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IAL 2020 Oct Q3

A Level / Edexcel / P1

IAL 2020 Oct Paper · Question 3

题目

Problem

A badge design consists of two congruent triangles, AOCAOC and BOCBOC, joined to a sector AOBAOB of a circle with centre OO.

The badge is shown in Figure 1.

Figure 1

Given that ∠AOB=α\angle AOB=\alpha radians, AO=OB=3AO=OB=3 cm, OC=5OC=5 cm, and the area of the sector AOBAOB is 7.2 cm27.2\text{ cm}^2,

(a) show that α=1.6\alpha=1.6.

(2)

(b) Find

(i) the total area of the badge, to 2 significant figures,

(ii) the total perimeter of the badge, to 1 decimal place.

(8)

解答

(a)

解法一

思路

展开

扇形面积公式是 12r2θ\frac12r^2\theta。这里半径是 33，扇形面积是 7.27.2，直接代入即可。

答题过程

展开 12(3)2α= 7.292α= 7.2α= 7.24.5= 1.6.\begin{align*} \frac12(3)^2\alpha=&\,7.2\\ \frac92\alpha=&\,7.2\\ \alpha=&\,\frac{7.2}{4.5}\\ =&\,1.6. \end{align*}

(b)(i)

解法一

思路

展开

两个三角形全等，所以只要求一个三角形的面积。围绕点 OO 的总角度是 2π2\pi，减去扇形角 1.61.6 后，剩下的角度平均分给 ∠COA\angle COA 和 ∠COB\angle COB。

答题过程

展开 ∠COA= 12(2π−1.6)= π−0.8.\begin{align*} \angle COA =&\,\frac12(2\pi-1.6)\\ =&\,\pi-0.8. \end{align*}

The area of one triangle is

12(5)(3)sin⁡(π−0.8).\begin{align*} \frac12(5)(3)\sin(\pi-0.8). \end{align*}

Hence the total area is

7.2+2{12(5)(3)sin⁡(π−0.8)}= 17.960…= 18 cm2\begin{align*} 7.2+2\left\{\frac12(5)(3)\sin(\pi-0.8)\right\} =&\,17.960\ldots\\ =&\,18\text{ cm}^2 \end{align*}

to 2 significant figures.

(b)(ii)

解法一

思路

展开

周长由弧 ABAB、边 ACAC 和边 BCBC 组成。由于两个三角形全等，AC=BCAC=BC。弧长用 rθr\theta，边 ACAC 用余弦定理。

答题过程

展开

The arc length ABAB is

3(1.6)=4.8.\begin{align*} 3(1.6)=4.8. \end{align*}

Using the cosine rule in triangle AOCAOC,

AC2= 52+32−2(5)(3)cos⁡(π−0.8)= 54.904…,\begin{align*} AC^2 =&\,5^2+3^2-2(5)(3)\cos(\pi-0.8)\\ =&\,54.904\ldots, \end{align*}

so

AC=7.409….\begin{align*} AC=7.409\ldots. \end{align*}

Therefore the perimeter is

4.8+2(7.409…)= 19.619…= 19.6 cm\begin{align*} 4.8+2(7.409\ldots) =&\,19.619\ldots\\ =&\,19.6\text{ cm} \end{align*}

to 1 decimal place.