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CIE 9709 2023 March Paper 12 Q8

A Level / CIE / P1

CIE 9709 2023 March Paper 12 Paper · Question 8

题目

Problem

The diagram shows triangle ABCABC in which angle BB is a right angle. The length of ABAB is 8 cm8\text{ cm} and the length of BCBC is 4 cm4\text{ cm}. The point DD on ABAB is such that AD=5 cmAD=5\text{ cm}. The sector DACDAC is part of a circle with centre DD.

(a) Find the perimeter of the shaded region.

(5)

(b) Find the area of the shaded region.

(3)
题目中文翻译

图中三角形 ABCABC 的角 BB 为直角。ABAB 长为 8 cm8\text{ cm},BCBC 长为 4 cm4\text{ cm}。点 DD 位于 ABAB 上,且 AD=5 cmAD=5\text{ cm}。扇形 DACDAC 是圆心为 DD 的圆的一部分。

(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

阴影周长由弧 ACAC 和弦 ACAC 组成。在小直角三角形中求角 BDCBDC,再取补角得到圆心角。

答题过程

展开

Since DB=8−5=3DB=8-5=3, the angle BDCBDC is arctan⁡(4/3)\arctan(4/3). As DADA and DBDB point in opposite directions, the sector angle is ϕ=π−arctan⁡(4/3)\phi=\pi-\arctan(4/3) radians. The arc has length 5ϕ5\phi and the chord has length 82+42=45\sqrt{8^2+4^2}=4\sqrt5. The perimeter is 5ϕ+45=20.05\phi+4\sqrt5=20.0 cm to three significant figures.

(b)

解法一

思路

展开

用扇形减去三角形 ADCADC。三角形的底边 ADAD 与 ABAB 共线,故可用 BCBC 作为高。

答题过程

展开

The sector has area 12(52)ϕ\tfrac12(5^2)\phi and triangle ADCADC has area 12(5)(4)=10\tfrac12(5)(4)=10. The shaded area is

252(π−arctan⁡43)−10= 17.7 cm2\begin{align*} \frac{25}{2}\Big(\pi-\arctan\frac43\Big)-10 =&\,17.7\,\mathrm{cm}^2 \end{align*}

to three significant figures.