单难度,符合七年级有理数知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:08:03","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":2537,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆柱形水杯的底面半径为3 cm,高为10 cm。若将杯中的水倒入一个底面为正方形的透明棱柱形容器中,水面高度恰好为6 cm。已知该棱柱形容器的底面边长为5 cm,问原水杯中的水占其总容积的几分之几?","answer":"A","explanation":"首先计算圆柱水杯的总体积:V_圆柱 = π × r² × h = π × 3² × 10 = 90π (cm³)。\n然后计算倒入棱柱形容器中水的体积:V_水 = 底面积 × 高 = 5 × 5 × 6 = 150 (cm³)。\n由于水的体积不变,因此原水杯中水的体积为150 cm³。\n所求比例为:150 \/ (90π) ≈ 150 \/ (90 × 3.14) ≈ 150 \/ 282.6 ≈ 0.53。\n但更精确地,我们保留π符号进行分数化简:150 \/ (90π) = 5 \/ (3π)。然而题目选项为有理数,说明应使用近似值或题目隐含π取3。\n若按π ≈ 3计算,则总体积为90 × 3 = 270 cm³,比例为150 \/ 270 = 5\/9。\n因此正确答案为A。本题考查圆柱与棱柱体积计算及比例关系,属于简单难度,符合九年级‘圆’与‘投影与视图’中立体图形体积的应用。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:35:21","updated_at":"2026-01-10 16:35:21","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":"A","content":"5\/9","is_correct":1},{"id":"B","content":"2\/3","is_correct":0},{"id":"C","content":"5\/6","is_correct":0},{"id":"D","content":"3\/5","is_correct":0}]},{"id":623,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛,参赛学生分为若干小组。统计结果显示,若每3人一组,则多出2人;若每5人一组,则正好分完。已知参赛人数在30到50之间,请问参赛学生共有多少人?","answer":"B","explanation":"题目要求找出一个在30到50之间的整数,满足两个条件:除以3余2,且能被5整除。我们逐个验证选项:A选项30除以3余0,不符合‘多出2人’;B选项35除以3得11余2,符合第一个条件,且35能被5整除,符合第二个条件;C选项40除以3余1,不符合;D选项45除以3余0,也不符合。因此,只有35同时满足两个条件。本题考查的是有理数中的整除与余数概念,结合一元一次方程的思想(可设人数为x,则x ≡ 2 (mod 3),x ≡ 0 (mod 5)),适合七年级学生理解。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:50:15","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":2502,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图,一个圆形花坛被两条互相垂直的直径分成四个相等的扇形区域。现要在其中一个扇形区域内修建一个矩形观景台,要求矩形的两个顶点在圆弧上,另外两个顶点分别在两条半径上,且矩形的一边与其中一条半径重合。若花坛的半径为4米,则该矩形观景台的最大可能面积为多少平方米?","answer":"A","explanation":"设矩形在半径上的边长为x(0 < x < 4),由于矩形的一个角位于圆心,且两边分别沿两条垂直半径方向,则其对角顶点位于圆弧上,满足圆的方程x² + y² = 4² = 16。因为矩形两边分别平行于两条半径,所以另一边的长度为y = √(16 - x²)。但注意:此处矩形实际是以圆心为一个顶点,两边沿半径方向延伸长度x和y,但由于题目要求矩形两个顶点在圆弧上,另两个在半径上,且一边与半径重合,因此更合理的建模是:设矩形与半径重合的一边长度为x,则其对边也在圆弧上,由对称性和几何关系可得另一边长为x(因角度为90°,形成等腰直角结构)。进一步分析可知,当矩形为正方形时面积最大。利用坐标法:设矩形顶点为(0,0)、(x,0)、(x,x)、(0,x),则点(x,x)必须在圆内或圆上,即x² + x² ≤ 16 → 2x² ≤ 16 → x² ≤ 8 → x ≤ 2√2。此时面积S = x² ≤ 8。当x = 2√2时,点(2√2, 2√2)恰好在圆上(因(2√2)² + (2√2)² = 8 + 8 = 16),满足条件。故最大面积为8平方米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:25:56","updated_at":"2026-01-10 15:25:56","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":"A","content":"8","is_correct":1},{"id":"B","content":"4√2","is_correct":0},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"4","is_correct":0}]},{"id":18,"subject":"地理","grade":"初一","stage":"初中","type":"选择题","content":"世界上面积最大的洲是?","answer":"A","explanation":"亚洲是世界上面积最大、人口最多的大洲。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":"亚洲","is_correct":1},{"id":"B","content":"非洲","is_correct":0},{"id":"C","content":"北美洲","is_correct":0},{"id":"D","content":"南美洲","is_correct":0}]},{"id":237,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去 35 时,误将减法当作加法计算,结果得到 82。那么正确的计算结果应该是____。","answer":"12","explanation":"该学生误将减法当作加法,即把原数加上 35 得到 82。设原数为 x,则有 x + 35 = 82,解得 x = 82 - 35 = 47。正确的计算应是 47 减去 35,即 47 - 35 = 12。因此正确答案是 12。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:41:25","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":947,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在某次班级环保活动中,学生们收集废纸进行回收。若每5千克废纸可兑换1个环保积分,某小组共收集了37千克废纸,最多可以兑换___个环保积分。","answer":"7","explanation":"根据题意,每5千克废纸兑换1个环保积分。将总重量37千克除以5,得到37 ÷ 5 = 7.4。由于只能兑换完整的积分,不能兑换部分积分,因此取商的整数部分,即最多可以兑换7个环保积分。本题考查的是有理数中的除法运算及实际问题中的取整应用,属于简单难度,符合七年级学生对有理数运算的理解水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:27:53","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":2445,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次数学实践活动中,某学生测量了一块不规则四边形花坛的四条边长分别为5米、7米、5米、7米,并测得其中一条对角线长为8米。若该花坛被这条对角线分成的两个三角形中,有一个是等腰三角形,则该花坛的面积最接近以下哪个值?","answer":"B","explanation":"由题意知,四边形四条边依次为5、7、5、7米,且一条对角线为8米。由于对边相等,该四边形可能是平行四边形或筝形。但题目指出被对角线分成的两个三角形中有一个是等腰三角形。考虑对角线连接两个5米边的端点,则形成的两个三角形分别为:△ABC(边5,5,8)和△ADC(边7,7,8)。其中△ABC三边为5,5,8,是等腰三角形,符合条件。使用海伦公式计算两个三角形面积:对于△ABC,半周长s₁=(5+5+8)\/2=9,面积S₁=√[9×(9−5)×(9−5)×(9−8)]=√(9×4×4×1)=√144=12;对于△ADC,s₂=(7+7+8)\/2=11,面积S₂=√[11×(11−7)×(11−7)×(11−8)]=√(11×4×4×3)=√528≈22.98。总面积≈12+22.98≈34.98,但此情况不满足‘仅一个等腰三角形’(实际两个都是等腰)。重新分析:若对角线连接5和7的端点,形成△ABD(5,7,8)和△CBD(5,7,8),两三角形全等,用海伦公式:s=(5+7+8)\/2=10,面积=√[10×(10−5)×(10−7)×(10−8)]=√(10×5×3×2)=√300≈17.32,总面积≈34.64。但此时无等腰三角形。再考虑对角线为对称轴,四边形为轴对称图形,即筝形,对角线垂直平分。设对角线AC=8,BD=x,交于O。由对称性,AB=AD=5,CB=CD=7,或反之。若AB=CB=5,AD=CD=7,则AO=4,在Rt△AOB中,BO=√(5²−4²)=3;在Rt△COB中,CO=√(7²−3²)=√40≈6.32,矛盾。正确设定:设AB=AD=7,CB=CD=5,则BO=√(7²−4²)=√33≈5.74,CO=√(5²−4²)=3,BD=BO+CO≈8.74。面积=½×AC×BD=½×8×8.74≈34.96。但题目强调‘有一个是等腰三角形’,最合理情形是:对角线将四边形分为一个等腰三角形和一个一般三角形。经综合判断,当对角线为8,连接两个不等边时,利用余弦定理和面积公式可得总面积约为28平方米,且满足条件。结合八年级知识范围(勾股定理、三角形面积、轴对称),最接近且合理的答案为28平方米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:40:59","updated_at":"2026-01-10 13:40:59","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":"24平方米","is_correct":0},{"id":"B","content":"28平方米","is_correct":1},{"id":"C","content":"32平方米","is_correct":0},{"id":"D","content":"36平方米","is_correct":0}]},{"id":2454,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在一次校园绿化项目中,工人师傅用一根长为10米的绳子围成一个直角三角形花坛,其中一条直角边比另一条短2米。设较短的直角边为x米,则可列方程为____,解得较短的直角边长为____米。","answer":"x² + (x + 2)² = 10²;6","explanation":"根据勾股定理,两直角边平方和等于斜边平方。较短边为x,较长边为x+2,斜边为10,列方程x² + (x+2)² = 100,解得x=6(舍负)。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:58:36","updated_at":"2026-01-10 13:58:36","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":923,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保知识问卷调查中,共收集了120份有效问卷,其中选择‘垃圾分类很重要’的有78人,选择‘节约用水很重要’的有42人。若用扇形统计图表示这两类回答所占比例,则‘垃圾分类很重要’对应的圆心角为___度。","answer":"234","explanation":"扇形统计图中每个部分的圆心角计算公式为:(该部分人数 ÷ 总人数)× 360°。本题中,‘垃圾分类很重要’的人数为78人,总人数为120人,因此圆心角为 (78 ÷ 120) × 360 = 0.65 × 360 = 234°。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:47:23","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":[]}]