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IAL 2022 May Q8

A Level / Edexcel / P1

IAL 2022 May Paper · Question 8

题目

Problem

Figure 3 shows a sketch of the outline of the face of a ceiling fan viewed from below.

Figure 3

The fan consists of three identical sections congruent to OABCDOOABCDO, shown in Figure 3, where

  • OABOOABO is a sector of a circle with centre OO and radius 9 cm9\,\text{cm}
  • OBCDOOBCDO is a sector of a circle with centre OO and radius 84 cm84\,\text{cm}
  • angle AOD=2π3AOD=\dfrac{2\pi}{3} radians

Given that the length of the arc ABAB is 15 cm15\,\text{cm},

(a) show that the length of the arc CDCD is 35.9 cm35.9\,\text{cm} to one decimal place.

(3)

The face of the fan is modelled to be a flat surface.

Find, according to the model,

(b) the perimeter of the face of the fan, giving your answer to the nearest cm,

(2)

(c) the surface area of the face of the fan.

Give your answer to 3 significant figures and make your units clear.

(5)

解答

(a)

解法一

思路

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同一片扇叶里，内外两段弧对应的圆心角加起来是 2π3\frac{2\pi}{3}。先用内弧 AB=15AB=15 和半径 99 求出它的圆心角，再求外弧 CDCD 的圆心角。

答题过程

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Let the angle subtended by arc ABAB be θ\theta.

Using s=rθs=r\theta,

15= 9θθ= 159= 53.\begin{align*} 15=&\,9\theta\\ \theta=&\,\frac{15}{9}\\ =&\,\frac53. \end{align*}

So the angle subtended by arc CDCD is

2π3−53.\begin{align*} \frac{2\pi}{3}-\frac53. \end{align*}

Therefore the length of arc CDCD is

84(2π3−53)= 35.929…= 35.9 cmto 1 d.p.\begin{align*} 84\left(\frac{2\pi}{3}-\frac53\right) =&\,35.929\ldots\\ =&\,35.9\text{ cm}\quad\text{to 1 d.p.} \end{align*}

(b)

解法一

思路

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整个风扇有三片相同部分。每片外边界贡献一段内弧、一段外弧，以及两条径向边。径向边长度是 84−9=7584-9=75。

答题过程

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Each section contributes:

15+35.929…+2(84−9).\begin{align*} 15+35.929\ldots+2(84-9). \end{align*}

So the total perimeter is

3(15+35.929…+2(75))= 602.787…= 603 cm\begin{align*} 3\left(15+35.929\ldots+2(75)\right) =&\,602.787\ldots\\ =&\,603\text{ cm} \end{align*}

to the nearest cm.

(c)

解法一

思路

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一片区域可以看成外大扇形减去内小扇形，但这里内小扇形对应的角是 53\frac53，外大扇形对应的角是 2π3−53\frac{2\pi}{3}-\frac53。算出一片面积后乘以 33。

答题过程

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The area of one outer sector is

12(84)2(2π3−53).\begin{align*} \frac12(84)^2\left(\frac{2\pi}{3}-\frac53\right). \end{align*}

The area of the corresponding inner sector is

12(9)2(53).\begin{align*} \frac12(9)^2\left(\frac53\right). \end{align*}

So the total area of the fan is

3[12(84)2(2π3−53)+12(9)2(53)]= 4729.577… cm2= 4730 cm2= 0.473 m2.\begin{align*} 3\left[ \frac12(84)^2\left(\frac{2\pi}{3}-\frac53\right) +\frac12(9)^2\left(\frac53\right) \right] =&\,4729.577\ldots\text{ cm}^2\\ =&\,4730\text{ cm}^2\\ =&\,0.473\text{ m}^2. \end{align*}

Therefore the surface area is

0.473 m2\begin{align*} 0.473\text{ m}^2 \end{align*}

to 3 significant figures.