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CIE 9709 2025 June Paper 11 Q9

A Level / CIE / P1

CIE 9709 2025 June Paper 11 Paper · Question 9

题目

Problem

The diagram shows a sector ABCABC of a circle with centre AA and radius rr cm. The angle BACBAC is α\alpha radians, where 0<α<12π0 < \alpha < \frac{1}{2}\pi.

(a) It is given that the area of the triangle ABCABC is 4 cm24\text{ cm}^2 and the area of the sector ABCABC is 8α cm28\alpha\text{ cm}^2.

Find the exact area of the shaded segment.

[4]

(b) It is given instead that the length of the chord BCBC is 12r\frac{1}{\sqrt{2}}r cm but the area of the triangle ABCABC is still 4 cm24\text{ cm}^2.

Find the area of the shaded segment. Give your answer correct to 3 significant figures.

[4]
题目中文翻译

图中显示了一个圆心为 AA、半径为 rr cm 的扇形 ABCABC。角 BACBAC 为 α\alpha 弧度,其中 0<α<12π0 < \alpha < \frac{1}{2}\pi。

(a) 已知三角形 ABCABC 的面积为 4 cm24\text{ cm}^2,扇形 ABCABC 的面积为 8α cm28\alpha\text{ cm}^2。

求阴影弓形面积的精确值。

[4]

(b) 改为已知弦 BCBC 的长度为 12r\frac{1}{\sqrt{2}}r cm,但三角形 ABCABC 的面积仍为 4 cm24\text{ cm}^2。

求阴影弓形的面积。答案精确到 3 位有效数字。

[4]

解答

(a)

解法一

思路

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先用扇形面积确定 r2r^2,再用三角形面积确定锐角 α\alpha。弓形面积等于扇形减三角形。

答题过程

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The sector area satisfies 12r2α=8α\frac12r^2\alpha=8\alpha. As α>0\alpha>0, r2=16r^2=16. The triangle area gives 12(16)sin⁡α=4\frac12(16)\sin\alpha=4, so sin⁡α=12\sin\alpha=\frac12. Since α\alpha is acute, α=π6\alpha=\frac\pi6. The shaded area is

8⋅π6−4=(4π3−4) cm2.8\cdot\frac\pi6-4=\Big(\frac{4\pi}3-4\Big)\text{ cm}^2.

(b)

解法一

思路

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弦长给出了等腰三角形的三边比例,可用余弦定理求角。再用固定的三角形面积求 r2r^2。

答题过程

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The cosine rule gives

r22=r2+r2−2r2cos⁡α,\frac{r^2}{2}=r^2+r^2-2r^2\cos\alpha,

so cos⁡α=34\cos\alpha=\frac34 and sin⁡α=74\sin\alpha=\frac{\sqrt7}{4}. Since the triangle area is 44,

12r2⋅74=4⟹r2=327.\frac12r^2\cdot\frac{\sqrt7}4=4 \quad\Longrightarrow\quad r^2=\frac{32}{\sqrt7}.

The shaded area is therefore

167cos⁡−1(34)−4≈0.371 cm2\frac{16}{\sqrt7}\cos^{-1}\Big(\frac34\Big)-4 \approx0.371\text{ cm}^2

to three significant figures.