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CIE 9709 2023 November Paper 12 Q4

A Level / CIE / P1

CIE 9709 2023 Nov Paper 12 Paper · Question 4

题目

Problem

The diagram shows the shape of a coin. The three arcs ABAB, BCBC and CACA are parts of circles with centres CC, AA and BB respectively. ABCABC is an equilateral triangle with sides of length 2 cm2\text{ cm}.

(a) Find the perimeter of the coin.

(2)

(b) Find the area of the face ABCABC of the coin, giving the answer in terms of π\pi and 3\sqrt{3}.

(4)
题目中文翻译

图中显示一枚硬币的形状。三段弧 ABAB、BCBC 和 CACA 分别属于以 CC、AA 和 BB 为圆心的圆。ABCABC 是边长为 2 cm2\text{ cm} 的等边三角形。

(a) 求硬币的周长。

(2)

(b) 求硬币的 ABCABC 面的面积,答案用含 π\pi 和 3\sqrt{3} 的式子表示。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

每条弧的半径都是等边三角形的边长,圆心角都是 π/3\pi/3。

答题过程

展开

Each arc has length 2(π/3)2(\pi/3) cm. The three arcs therefore have total length 2π2\pi cm.

(b)

解法一

思路

展开

把币面分成等边三角形和三个全等弓形。每个弓形的面积是一个扇形减去三角形。

答题过程

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The equilateral triangle has area 12(2)(2)sin⁡(π/3)=3 cm2\tfrac12(2)(2)\sin(\pi/3)=\sqrt3\,\mathrm{cm}^2. Each sector has area 12(22)(π/3)=2π/3 cm2\tfrac12(2^2)(\pi/3)=2\pi/3\,\mathrm{cm}^2. Thus the coin face has area

3+3(2π3−3)= 2π−23 cm2.\begin{align*} \sqrt3+3\Big(\frac{2\pi}{3}-\sqrt3\Big) =&\,2\pi-2\sqrt3\ \mathrm{cm}^2. \end{align*}