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CIE 9709 2023 November Paper 11 Q6

A Level / CIE / P1

CIE 9709 2023 Nov Paper 11 Paper · Question 6

题目

Problem

The diagram shows a motif formed by the major arc ABAB of a circle with radius rr and centre OO, and the minor arc AOBAOB of a circle, also with radius rr but with centre CC. The point CC lies on the circle with centre OO.

(a) Given that angle ACB=kπACB=k\pi radians, state the value of the fraction kk.

(1)

(b) State the perimeter of the shaded motif in terms of π\pi and rr.

(1)

(c) Find the area of the shaded motif, giving your answer in terms of π\pi, rr and 3\sqrt{3}.

(5)
题目中文翻译

图中所示的图案由两条圆弧构成:一条是圆心为 OO、半径为 rr 的圆的优弧 ABAB;另一条是圆心为 CC、半径也为 rr 的圆的劣弧 AOBAOB。点 CC 位于以 OO 为圆心的圆上。

(a) 已知角 ACB=kπACB=k\pi 弧度,写出分数 kk 的值。

(1)

(b) 用 π\pi 和 rr 表示阴影图案的周长。

(1)

(c) 求阴影图案的面积,答案用 π\pi、rr 和 3\sqrt{3} 表示。

(5)
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

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连结两圆的圆心。它与各交点组成边长都为 rr 的等边三角形。

答题过程

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The triangles ACOACO and BCOBCO are equilateral because AC=BC=AO=BO=CO=rAC=BC=AO=BO=CO=r. Thus ∠ACB=2π/3\angle ACB=2\pi/3, giving k=2/3k=2/3.

(b)

解法一

思路

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周长由同半径的两段弧组成。它们各自的圆心角相加恰好为一周。

答题过程

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At OO the minor angle AOBAOB is also 2π/32\pi/3, so the major arc at OO subtends 4π/34\pi/3. The minor arc at CC subtends 2π/32\pi/3. The perimeter is r(4π/3)+r(2π/3)=2πrr(4\pi/3)+r(2\pi/3)=2\pi r.

(c)

解法一

思路

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把阴影视为以 OO 为圆心的整圆减去两圆的重叠部分。重叠部分被弦 ABAB 分成两个全等弓形。

答题过程

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Each segment is the difference between a sector of angle 2π/32\pi/3 and its isosceles triangle, so it has area

 12r22π3−12r2sin⁡2π3= r2(π3−34).\begin{align*} &\,\frac12r^2\frac{2\pi}{3}-\frac12r^2\sin\frac{2\pi}{3}\\ =&\,r^2\Big(\frac\pi3-\frac{\sqrt3}{4}\Big). \end{align*}

The shaded area is

 πr2−2r2(π3−34)= r2(π3+32).\begin{align*} &\,\pi r^2-2r^2\Big(\frac\pi3-\frac{\sqrt3}4\Big)\\ =&\,r^2\Big(\frac\pi3+\frac{\sqrt3}2\Big). \end{align*}

解法二

思路

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也可以从 OO 的大扇形出发,减去弧 AOAO 和 OBOB 各自与半径围成的小弓形。

答题过程

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The major sector at OO has area 12r2(4π/3)\frac12r^2(4\pi/3). The chord AOAO subtends π/3\pi/3 at CC, and the chord OBOB does the same. The two small segments removed from this sector each have area 12r2(π/3−sin⁡(π/3))\frac12r^2(\pi/3-\sin(\pi/3)). Hence the shaded area is

 2πr23−r2(π3−32)= r2(π3+32).\begin{align*} &\,\frac{2\pi r^2}{3}-r^2\Big(\frac\pi3-\frac{\sqrt3}{2}\Big)\\ =&\,r^2\Big(\frac\pi3+\frac{\sqrt3}{2}\Big). \end{align*}