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CIE 9709 2026 June Paper 11 Q7

A Level / CIE / P1

CIE 9709 2026 June Paper 11 Paper · Question 7

题目

Problem

The diagram shows a sector ACDACD of a circle with centre CC and radius 2r2r. The midpoint of ACAC is BB. The length of BDBD is r7r\sqrt{7}.

(a) Find the exact perimeter of the shaded region. Give your answer in terms of π\pi and rr.

(4)

(b) Find the exact area of the shaded region. Give your answer in terms of π\pi and rr.

(4)
题目中文翻译

图中所示的扇形 ACDACD 所在圆的圆心为 CC,半径为 2r2r。BB 是 ACAC 的中点,BDBD 的长度为 r7r\sqrt{7}。

(a) 求阴影区域周长的精确值,答案用 π\pi 和 rr 表示。

(4)

(b) 求阴影区域面积的精确值,答案用 π\pi 和 rr 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

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用三角形 BCDBCD 的三边及余弦定理求圆心角。阴影的周长包含弧 ADAD、直边 ABAB 和 BDBD。

答题过程

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Since BB is the midpoint of ACAC, we have CB=AB=rCB=AB=r. Let ϕ=∠BCD=∠ACD\phi=\angle BCD=\angle ACD. The cosine rule gives

7r2=r2+(2r)2−2(r)(2r)cos⁡ϕ,7r^2=r^2+(2r)^2-2(r)(2r)\cos\phi,

so cos⁡ϕ=−1/2\cos\phi=-1/2 and ϕ=2π/3\phi=2\pi/3. The arc length is (2r)(2π/3)=4πr/3(2r)(2\pi/3)=4\pi r/3. Adding the two straight edges gives the exact perimeter

r(4π3+1+7).r\Big(\frac{4\pi}{3}+1+\sqrt7\Big).

(b)

解法一

思路

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阴影是整个扇形 ACDACD 去掉三角形 BCDBCD。两个面积都使用 (a) 的圆心角。

答题过程

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The sector area is 12(2r)2(2π/3)=4πr2/3\frac12(2r)^2(2\pi/3)=4\pi r^2/3. The triangle area is 12(r)(2r)sin⁡(2π/3)=3r2/2\frac12(r)(2r)\sin(2\pi/3)=\sqrt3r^2/2. The shaded area is therefore

r2(4π3−32).r^2\Big(\frac{4\pi}{3}-\frac{\sqrt3}{2}\Big).