题目

Problem

Figure 2 shows a plan view of a semicircular garden $ABCDEOA$ .

The semicircle has

centre $O$
diameter $AOE$
radius $3$ m

Figure 2

The straight line $BD$ is parallel to $AE$ and angle $BOA$ is $0.7$ radians.

(a) Show that, to 4 significant figures, angle $BOD$ is $1.742$ radians.

(1)

The flowerbed $R$ , shown shaded in Figure 2, is bounded by $BD$ and the arc $BCD$ .

(b) Find the area of the flowerbed, giving your answer in square metres to one decimal place.

(3)

解答

(a)

解法一

思路

展开

$BD\parallel AE$ ，所以左右两边被切掉的两个小角相等，都是 $0.7$ radians。整个半圆对应的中心角是 $\pi$ ，因此中间角就是 $\pi-2(0.7)$ 。

答题过程

展开

Since $BD$ is parallel to $AE$ ,

\begin{align*} \angle BOA=\angle EOD=0.7. \end{align*}

The angle on a semicircle at the centre is $\pi$ radians, so

\begin{align*} \angle BOD =&\,\pi-0.7-0.7\\ =&\,\pi-1.4\\ =&\,1.74159\ldots\\ =&\,1.742\quad\text{to 4 significant figures.} \end{align*}

(b)

解法一

思路

展开

花圃面积是扇形 $BOD$ 减去三角形 $BOD$ 。半径是 $3$ ，中心角用 (a) 的 $1.742$ radians。

答题过程

展开

The area of sector $BOD$ is

\begin{align*} \frac12r^2\theta =&\,\frac12(3)^2(1.742). \end{align*}

The area of triangle $BOD$ is

\begin{align*} \frac12r^2\sin\theta =&\,\frac12(3)^2\sin(1.742). \end{align*}

Therefore the area of $R$ is

\begin{align*} \frac12(3)^2(1.742) -\frac12(3)^2\sin(1.742) =&\,3.401\ldots\\ =&\,3.4. \end{align*}

So the area of the flowerbed is

\begin{align*} 3.4\text{ m}^2. \end{align*}

(c)

解法一

思路

展开

周长由弧长 $BCD$ 和弦长 $BD$ 组成。弧长用 $r\theta$ 。弦长可以用等腰三角形的一半，也就是 $BD=2r\sin(\theta/2)$ 。

答题过程

展开

The arc length $BCD$ is

\begin{align*} 3(1.742)=5.226. \end{align*}

The chord length $BD$ is

\begin{align*} BD =&\,2(3)\sin\left(\frac{1.742}{2}\right)\\ =&\,4.590\ldots. \end{align*}

Therefore the perimeter of $R$ is

\begin{align*} 5.226+4.590\ldots =&\,9.816\ldots\\ =&\,9.8. \end{align*}

So the perimeter of the flowerbed is

\begin{align*} 9.8\text{ m}. \end{align*}

IAL 2022 Jan 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