某校七年级组织学生参加数学实践活动,要求每组学生设计一个矩形花坛,花坛的周长为20米。为了美观,要求花坛的长和宽都是正实数,并且长比宽多至少2米。同时,学校规定花坛的面积不能小于21平方米。现有一名学生设计了多个方案,其中长和宽满足上述所有条件。若该学生希望花坛的面积尽可能大,求此时花坛的长和宽各是多少米?并求出最大面积。
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[{"id":467,"content":"42","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"待完善","explanation":"解析待完善","options":[]},{"id":2266,"content":"在数轴上,点A表示的数是-3,点B表示的数是5。若点C位于点A和点B之间,且AC的长度是CB长度的2倍,那么点C表示的数是多少?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"中等","answer":"D","explanation":"点A为-3,点B为5,AB之间的距离为5 - (-3) = 8。设CB的长度为x,则AC = 2x,由AC + CB = AB得2x + x = 8,解得x = 8\/3。因此AC = 16\/3。从点A向右移动16\/3个单位,得到点C的坐标为-3 + 16\/3 = (-9 + 16)\/3 = 7\/3。故点C表示的数是7\/3。","options":[{"id":"A","content":"1"},{"id":"B","content":"-1"},{"id":"C","content":"3"},{"id":"D","content":"7\/3"}]},{"id":2023,"content":"在一次校园植物测量活动中,一名学生测得一棵树底部到地面的垂直高度为4米,同时测得从树顶到地面某固定标志点的水平距离为3米。若该学生站在标志点处,视线与地面成直角三角形的斜边,则树顶到该标志点的直线距离是多少米?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"根据题意,树高4米为直角三角形的一条直角边,水平距离3米为另一条直角边,所求的直线距离为斜边。应用勾股定理:斜边² = 3² + 4² = 9 + 16 = 25,因此斜边 = √25 = 5(米)。故正确答案为A。","options":[{"id":"A","content":"5"},{"id":"B","content":"6"},{"id":"C","content":"7"},{"id":"D","content":"√7"}]},{"id":152,"content":"下列各数中,属于无理数的是( )","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"C","explanation":"无理数是指不能写成两个整数之比的实数,其小数部分无限不循环。选项A(0.5)可化为1\/2,是有理数;选项B(√4 = 2)是整数,属于有理数;选项D(1\/3)是分数,也是有理数;而选项C(π)是一个著名的无理数,其小数无限不循环,不能表示为分数。因此正确答案是C。","options":[{"id":"A","content":"0.5"},{"id":"B","content":"√4"},{"id":"C","content":"π"},{"id":"D","content":"1\/3"}]},{"id":371,"content":"某班级组织了一次环保知识竞赛,共20道题,答对一题得5分,答错或不答扣2分。一名学生最终得分为65分,请问他答对了多少道题?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"设这名学生答对了x道题,则答错或不答的题数为(20 - x)道。根据题意,答对一题得5分,答错或不答扣2分,总得分为65分,可列出一元一次方程:5x - 2(20 - x) = 65。展开并化简:5x - 40 + 2x = 65,合并同类项得7x - 40 = 65,移项得7x = 105,解得x = 15。因此,该学生答对了15道题。","options":[{"id":"A","content":"15"},{"id":"B","content":"14"},{"id":"C","content":"13"},{"id":"D","content":"12"}]},{"id":854,"content":"在一次班级环保活动中,某学生收集了可回收物品的数据,其中废纸的重量是塑料瓶重量的2倍少3千克。如果塑料瓶重x千克,那么废纸的重量可以表示为______千克。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"2x - 3","explanation":"根据题意,废纸的重量是塑料瓶重量的2倍少3千克。塑料瓶重量为x千克,其2倍就是2x千克,再减去3千克,得到废纸重量为(2x - 3)千克。本题考查整式的加减中用代数式表示数量关系,属于简单难度的列代数式问题,符合七年级数学课程要求。","options":[]},{"id":2214,"content":"某学生在记录一周内每天气温变化时,发现某天的气温比前一天上升了5℃,记作+5℃;第二天又下降了8℃,应记作____℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"-8","explanation":"根据正数和负数表示相反意义的量的知识点,气温上升用正数表示,下降则用负数表示。因此,气温下降8℃应记作-8℃。","options":[]},{"id":805,"content":"某学生在整理班级同学的课外阅读情况时,收集了每位同学每月阅读的书籍数量,并将数据按从小到大的顺序排列。已知这组数据的中位数是4,且数据个数为奇数。如果去掉最大的一个数据后,新的中位数变为3.5,那么原数据中最少有多少个数据?____","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"7","explanation":"设原数据有n个,且n为奇数。中位数为第(n+1)\/2个数,已知为4。去掉最大的一个数据后,剩下n-1个数据(偶数个),中位数为中间两个数的平均数,即第(n-1)\/2个和第(n+1)\/2个数据的平均值为3.5。由于原数据有序,去掉最大值后,中间两个数应分别为3和4(因为(3+4)\/2=3.5)。为了使这种情况成立,原数据中第(n+1)\/2个数必须是4,且其前一个数为3。当n=7时,原数据第4个数为4,去掉最大值后剩下6个数,第3和第4个数分别为3和4,满足新中位数为3.5。若n<7(如n=5),则无法满足去掉最大值后中间两数为3和4的条件。因此原数据最少有7个。","options":[]},{"id":1808,"content":"某学生测量了一个等腰三角形的底边长为6厘米,两腰各为5厘米。若以该三角形的底边为轴进行轴对称变换,得到的新三角形与原三角形组成的图形是什么?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"D","explanation":"原三角形是等腰三角形,底边为6厘米,两腰为5厘米。以底边为轴作轴对称变换后,会得到一个与原三角形完全对称的新三角形,两个三角形共用底边,顶点分别在底边两侧。这样形成的四边形有两组对边分别相等(每条腰5厘米,底边6厘米被对称复制),且由于对称性,对边平行,因此构成一个平行四边形。由于边长不等(5≠6),不是菱形;角度不是直角,也不是矩形或正方形。故正确答案为D。","options":[{"id":"A","content":"菱形"},{"id":"B","content":"矩形"},{"id":"C","content":"正方形"},{"id":"D","content":"平行四边形"}]},{"id":610,"content":"某学生在整理班级同学的课外阅读时间时,记录了5位同学每周阅读的小时数分别为:3,5,4,6,2。如果老师要求每位同学的阅读时间都增加相同的整数小时,使得新的数据中位数变为5,那么每位同学至少需要增加多少小时?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"原始数据为:3,5,4,6,2。先将数据从小到大排序:2,3,4,5,6。当前中位数是中间的数,即4。设每位同学增加x小时(x为正整数),则新数据为:2+x,3+x,4+x,5+x,6+x。排序后仍为:2+x,3+x,4+x,5+x,6+x,中位数是4+x。要求中位数为5,即4 + x = 5,解得x = 1。因此,每位同学至少需要增加1小时。验证:增加1小时后数据为3,4,5,6,7,排序后中位数为5,符合条件。故正确答案为A。","options":[{"id":"A","content":"1"},{"id":"B","content":"2"},{"id":"C","content":"3"},{"id":"D","content":"4"}]}]