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IAL 2023 May Q5

A Level / Edexcel / P1

IAL 2023 May Paper · Question 5

题目

Problem

Figure 1 shows the plan for a garden.

Figure 1

In the plan

  • OAOA and CDCD are perpendicular to ODOD
  • ABAB is an arc of the circle with centre OO and radius 44 metres
  • BCBC is parallel to ODOD
  • ODOD is 66 metres, OAOA is 44 metres and CDCD is 1.51.5 metres

(a) Show that angle AOBAOB is 1.1861.186 radians to 4 significant figures.

(2)

(b) Find the perimeter of the garden, giving your answer in metres to 3 significant figures.

(4)

(c) Find the area of the garden, giving your answer in square metres to 3 significant figures.

(4)

解答

(a)

解法一

思路

展开

BB 在半径为 44 的圆上,且 BCBC 水平,所以 BB 的高度与 CC 相同,都是 1.51.5。在直角三角形中,OB=4OB=4,竖直高度是 1.51.5。可先求 BOD\angle BOD,再用 π2BOD\frac\pi2-\angle BOD 得到 AOB\angle AOB

答题过程

展开

Since BCBC is parallel to ODOD, point BB has height 1.51.5 above ODOD.

In the right triangle with hypotenuse OB=4OB=4,

sinBOD=1.54.\begin{align*} \sin\angle BOD=\frac{1.5}{4}. \end{align*}

So

AOB=π2sin1(1.54)=1.186=1.186\begin{align*} \angle AOB =&\,\frac{\pi}{2}-\sin^{-1}\left(\frac{1.5}{4}\right)\\ =&\,1.186\ldots\\ =&\,1.186 \end{align*}

to 4 significant figures.

(b)

解法一

思路

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周长由 OAOA、弧 ABABBCBCCDCDDODO 组成。弧长用 rθr\theta,而 BCBC 可由 DDBB 的水平距离算出。

答题过程

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The arc length ABAB is

4(1.186)=4.744.\begin{align*} 4(1.186\ldots)=4.744\ldots. \end{align*}

The horizontal distance from OO to BB is

421.52=3.708.\begin{align*} \sqrt{4^2-1.5^2}=3.708\ldots. \end{align*}

Hence

BC=63.708=2.291.\begin{align*} BC =&\,6-3.708\ldots\\ =&\,2.291\ldots. \end{align*}

Therefore the perimeter is

4+4.744+2.291+1.5+6=18.535=18.5 m\begin{align*} 4+4.744\ldots+2.291\ldots+1.5+6 =&\,18.535\ldots\\ =&\,18.5\text{ m} \end{align*}

to 3 significant figures.

(c)

解法一

思路

展开

花园面积可以拆成扇形 OABOAB 加上四边形 OBCDOBCD。四边形 OBCDOBCD 是一个宽为 66、高为 1.51.5 的矩形减去右上方的直角三角形。

答题过程

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Area of sector OABOAB:

12(4)2(1.186)=9.491.\begin{align*} \frac12(4)^2(1.186\ldots)=9.491\ldots. \end{align*}

For OBCDOBCD, use the rectangle of area 6(1.5)6(1.5) and subtract the triangle with base 3.7083.708\ldots and height 1.51.5:

Area of OBCD=6(1.5)12(3.708)(1.5)=6.218.\begin{align*} \text{Area of }OBCD =&\,6(1.5)-\frac12(3.708\ldots)(1.5)\\ =&\,6.218\ldots. \end{align*}

So the total area is

9.491+6.218=15.710=15.7 m2\begin{align*} 9.491\ldots+6.218\ldots =&\,15.710\ldots\\ =&\,15.7\text{ m}^2 \end{align*}

to 3 significant figures.