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CIE 9709 2022 June Paper 11 Q5

A Level / CIE / P1

CIE 9709 2022 June Paper 11 Paper · Question 5

题目

Problem

The diagram shows a sector ABCABC of a circle with centre AA and radius rr. The line BDBD is perpendicular to ACAC. Angle CABCAB is θ\theta radians.

(a) Given that θ=16π\theta=\frac16\pi, find the exact area of BCDBCD in terms of rr.

(3)

(b) Given instead that the length of BDBD is 32r\frac{\sqrt3}{2}r, find the exact perimeter of BCDBCD in terms of rr.

(4)
题目中文翻译

图中所示为圆心为 AA、半径为 rr 的圆的扇形 ABCABC。线段 BDBD 垂直于 ACAC。角 CABCAB 为 θ\theta 弧度。

(a) 已知 θ=16π\theta=\frac16\pi,用 rr 表示区域 BCDBCD 面积的精确值。

(3)

(b) 改为已知 BDBD 的长度为 32r\frac{\sqrt3}{2}r,用 rr 表示区域 BCDBCD 周长的精确值。

(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

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用扇形面积减去直角三角形 ABDABD。由半径和角度求三角形的两条直角边。

答题过程

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The sector area is 12r2(π/6)=πr2/12\tfrac12r^2(\pi/6)=\pi r^2/12. In triangle ABDABD, the two legs have lengths rcos⁡(π/6)=3r/2r\cos(\pi/6)=\sqrt3r/2 and rsin⁡(π/6)=r/2r\sin(\pi/6)=r/2. Hence the required area is

r2(π12−38).r^2\Big(\frac{\pi}{12}-\frac{\sqrt3}{8}\Big).

(b)

解法一

思路

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由新的 BDBD 求角度和 ADAD。周长由 BDBD、DCDC 和弧 BCBC 组成。

答题过程

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As BD/AB=3/2BD/AB=\sqrt3/2 and the angle at AA is acute in triangle ABDABD, θ=π/3\theta=\pi/3. Then AD=rcos⁡(π/3)=r/2AD=r\cos(\pi/3)=r/2, so DC=r/2DC=r/2. The arc length is rπ/3r\pi/3. The perimeter is

r(32+12+π3).r\Big(\frac{\sqrt3}{2}+\frac12+\frac{\pi}{3}\Big).