题目

Problem

Figure 2 shows the plan view of a garden.

Figure 2

The shape of the garden $ABCDEA$ consists of a triangle $ABE$ and a right-angled triangle $BCD$ joined to a sector $BDE$ of a circle with radius $6$ m and centre $B$ .

The points $A$ , $B$ and $C$ lie on a straight line with $AB=10.8$ m.

Angle $BCD=\dfrac{\pi}{2}$ radians, angle $EBD=1.3$ radians and $AE=12.2$ m.

(a) Find the area of the sector $BDE$ , giving your answer in $\text{m}^2$ .

(2)

(b) Find the size of angle $ABE$ , giving your answer in radians to $2$ decimal places.

(2)

(3)

解答

(a)

解法一

思路

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扇形面积公式是 $\frac12r^2\theta$ ，这里半径 $r=6$ ，圆心角 $\theta=1.3$ 。

答题过程

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\begin{align*} \text{Area of sector }BDE =&\,\frac12r^2\theta\\ =&\,\frac12(6)^2(1.3)\\ =&\,23.4. \end{align*}

Therefore the area is

\begin{align*} 23.4\text{ m}^2. \end{align*}

(b)

解法一

思路

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在三角形 $ABE$ 中，三边 $AB=10.8$ 、 $BE=6$ 、 $AE=12.2$ 都已知，所以用余弦定理求夹角 $\angle ABE$ 。

答题过程

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Using the cosine rule in triangle $ABE$ ,

\begin{align*} AE^2=AB^2+BE^2-2(AB)(BE)\cos(\angle ABE). \end{align*}

Substitute the values:

\begin{align*} 12.2^2 =&\,10.8^2+6^2-2(10.8)(6)\cos(\angle ABE). \end{align*}

\begin{align*} \cos(\angle ABE) =&\,\frac{10.8^2+6^2-12.2^2}{2(10.8)(6)}\\ =&\,\frac{19}{648}. \end{align*}

Hence

\begin{align*} \angle ABE =&\,\cos^{-1}\left(\frac{19}{648}\right)\\ =&\,1.54\text{ radians} \end{align*}

to $2$ decimal places.

(c)

解法一

思路

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花园总面积由三部分组成：

\begin{align*} \text{triangle }ABE+\text{sector }BDE+\text{triangle }BCD. \end{align*}

其中 $ABE$ 用 $\frac12ab\sin C$ ； $BCD$ 是直角三角形，先找出 $\angle DBC$ ，再用两条直角边求面积。

答题过程

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From part (b),

\begin{align*} \angle ABE=1.5414\ldots. \end{align*}

Area of triangle $ABE$ :

\begin{align*} \frac12(10.8)(6)\sin(1.5414\ldots) =&\,32.386\ldots. \end{align*}

Since $A$ , $B$ , $C$ are on a straight line,

\begin{align*} \angle DBC =&\,\pi-1.3-1.5414\ldots\\ =&\,0.3001\ldots. \end{align*}

In right-angled triangle $BCD$ ,

\begin{align*} BC=&\,6\cos(0.3001\ldots),\\ CD=&\,6\sin(0.3001\ldots). \end{align*}

\begin{align*} \text{Area of triangle }BCD =&\,\frac12(6\cos(0.3001\ldots))(6\sin(0.3001\ldots))\\ =&\,5.083\ldots. \end{align*}

Therefore the total area is

\begin{align*} 32.386\ldots+23.4+5.083\ldots =&\,60.869\ldots. \end{align*}

To $3$ significant figures,

\begin{align*} \text{area}=60.9\text{ m}^2. \end{align*}

IAL 2024 May Q5

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