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CIE 9709 2025 June Paper 13 Q8

A Level / CIE / P1

CIE 9709 2025 June Paper 13 Paper · Question 8

题目

Problem

The diagram shows a square ABCDABCD where each side has length 12 cm. Points EE and FF lie on the sides BCBC and CDCD respectively and are such that BE=13BCBE = \frac{1}{3}BC and DF=13DCDF = \frac{1}{3}DC. The arc EFEF is part of a circle with centre AA. The shaded region is bounded by the arc EFEF and the line segments ECEC and FCFC.

(a) Show that the size of angle EAFEAF is 0.9273 radians, correct to 4 significant figures.

[2]

(b) Find the perimeter of the shaded region.

[3]

(c) Find the area of the shaded region.

[3]
题目中文翻译

图中为正方形 ABCDABCD,每条边长为 12 cm。点 EE 和 FF 分别位于边 BCBC 和 CDCD 上,且 BE=13BCBE = \frac{1}{3}BC、DF=13DCDF = \frac{1}{3}DC。弧 EFEF 是以 AA 为圆心的圆的一部分。阴影区域由弧 EFEF 以及线段 ECEC、FCFC 围成。

(a) 证明角 EAFEAF 的大小为 0.9273 弧度,精确到 4 位有效数字。

[2]

(b) 求阴影区域的周长。

[3]

(c) 求阴影区域的面积。

[3]

解答

(a)

解法一

思路

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角 EAFEAF 是正方形的直角去掉两侧的两个相等小角。每个小角都能由直角三角形的正切求得。

答题过程

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Since BE=DF=4BE=DF=4, the two right triangles ABEABE and ADFADF are congruent. The angle at AA in each is tan⁡−1(4/12)\tan^{-1}(4/12). Hence

∠EAF= π2−2tan⁡−1(13)= 0.927295…=0.9273\begin{align*} \angle EAF=&\,\frac\pi2-2\tan^{-1}\Big(\frac13\Big)\\ =&\,0.927295\ldots=0.9273 \end{align*}

radians to four significant figures.

解法二

思路

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也可以先求弦 EFEF。从圆心向弦作垂线,就能用半弦与半径求半个圆心角。

答题过程

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The chord EFEF is the hypotenuse of the right triangle ECFECF, with EC=FC=8EC=FC=8. Thus EF=82EF=8\sqrt2. The radius is AE=122+42=410AE=\sqrt{12^2+4^2}=4\sqrt{10}. If ϕ=∠EAF\phi=\angle EAF, bisecting the chord gives

sin⁡ϕ2=42410=15.\sin\frac\phi2=\frac{4\sqrt2}{4\sqrt{10}}=\frac1{\sqrt5}.

Hence ϕ=2sin⁡−1(1/5)=0.9273\phi=2\sin^{-1}(1/\sqrt5)=0.9273 radians to four significant figures.

解法三

思路

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也可先从正方形扣除三个直角三角形,求出三角形 AEFAEF 的面积,再用两边夹角的面积公式求圆心角。该角小于直角,因此反正弦取主值。

答题过程

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Since BE=DF=4BE=DF=4, we have EC=FC=8EC=FC=8. The area of triangle AEFAEF is

122−12(12)(4)−12(12)(4)−12(8)(8)=64.12^2-\frac12(12)(4)-\frac12(12)(4)-\frac12(8)(8)=64.

Also AE=AF=122+42=410AE=AF=\sqrt{12^2+4^2}=4\sqrt{10}. Writing ϕ=∠EAF\phi=\angle EAF, the area formula gives

64=12(410)2sin⁡ϕ⟹sin⁡ϕ=45.64=\frac12(4\sqrt{10})^2\sin\phi \quad\Longrightarrow\quad \sin\phi=\frac45.

The angle lies inside the right angle BADBAD, so 0<ϕ<π/20<\phi<\pi/2. Hence ϕ=sin⁡−1(4/5)=0.927295…=0.9273\phi=\sin^{-1}(4/5)=0.927295\ldots=0.9273 radians to four significant figures.

(b)

解法一

思路

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阴影的周长由两条直边和一段弧组成。先由 AEAE 求半径,再用 (a) 的未舍入角度求弧长。

答题过程

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The radius is AE=122+42=410AE=\sqrt{12^2+4^2}=4\sqrt{10}. With ϕ=∠EAF\phi=\angle EAF, the arc length is 410 ϕ4\sqrt{10}\,\phi. Since EC=FC=8EC=FC=8, the perimeter is

16+410 ϕ≈27.7 cm.16+4\sqrt{10}\,\phi\approx27.7\text{ cm}.

(c)

解法一

思路

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把正方形分为两个边角三角形、扇形和阴影。扣除前三部分就得到阴影面积。

答题过程

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Each of ABEABE and ADFADF has area 12(12)(4)=24\frac12(12)(4)=24. The sector area is 12(410)2ϕ=80ϕ\frac12(4\sqrt{10})^2\phi=80\phi. Thus the shaded area is

144−24−24−80ϕ=96−80ϕ≈21.8 cm2.144-24-24-80\phi=96-80\phi\approx21.8\text{ cm}^2.

解法二

思路

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也可先取直角三角形 ECFECF,再减去弦 EFEF 与弧之间的弓形。

答题过程

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Triangle ECFECF has area 3232. The circular segment cut off by EFEF has area

12(410)2(ϕ−sin⁡ϕ)=80(ϕ−sin⁡ϕ).\frac12(4\sqrt{10})^2(\phi-\sin\phi)=80(\phi-\sin\phi).

Hence the shaded area is 32−80(ϕ−sin⁡ϕ)≈21.8 cm232-80(\phi-\sin\phi)\approx21.8\text{ cm}^2.