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CIE 9709 2025 June Paper 15 Q5

A Level / CIE / P1

CIE 9709 2025 June Paper 15 Paper · Question 5

题目

Problem

The diagram shows a sector ABDABD of a circle with centre AA and radius 1010 cm. The perpendicular bisector of ABAB passes through DD.

(a) Find the perimeter of the shaded region BCDBCD, giving your answer correct to 1 decimal place.

[4]

(b) Find the area of the shaded region BCDBCD, giving your answer correct to 1 decimal place.

[2]
题目中文翻译

图中显示一个圆心为 AA、半径为 1010 cm 的扇形 ABDABD。ABAB 的垂直平分线经过 DD。

(a) 求阴影区域 BCDBCD 的周长,答案精确到 1 位小数。

(b) 求阴影区域 BCDBCD 的面积,答案精确到 1 位小数。

解答

(a)

解法一

思路

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垂直平分线先确定 AC=CB=5AC=CB=5。利用直角三角形 ACDACD 求 CDCD 和圆心角,然后沿阴影边界求周长。

答题过程

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Since CC is the midpoint of ABAB, AC=CB=5AC=CB=5 cm. In the right triangle ACDACD with hypotenuse AD=10AD=10,

CD=102−52=53,cos⁡∠CAD=12.CD=\sqrt{10^2-5^2}=5\sqrt3,\qquad \cos\angle CAD=\frac12.

Thus the central angle BADBAD is π/3\pi/3. The perimeter of the shaded region is the arc BDBD plus CDCD and CBCB:

10⋅π3+53+5≈24.1 cm10\cdot\frac\pi3+5\sqrt3+5\approx24.1\text{ cm}

to one decimal place.

(b)

解法一

思路

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从整个扇形减去直角三角形 ACDACD。用 (a) 的精确边长和角度计算,再四舍五入。

答题过程

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The area of the shaded region is

12(10)2⋅π3−12(5)(53)=50π3−2532≈30.7 cm2\frac12(10)^2\cdot\frac\pi3-\frac12(5)(5\sqrt3) =\frac{50\pi}3-\frac{25\sqrt3}2 \approx30.7\text{ cm}^2

to one decimal place.