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CIE 9709 2022 June Paper 13 Q9

A Level / CIE / P1

CIE 9709 2022 June Paper 13 Paper · Question 9

题目

Problem

The diagram shows triangle ABCABC with AB=BC=6 cmAB=BC=6\text{ cm} and angle ABC=1.8ABC=1.8 radians. The arc CDCD is part of a circle with centre AA and ABDABD is a straight line.

(a) Find the perimeter of the shaded region.

(5)

(b) Find the area of the shaded region.

(3)
题目中文翻译

图中所示为三角形 ABCABC,其中 AB=BC=6 cmAB=BC=6\text{ cm},角 ABC=1.8ABC=1.8 弧度。弧 CDCD 是以 AA 为圆心的圆的一部分,且 AA、BB、DD 在同一直线上。

(a) 求阴影区域的周长。

(5)

(b) 求阴影区域的面积。

(3)
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

先求圆的半径 ACAC 与圆心角 CADCAD。阴影的周长由弧 CDCD、DBDB 和 BCBC 组成。

答题过程

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Bisecting the isosceles triangle at BB gives AC=12sin⁡(0.9)AC=12\sin(0.9). The sector angle is the angle at AA, so ϕ=(π−1.8)/2\phi=(\pi-1.8)/2 radians. Since AD=ACAD=AC and AB=BC=6AB=BC=6, we have DB+BC=AC−6+6=ACDB+BC=AC-6+6=AC. The perimeter is therefore

ACϕ+AC=12sin⁡(0.9)(1+ϕ)=15.7 cmAC\phi+AC=12\sin(0.9)(1+\phi)=15.7\,\mathrm{cm}

to three significant figures.

(b)

解法一

思路

展开

用扇形 ADCADC 减去三角形 ABCABC。扇形沿用 (a) 的精确半径和角度,三角形用两边与夹角。

答题过程

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The shaded area is

 12(12sin⁡(0.9))2ϕ−12(6)(6)sin⁡(1.8)= 12.1060… cm2.\begin{align*} &\,\frac12\bigl(12\sin(0.9)\bigr)^2\phi-\frac12(6)(6)\sin(1.8)\\ =&\,12.1060\ldots\,\mathrm{cm}^2. \end{align*}

Hence it is 12.1 cm212.1\,\mathrm{cm}^2 to three significant figures.