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IAL 2023 Jan Q6

A Level / Edexcel / P1

IAL 2023 Jan Paper · Question 6

题目

Problem

Figure 1 shows the plan view for the design of a stage.

Figure 1

The design consists of a sector OBCOBC of a circle, with centre OO, joined to two congruent triangles OABOAB and ODCODC.

Given that

  • angle BOC=2.4BOC=2.4 radians
  • area of sector BOC=40 m2BOC=40\,\text{m}^2
  • AODAOD is a straight line of length 12.5 m12.5\,\text{m}

(a) find the radius of the sector, giving your answer, in m, to 2 decimal places,

(2)

(b) find the size of angle AOBAOB, in radians, to 2 decimal places.

(1)

Hence find

(c) the total area of the stage, giving your answer, in m2\text{m}^2, to one decimal place,

(3)

(d) the total perimeter of the stage, giving your answer, in m, to one decimal place.

(4)

解答

(a)

解法一

思路

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扇形面积公式是 A=12r2θA=\frac12r^2\theta，其中 θ\theta 必须用弧度。题目已经给了面积和圆心角，直接代入求半径。

答题过程

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Use the sector area formula:

A=12r2θ.\begin{align*} A=\frac12r^2\theta. \end{align*}

Here A=40A=40 and θ=2.4\theta=2.4, so

40= 12r2(2.4)40= 1.2r2r2= 401.2= 802.4.\begin{align*} 40=&\,\frac12r^2(2.4)\\ 40=&\,1.2r^2\\ r^2=&\,\frac{40}{1.2}\\ =&\,\frac{80}{2.4}. \end{align*}

Therefore

r= 802.4= 5.77 mto 2 d.p.\begin{align*} r=&\,\sqrt{\frac{80}{2.4}}\\ =&\,5.77\text{ m}\quad\text{to 2 d.p.} \end{align*}

(b)

解法一

思路

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AODAOD 是一条直线，所以从 OAOA 到 ODOD 的总角度是 π\pi。两边三角形全等，因此 ∠AOB=∠COD\angle AOB=\angle COD。

答题过程

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Since AODAOD is a straight line,

∠AOB+∠BOC+∠COD=π.\begin{align*} \angle AOB+\angle BOC+\angle COD=\pi. \end{align*}

The two triangles are congruent, so

∠AOB=∠COD.\begin{align*} \angle AOB=\angle COD. \end{align*}

Therefore

2∠AOB+2.4= π∠AOB= π−2.42= 0.37 radiansto 2 d.p.\begin{align*} 2\angle AOB+2.4=&\,\pi\\ \angle AOB=&\,\frac{\pi-2.4}{2}\\ =&\,0.37\text{ radians}\quad\text{to 2 d.p.} \end{align*}

(c)

解法一

思路

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舞台面积由中间扇形加上两个全等三角形组成。三角形 OABOAB 的两边是 OA=6.25OA=6.25 和 OB=rOB=r，夹角是上一小题的角。

答题过程

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Since AOD=12.5AOD=12.5 and the two triangles are congruent,

OA=OD=12.52=6.25.\begin{align*} OA=OD=\frac{12.5}{2}=6.25. \end{align*}

Using r=5.77r=5.77 and ∠AOB=0.37\angle AOB=0.37,

area of triangle OAB= 12(6.25)(5.77)sin⁡(0.37)= 6.5377…\begin{align*} \text{area of triangle }OAB =&\,\frac12(6.25)(5.77)\sin(0.37)\\ =&\,6.5377\ldots \end{align*}

So the total area is

40+2(6.5377…)= 53.075…= 53.1 m2to 1 d.p.\begin{align*} 40+2(6.5377\ldots) =&\,53.075\ldots\\ =&\,53.1\text{ m}^2\quad\text{to 1 d.p.} \end{align*}

(d)

解法一

思路

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外周长由 ADAD、两条相等的边 ABAB 和 DCDC、以及弧 BCBC 组成。弧长用 rθr\theta，边 ABAB 用余弦定理。

答题过程

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The arc length BCBC is

rθ= 5.77(2.4)= 13.848.\begin{align*} r\theta =&\,5.77(2.4)\\ =&\,13.848. \end{align*}

Find ABAB using the cosine rule in triangle AOBAOB:

AB2= 6.252+5.772−2(6.25)(5.77)cos⁡(0.37)= 5.105…\begin{align*} AB^2 =&\,6.25^2+5.77^2\\ &\quad -2(6.25)(5.77)\cos(0.37)\\ =&\,5.105\ldots \end{align*}

Hence

AB= 5.105…= 2.259…\begin{align*} AB=&\,\sqrt{5.105\ldots}\\ =&\,2.259\ldots \end{align*}

The total perimeter is

12.5+2(2.259…)+13.848= 30.866…= 30.9 mto 1 d.p.\begin{align*} 12.5+2(2.259\ldots)+13.848 =&\,30.866\ldots\\ =&\,30.9\text{ m}\quad\text{to 1 d.p.} \end{align*}