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IAL 2022 Oct Q8

A Level / Edexcel / P1

IAL 2022 Oct Paper · Question 8

题目

Problem

Figure 2 shows the plan view of a design for a pond.

Figure 2

The design consists of a sector AOBXAOBX of a circle centre OO joined to a quadrilateral AOBCAOBC.

  • BC=8 mBC=8\,\text{m}
  • OA=OB=3 mOA=OB=3\,\text{m}
  • angle AOB=2π3AOB=\dfrac{2\pi}{3} radians
  • angle BCA=π6BCA=\dfrac{\pi}{6} radians

(a) Calculate

(i) the exact area of the sector AOBXAOBX,

(ii) the exact perimeter of the sector AOBXAOBX.

(5)

(b) Calculate the exact area of the triangle AOBAOB.

(2)

(c) Show that the length ABAB is 33 m3\sqrt3\,\text{m}.

(2)

(d) Find the total surface area of the pond. Give your answer in m2\text{m}^2 correct to 2 significant figures.

(5)

解答

(a)(i)

解法一

思路

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扇形 AOBXAOBX 是图中外侧的大扇形。已知小角 ∠AOB=2π3\angle AOB=\frac{2\pi}{3}，所以大扇形的圆心角是 2π−2π3=4π32\pi-\frac{2\pi}{3}=\frac{4\pi}{3}。

答题过程

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The angle of the sector AOBXAOBX is

2π−2π3=4π3.\begin{align*} 2\pi-\frac{2\pi}{3} =\frac{4\pi}{3}. \end{align*}

Using A=12r2θA=\frac12r^2\theta,

area= 12(3)2(4π3)= 6π.\begin{align*} \text{area} =&\,\frac12(3)^2\left(\frac{4\pi}{3}\right)\\ =&\,6\pi. \end{align*}

So the exact area is

6π m2.\begin{align*} 6\pi\text{ m}^2. \end{align*}

(a)(ii)

解法一

思路

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扇形周长由两条半径和大弧长组成。弧长公式是 rθr\theta。

答题过程

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The arc length is

rθ= 3(4π3)= 4π.\begin{align*} r\theta =&\,3\left(\frac{4\pi}{3}\right)\\ =&\,4\pi. \end{align*}

The sector perimeter is

4π+3+3=4π+6.\begin{align*} 4\pi+3+3 =4\pi+6. \end{align*}

So the exact perimeter is

(4π+6) m.\begin{align*} (4\pi+6)\text{ m}. \end{align*}

(b)

解法一

思路

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三角形 AOBAOB 的两边 OAOA 和 OBOB 都是 33，夹角是 2π3\frac{2\pi}{3}，直接用面积公式 12absin⁡C\frac12ab\sin C。

答题过程

展开 area of triangle AOB= 12(3)(3)sin⁡(2π3)= 92⋅32= 934.\begin{align*} \text{area of triangle }AOB =&\,\frac12(3)(3)\sin\left(\frac{2\pi}{3}\right)\\ =&\,\frac92\cdot \frac{\sqrt3}{2}\\ =&\,\frac{9\sqrt3}{4}. \end{align*}

So the exact area is

934 m2.\begin{align*} \frac{9\sqrt3}{4}\text{ m}^2. \end{align*}

(c)

解法一

思路

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在三角形 AOBAOB 中已知两边和夹角，用余弦定理求 ABAB。这是 show that 题，要自然推出 333\sqrt3。

答题过程

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Using the cosine rule in triangle AOBAOB:

AB2= 32+32−2(3)(3)cos⁡(2π3)= 9+9−18(−12)= 27.\begin{align*} AB^2 =&\,3^2+3^2-2(3)(3)\cos\left(\frac{2\pi}{3}\right)\\ =&\,9+9-18\left(-\frac12\right)\\ =&\,27. \end{align*}

Since AB>0AB>0,

AB=27=33.\begin{align*} AB=\sqrt{27}=3\sqrt3. \end{align*}

Hence

AB=33 m.\begin{align*} AB=3\sqrt3\text{ m}. \end{align*}

(d)

解法一

思路

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总面积等于大扇形面积、三角形 AOBAOB 面积、三角形 ABCABC 面积之和。为了求三角形 ABCABC，先用正弦定理求角 BACBAC，再用夹角面积公式。

答题过程

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In triangle ABCABC,

AB=33,BC=8,∠BCA=π6.\begin{align*} AB=3\sqrt3, \qquad BC=8, \qquad \angle BCA=\frac{\pi}{6}. \end{align*}

Using the sine rule,

sin⁡BAC8= sin⁡(π6)33sin⁡BAC= 8⋅1233= 433.\begin{align*} \frac{\sin BAC}{8} =&\,\frac{\sin\left(\frac{\pi}{6}\right)}{3\sqrt3}\\ \sin BAC =&\,\frac{8\cdot \frac12}{3\sqrt3}\\ =&\,\frac4{3\sqrt3}. \end{align*}

So

∠BAC=0.8785…\begin{align*} \angle BAC=0.8785\ldots \end{align*}

and hence

∠ABC= π−π6−0.8785…\begin{align*} \angle ABC =&\,\pi-\frac{\pi}{6}-0.8785\ldots \end{align*}

Now

area of triangle ABC= 12(33)(8)sin⁡(π−π6−0.8785…)= 20.4896…\begin{align*} \text{area of triangle }ABC =&\,\frac12(3\sqrt3)(8) \sin\left(\pi-\frac{\pi}{6}-0.8785\ldots\right)\\ =&\,20.4896\ldots \end{align*}

Therefore the total surface area is

6π+934+20.4896…= 43.220…= 43 m2\begin{align*} 6\pi+\frac{9\sqrt3}{4}+20.4896\ldots =&\,43.220\ldots\\ =&\,43\text{ m}^2 \end{align*}

to 2 significant figures.

解法二

思路

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三角形 ABCABC 也可以不用先求角。过 BB 向 ACAC 作垂线，高是 8sin⁡π6=48\sin\frac{\pi}{6}=4。再分别求底边两段，得到三角形 ABCABC 的精确面积。

答题过程

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Drop a perpendicular from BB to ACAC, meeting ACAC at YY.

In triangle BCYBCY,

BY= 8sin⁡(π6)= 4,\begin{align*} BY =&\,8\sin\left(\frac{\pi}{6}\right)\\ =&\,4, \end{align*}

and

YC= 8cos⁡(π6)= 43.\begin{align*} YC =&\,8\cos\left(\frac{\pi}{6}\right)\\ =&\,4\sqrt3. \end{align*}

In triangle ABYABY,

AY= AB2−BY2= (33)2−42= 27−16= 11.\begin{align*} AY =&\,\sqrt{AB^2-BY^2}\\ =&\,\sqrt{(3\sqrt3)^2-4^2}\\ =&\,\sqrt{27-16}\\ =&\,\sqrt{11}. \end{align*}

So

AC=11+43.\begin{align*} AC=\sqrt{11}+4\sqrt3. \end{align*}

The area of triangle ABCABC is

12(AC)(BY)= 12(11+43)(4)= 211+83.\begin{align*} \frac12(AC)(BY) =&\,\frac12(\sqrt{11}+4\sqrt3)(4)\\ =&\,2\sqrt{11}+8\sqrt3. \end{align*}

Therefore the total surface area is

6π+934+211+83= 6π+211+4134= 43.220…= 43 m2\begin{align*} 6\pi+\frac{9\sqrt3}{4} +2\sqrt{11}+8\sqrt3 =&\,6\pi+2\sqrt{11}+\frac{41\sqrt3}{4}\\ =&\,43.220\ldots\\ =&\,43\text{ m}^2 \end{align*}

to 2 significant figures.

解法三

思路

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利用余弦定理直接建立一元二次方程求 ACAC 的方法。 在三角形 ABCABC 中，已知 BC=8BC = 8，由 (c) 可知 AB=33AB = 3\sqrt3，且角 ∠BCA=π6\angle BCA = \frac{\pi}{6}。 我们可以利用余弦定理直接建立关于 ACAC 的二次方程：AB2=AC2+BC2−2⋅AC⋅BCcos⁡∠BCAAB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cos\angle BCA。 代入已知数据化简整理后，得到一个一元二次方程：AC2−83AC+37=0AC^2 - 8\sqrt3 AC + 37 = 0。 解该方程求出 AC=43±11AC = 4\sqrt3 \pm \sqrt{11}。因为 ACAC 在图形中是钝角三角形中最长的边，且 AC>BC=8AC > BC = 8，所以我们舍去较小根，取 AC=43+11AC = 4\sqrt3 + \sqrt{11}。 接着利用面积公式 Area=12⋅BC⋅ACsin⁡∠BCA\text{Area} = \frac{1}{2} \cdot BC \cdot AC \sin\angle BCA 即可求出三角形 ABCABC 的面积，最后累加三部分面积得出总表面积。

答题过程

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Using the cosine rule in triangle ABCABC:

AB2=AC2+BC2−2⋅AC⋅BCcos⁡∠BCA.\begin{align*} AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cos \angle BCA. \end{align*}

Substitute the given values AB=33AB = 3\sqrt3, BC=8BC = 8, and ∠BCA=π6\angle BCA = \frac{\pi}{6}:

(33)2=  AC2+82−2⋅AC⋅8cos⁡(π6)27=  AC2+64−16⋅AC(32)AC2−83AC+37=  0.\begin{align*} (3\sqrt3)^2 =&\,\, AC^2 + 8^2 - 2 \cdot AC \cdot 8 \cos\left(\frac{\pi}{6}\right)\\[3mm] 27 =&\,\, AC^2 + 64 - 16 \cdot AC \left(\frac{\sqrt3}{2}\right)\\[3mm] AC^2 - 8\sqrt3 AC + 37 =&\,\, 0. \end{align*}

Solve for ACAC using the quadratic formula:

AC=  83±(−83)2−4(1)(37)2=  83±192−1482=  83±442=  83±2112=  43±11.\begin{align*} AC =&\,\, \frac{8\sqrt3 \pm \sqrt{(-8\sqrt3)^2 - 4(1)(37)}}{2}\\[3mm] =&\,\, \frac{8\sqrt3 \pm \sqrt{192 - 148}}{2}\\[3mm] =&\,\, \frac{8\sqrt3 \pm \sqrt{44}}{2}\\[3mm] =&\,\, \frac{8\sqrt3 \pm 2\sqrt{11}}{2}\\[3mm] =&\,\, 4\sqrt3 \pm \sqrt{11}. \end{align*}

\begin{align*} Since ACAC is the longest side of the triangle (AC>BC=8AC > BC = 8), we choose the larger root: \end{align*}

AC=43+11.AC = 4\sqrt3 + \sqrt{11}.

Now calculate the area of triangle ABCABC:

AreaABC=  12⋅BC⋅ACsin⁡∠BCA=  12(8)(43+11)sin⁡(π6)=  4(43+11)(12)=  83+211.\begin{align*} \text{Area}_{ABC} =&\,\, \frac{1}{2} \cdot BC \cdot AC \sin \angle BCA\\[3mm] =&\,\, \frac{1}{2}(8)(4\sqrt3 + \sqrt{11})\sin\left(\frac{\pi}{6}\right)\\[3mm] =&\,\, 4(4\sqrt3 + \sqrt{11})\left(\frac{1}{2}\right)\\[3mm] =&\,\, 8\sqrt3 + 2\sqrt{11}. \end{align*}

The total surface area of the pond is:

Total Area=  Areasector+AreaAOB+AreaABC=  6π+934+83+211=  6π+211+4134≈  43.220…≈  43 m2\begin{align*} \text{Total Area} =&\,\, \text{Area}_{\text{sector}} + \text{Area}_{AOB} + \text{Area}_{ABC}\\[3mm] =&\,\, 6\pi + \frac{9\sqrt3}{4} + 8\sqrt3 + 2\sqrt{11}\\[3mm] =&\,\, 6\pi + 2\sqrt{11} + \frac{41\sqrt3}{4}\\[3mm] \approx&\,\, 43.220\ldots\\[3mm] \approx&\,\, 43\text{ m}^2 \end{align*}

to 2 significant figures.