题目

Problem

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $r$ cm. Angle $AOB = \theta$ radians. It is given that the length of the arc $AB$ is $9.6$ cm and that the area of the sector $OAB$ is $76.8$ cm $^2$ .

(a) Find the area of the shaded region.

[5]

(b) Find the perimeter of the shaded region.

[2]

题目中文翻译

图中显示圆心为 $O$ 、半径为 $r$ cm 的圆的扇形 $OAB$ 。角 $AOB = \theta$ 弧度。已知弧 $AB$ 的长度为 $9.6$ cm，扇形 $OAB$ 的面积为 $76.8$ cm $^2$ 。

(a) 求阴影区域的面积。

[5]

(b) 求阴影区域的周长。

[2]

解答

(a)

解法一

思路

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弧长和扇形面积同时包含 $r\theta$ ，把两式相除可以先求半径。阴影是弓形，面积等于扇形面积减去两条半径与弦围成的三角形面积。

答题过程

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The arc length and sector area give

r\theta=9.6, \qquad \frac12r^2\theta=76.8.

Since $r$ and $\theta$ are positive, dividing the second equation by the first gives

\frac r2=\frac{76.8}{9.6}=8.

Hence $r=16$ cm and $\theta=9.6/16=0.6$ radians. The area of triangle $OAB$ is

\frac12r^2\sin\theta=128\sin0.6.

Subtracting this from the sector area, the shaded segment has area

76.8-128\sin0.6\approx4.53\text{ cm}^2.

(b)

解法一

思路

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弓形的周长由弧 $AB$ 和弦 $AB$ 组成，不包含两条半径。平分等腰三角形后，用半角的正弦求半弦长。

答题过程

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Bisecting the isosceles triangle $OAB$ gives the chord length

AB=2r\sin\frac\theta2=32\sin0.3.

Adding the arc length gives the perimeter of the shaded segment:

9.6+32\sin0.3\approx19.1\text{ cm}.

CIE 9709 2023 June Paper 13 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程