某学生在研究一个实际问题时,发现一个等腰三角形的底边长为 6 cm,腰长为 5 cm。若以该三角形的底边为边长构造一个正方形,并以该三角形的腰为半径画一个扇形,扇形的圆心角为 60°,则正方形面积与扇形面积的比值最接近下列哪个数值?(取 π ≈ 3.14)
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观察已知三个点 A(1, 2)、B(3, 4)、C(5, 6),可以看出横坐标每次增加 2,纵坐标也每次增加 2,说明这些点位于一条斜率为 1 的直线上。进一步分析可知,每个点的纵坐标都比横坐标大 1,即满足关系式 y = x + 1。当横坐标为 7 时,代入得 y = 7 + 1 = 8。因此,点 D 的纵坐标是 8。
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[{"id":368,"content":"某学生在整理班级同学的身高数据时,随机抽取了10名同学的身高(单位:厘米),分别为:152,158,160,155,162,158,156,160,158,161。这组数据的众数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"众数是一组数据中出现次数最多的数。观察数据:152出现1次,158出现3次,160出现2次,155出现1次,162出现1次,156出现1次,161出现1次。其中158出现的次数最多,共3次,因此这组数据的众数是158。","options":[{"id":"A","content":"158"},{"id":"B","content":"160"},{"id":"C","content":"155"},{"id":"D","content":"162"}]},{"id":2500,"content":"某学生用三根木棒搭建一个直角三角形支架,其中两根木棒的长度分别为3cm和4cm。若他将这个三角形绕长度为4cm的木棒所在直线旋转一周,所形成的几何体的俯视图是以下哪种图形?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"根据勾股定理,第三边长度为√(4² - 3²) = √7 cm 或 √(3² + 4²) = 5 cm。由于题目说明是直角三角形且已知两边为3cm和4cm,可判断第三边为5cm(斜边)或√7 cm(当4cm为斜边时)。但无论哪种情况,绕长度为4cm的直角边旋转时,另一条直角边(3cm)将作为旋转半径,形成一个圆锥体。圆锥的俯视图是从上往下看,呈现为一个完整的圆。因此正确答案是A。本题考查旋转形成的几何体及其视图,属于投影与视图和旋转知识点的综合应用,难度适中,符合九年级学生认知水平。","options":[{"id":"A","content":"一个圆"},{"id":"B","content":"一个矩形"},{"id":"C","content":"一个三角形"},{"id":"D","content":"一个扇形"}]},{"id":227,"content":"某学生计算一个长方形花坛的面积,已知长为8米,宽为5米,那么这个花坛的面积是_平方米。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"40","explanation":"长方形的面积计算公式是:面积 = 长 × 宽。题目中给出的长是8米,宽是5米,因此面积为 8 × 5 = 40 平方米。","options":[]},{"id":249,"content":"某学生用一根长为48厘米的铁丝围成一个长方形,若长方形的长比宽多6厘米,则这个长方形的面积是___平方厘米。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"中等","answer":"135","explanation":"设长方形的宽为x厘米,则长为(x + 6)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),可得方程:2 × (x + x + 6) = 48。化简得:2 × (2x + 6) = 48,即4x + 12 = 48。解得4x = 36,x = 9。因此宽为9厘米,长为15厘米。面积为长 × 宽 = 15 × 9 = 135平方厘米。","options":[]},{"id":460,"content":"144度","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":262,"content":"某学生在解方程 3(x - 4) + 2 = 5x - 10 时,第一步将括号展开后得到 3x - 12 + 2 = 5x - 10,合并同类项后得到 3x - 10 = 5x - 10。接下来,他应该将含 x 的项移到等式的一边,常数项移到另一边,于是他将 3x 移到右边,得到 -10 = 2x - 10。然后,他将 -10 移到左边,得到 ___ = 2x。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"中等","answer":"0","explanation":"从步骤 -10 = 2x - 10 开始,要将常数项移到等式左边,需在等式两边同时加上 10:-10 + 10 = 2x - 10 + 10,化简后得到 0 = 2x。因此,空白处应填 0。此题考查一元一次方程的移项与合并同类项能力,要求学生掌握等式的基本性质,属于中等难度,符合七年级数学课程内容。","options":[]},{"id":319,"content":"8人","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"答案待完善","explanation":"解析待完善","options":[]},{"id":339,"content":"20","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"答案待完善","explanation":"解析待完善","options":[]},{"id":2520,"content":"某学生设计了一个圆形花坛,其边缘由一段圆弧和两条半径围成,形成一个扇形区域。已知该扇形的圆心角为60°,面积为6π平方米。若在该扇形区域内接一个最大的等边三角形(三个顶点均在扇形边界上),则这个等边三角形的边长是多少?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先,根据扇形面积公式 S = (θ\/360°) × πr²,其中θ = 60°,S = 6π,代入得:6π = (60\/360) × πr² → 6π = (1\/6)πr² → r² = 36 → r = 6米。因此扇形半径为6米。由于圆心角为60°,若将扇形的两条半径端点与圆心连接,可构成一个边长为6米的等边三角形(因为两边为半径,夹角60°,三边相等)。此三角形完全位于扇形内,且是内接于该扇形中最大的等边三角形(任何其他构造都会导致边长更短或超出边界)。故该等边三角形边长为6米。","options":[{"id":"A","content":"6米"},{"id":"B","content":"3√3米"},{"id":"C","content":"4√3米"},{"id":"D","content":"2√6米"}]},{"id":2286,"content":"在数轴上,点A表示的数是-3,点B与点A之间的距离是7个单位长度,且点B在原点右侧,则点B表示的数是____。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"4","explanation":"点A表示-3,点B与点A相距7个单位长度,且在原点右侧。从-3向右移动7个单位,即计算 -3 + 7 = 4。因此点B表示的数是4。","options":[]}]