题目

Problem

Figure 3 shows a sketch of the plan view of a platform.

Figure 3

The plan view of the platform consists of a sector $DOC$ of a circle centre $O$ joined to a sector $AOBEA$ of a different circle, also with centre $O$ .

Given that

angle $AOB=0.8$ radians
arc length $CD=9$ m
$DA:AO=3:5$

(a) show that $AO=7.03$ m to 3 significant figures.

(3)

(b) Find the perimeter of the platform, in m, to 3 significant figures.

(3)

解答

(a)

解法一

思路

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外层小扇形的弧长 $CD=9$ ，圆心角也是 $0.8$ radians，所以先用 $s=r\theta$ 求外半径 $OD$ 。又因为 $DA:AO=3:5$ ，所以 $AO$ 是整个 $OD$ 的 $\frac58$ 。

答题过程

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Using $s=r\theta$ for arc $CD$ ,

\begin{align*} 9=&\,OD(0.8)\\ OD=&\,\frac{9}{0.8}\\ =&\,11.25. \end{align*}

Since

\begin{align*} DA:AO=3:5, \end{align*}

the whole length $DO$ is $8$ parts, and $AO$ is $5$ parts. Therefore

\begin{align*} AO =&\,\frac58(11.25)\\ =&\,7.03125\\ =&\,7.03\text{ m}\quad\text{to 3 significant figures.} \end{align*}

(b)

解法一

思路

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周长由三部分组成：外层弧 $CD$ 、两条直边 $DA$ 与 $CB$ 、以及内层的大弧 $AEB$ 。内层大弧的圆心角是 $2\pi-0.8$ 。

答题过程

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From part (a),

\begin{align*} OD=OC=11.25,\qquad AO=BO=7.03125. \end{align*}

\begin{align*} DA=CB =&\,11.25-7.03125\\ =&\,4.21875. \end{align*}

The major arc $AEB$ has angle

\begin{align*} 2\pi-0.8. \end{align*}

So its length is

\begin{align*} 7.03125(2\pi-0.8). \end{align*}

Therefore the perimeter is

\begin{align*} 9+2(4.21875)+7.03125(2\pi-0.8) =&\,55.982\ldots\\ =&\,56.0\text{ m} \end{align*}

to 3 significant figures.

(c)

解法一

思路

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总面积是外层小扇形 $DOC$ 加上内层大扇形 $AOBEA$ 。两个扇形都用 $\frac12r^2\theta$ ，注意内层用的是大角 $2\pi-0.8$ 。

答题过程

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The area of sector $DOC$ is

\begin{align*} \frac12(11.25)^2(0.8) =&\,50.625. \end{align*}

The area of sector $AOBEA$ is

\begin{align*} \frac12(7.03125)^2(2\pi-0.8) =&\,135.62\ldots. \end{align*}

Therefore the total area is

\begin{align*} 50.625+135.62\ldots =&\,186.24\ldots\\ =&\,186\text{ m}^2 \end{align*}

to the nearest whole number.

IAL 2024 Jan Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程