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CIE 9709 2025 March Paper 12 Q4

A Level / CIE / P1

CIE 9709 2025 March Paper 12 Paper · Question 4

题目

Problem

The diagram shows a triangle OABOAB where OA=OB=10OA = OB = 10 cm and angle AOB=0.8AOB = 0.8 radians. Points CC and DD on OAOA and OBOB respectively are such that the arc CDCD is part of a circle with centre OO and radius 6 cm. The shaded region is bounded by the arc CDCD and the line segments CACA, ABAB and BDBD.

(a) Find the perimeter of the shaded region.

[3]

(b) Find the area of the shaded region.

[3]
题目中文翻译

图中为三角形 OABOAB,其中 OA=OB=10OA = OB = 10 cm,角 AOB=0.8AOB = 0.8 弧度。点 CC 和 DD 分别位于 OAOA 和 OBOB 上,且弧 CDCD 是以 OO 为圆心、半径为 6 cm 的圆的一部分。阴影区域由弧 CDCD 以及线段 CACA、ABAB、BDBD 围成。

(a) 求阴影区域的周长。

[3]

(b) 求阴影区域的面积。

[3]

解答

(a)

解法一

思路

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外边界包括两段长 44 的直边、弧 CDCD 和弦 ABAB。等腰三角形的高把顶角与底边同时平分。

答题过程

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We have CA=BD=10−6=4CA=BD=10-6=4 cm and the arc length CD=6(0.8)=4.8CD=6(0.8)=4.8 cm. Bisecting the isosceles triangle gives AB=20sin⁡(0.4)AB=20\sin(0.4) cm. Thus the perimeter is

8+4.8+20sin⁡(0.4)≈20.6 cm.8+4.8+20\sin(0.4)\approx20.6\text{ cm}.

(b)

解法一

思路

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从大三角形 OABOAB 的面积减去小扇形 OCDOCD 的面积。

答题过程

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The triangle has area 12(10)(10)sin⁡(0.8)\frac12(10)(10)\sin(0.8). The sector has area 12(6)2(0.8)=14.4\frac12(6)^2(0.8)=14.4. Their difference is

50sin⁡(0.8)−14.4≈21.5 cm2.50\sin(0.8)-14.4\approx21.5\text{ cm}^2.