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IAL 2024 May R Q6

A Level / Edexcel / P1

IAL 2024 May (R) Paper · Question 6

题目

Problem

Figure 1 shows a sketch of the entrance to a tunnel.

Figure 1

The shape of the entrance consists of a sector BCDFBCDF, of a circle centre FF, joined to two congruent (identical) triangles ABFABF and EDFEDF.

Given that

AFE is a straight line\begin{align*} AFE\text{ is a straight line} \end{align*} AF=FE=6.4 m\begin{align*} AF=FE=6.4\text{ m} \end{align*} FB=FD=6.2 m\begin{align*} FB=FD=6.2\text{ m} \end{align*} angle BFD=2.275 radians\begin{align*} \text{angle }BFD=2.275\text{ radians} \end{align*}

(a) Show that angle AFB=0.433AFB=0.433 radians to 33 decimal places.

(1)

(b) Find the perimeter of the entrance to the tunnel, ABCDEFAABCDEFA, in metres, to one decimal place.

(4)

(c) Find the cross-sectional area of the entrance to the tunnel, ABCDEFAABCDEFA, in m2\text{m}^2, to one decimal place.

(4)

解答

(a)

解法一

思路

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AFEAFE 是直线,所以围绕点 FF 的上半部分总角度是 π\pi。中间扇形角是 2.2752.275,两侧三角形全等,所以两侧角相等。

答题过程

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Since AFEAFE is a straight line and the two triangles are congruent,

AFB=π2.2752=0.433296=0.433 radians\begin{align*} \angle AFB =&\,\frac{\pi-2.275}{2}\\ =&\,0.433296\ldots\\ =&\,0.433\text{ radians} \end{align*}

to 33 decimal places.

(b)

解法一

思路

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周长由底边 AF+FEAF+FE、两条斜边 AB,DEAB,DE 和弧长 BCDBCD 组成。弧长用 rθr\theta;斜边 ABAB 可在三角形 ABFABF 中用余弦定理求出。

答题过程

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The arc length BCDBCD is

rθ=6.2(2.275)=14.105.\begin{align*} r\theta =&\,6.2(2.275)\\ =&\,14.105. \end{align*}

Using the cosine rule in triangle ABFABF,

AB2=6.42+6.222(6.4)(6.2)cos(0.433296)=7.367.\begin{align*} AB^2 =&\,6.4^2+6.2^2-2(6.4)(6.2)\cos(0.433296\ldots)\\ =&\,7.367\ldots. \end{align*}

So

AB=2.7142.\begin{align*} AB=2.7142\ldots. \end{align*}

The perimeter is

2AB+arc BCD+AF+FE=2(2.7142)+14.105+12.8=32.333.\begin{align*} 2AB+\text{arc }BCD+AF+FE =&\,2(2.7142\ldots)+14.105+12.8\\ =&\,32.333\ldots. \end{align*}

Therefore the perimeter is

32.3 m.\begin{align*} 32.3\text{ m}. \end{align*}

(c)

解法一

思路

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总面积等于中间扇形面积加两侧全等三角形面积。扇形面积用 12r2θ\frac12r^2\theta,三角形面积用 12absinC\frac12ab\sin C

答题过程

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The area of sector BCDFBCDF is

12r2θ=12(6.2)2(2.275)=43.7255.\begin{align*} \frac12r^2\theta =&\,\frac12(6.2)^2(2.275)\\ =&\,43.7255. \end{align*}

The area of triangle ABFABF is

12(6.4)(6.2)sin(0.433296)=8.325.\begin{align*} \frac12(6.4)(6.2)\sin(0.433296\ldots) =8.325\ldots. \end{align*}

There are two congruent triangles, so the total area is

43.7255+2(8.325)=60.376.\begin{align*} 43.7255+2(8.325\ldots) =&\,60.376\ldots. \end{align*}

Therefore the cross-sectional area is

60.4 m2.\begin{align*} 60.4\text{ m}^2. \end{align*}