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CIE 9709 2025 Nov Paper 15 Q4

A Level / CIE / P1

CIE 9709 2025 Nov Paper 15 Paper · Question 4

题目

Problem

The diagram shows the design for a company’s new logo. The sector of the circle, centre OO, has radius rr cm. The acute angle AOC=13πAOC=\frac{1}{3}\pi radians. The quadrilateral OABCOABC is a rhombus.

(a) Find an expression for the perimeter of the design. Give your answer in terms of π\pi and rr.

[2]

(b) It is now given that the perimeter of the design is 200200 cm.

Find the area of the design. Give your answer to 33 significant figures.

[5]
题目中文翻译

图中显示的是一家公司新标志的设计。圆心为 OO 的圆扇形半径为 rr cm。锐角 AOC=13πAOC=\frac{1}{3}\pi 弧度。四边形 OABCOABC 是菱形。

(a) 求这个设计的周长表达式。答案用 π\pi 和 rr 表示。

(b) 现在已知这个设计的周长为 200200 cm。

求这个设计的面积。答案保留 33 位有效数字。

解答

(a)

解法一

思路

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沿图形外边界走一圈,只经过优弧和菱形的两条外边。OAOA、OCOC 是内部的接合边。

答题过程

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The angle of the major sector is 2π−π3=5π32\pi-\frac\pi3=\frac{5\pi}{3}. Its arc has length 5πr3\frac{5\pi r}{3}. Since all sides of the rhombus have length rr, the perimeter is

P=5πr3+r+r=(5π3+2)r.P=\frac{5\pi r}{3}+r+r=\Big(\frac{5\pi}{3}+2\Big)r.

(b)

解法一

思路

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先用周长求半径。图形由大扇形和菱形拼成,菱形的面积可用相邻两边及其夹角求出。

答题过程

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Using (a),

200=(5π3+2)r⟹r=6005π+6.200=\Big(\frac{5\pi}{3}+2\Big)r \quad\Longrightarrow\quad r=\frac{600}{5\pi+6}.

The area of the major sector is 12r2⋅5π3\frac12r^2\cdot\frac{5\pi}{3}. The rhombus has area r2sin⁡(π/3)=32r2r^2\sin(\pi/3)=\frac{\sqrt3}{2}r^2. Thus the total area is

(6005π+6)2(5π6+32)≈2.66×103 cm2\Big(\frac{600}{5\pi+6}\Big)^2 \Big(\frac{5\pi}{6}+\frac{\sqrt3}{2}\Big) \approx 2.66\times10^3\text{ cm}^2

to three significant figures.

解法二

思路

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也可以从整圆出发。先去掉弦 ACAC 与劣弧 ACAC 之间的弓形,再加上三角形 ABCABC。

答题过程

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From the perimeter, r=6005π+6r=\frac{600}{5\pi+6}. The area between the minor arc and the chord ACAC is

12r2⋅π3−12r2sin⁡π3.\frac12r^2\cdot\frac\pi3-\frac12r^2\sin\frac\pi3.

The diagonal ACAC divides the rhombus into two congruent triangles, so triangle ABCABC has area 12r2sin⁡π3\frac12r^2\sin\frac\pi3. The required area is

 πr2−(πr26−3r24)+3r24= r2(5π6+32)≈ 2.66×103 cm2\begin{align*} &\,\pi r^2-\Big(\frac{\pi r^2}{6}-\frac{\sqrt3r^2}{4}\Big)+\frac{\sqrt3r^2}{4}\\ =&\,r^2\Big(\frac{5\pi}{6}+\frac{\sqrt3}{2}\Big)\\ \approx&\, 2.66\times10^3\text{ cm}^2 \end{align*}

to three significant figures.