题目

Problem

A sector $AOB$ , of a circle centre $O$ , has radius $r$ cm and angle $\theta$ radians.

Given that the area of the sector is $6\text{ cm}^2$ and that the perimeter of the sector is $10$ cm,

(a) show that

\begin{align*} 3\theta^2-13\theta+12=0. \end{align*}

(4)

(b) Hence find possible values of $r$ and $\theta$ .

(3)

解答

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用扇形面积和周长各写一个方程，然后由周长式写出 $r=\frac{10}{2+\theta}$ ，代入面积式。

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\begin{align*} \frac12r^2\theta=&\,6,\\ 2r+r\theta=&\,10. \end{align*}

From the perimeter equation,

\begin{align*} r(2+\theta)=&\,10\\ r=&\,\frac{10}{2+\theta}. \end{align*}

Substitute into the area equation:

\begin{align*} \frac12\left(\frac{10}{2+\theta}\right)^2\theta=&\,6\\ 50\theta=&\,6(2+\theta)^2\\ 50\theta=&\,6(\theta^2+4\theta+4)\\ 50\theta=&\,6\theta^2+24\theta+24\\ 3\theta^2-13\theta+12=&\,0. \end{align*}

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先解二次方程，再代回周长式求 $r$ 。

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\begin{align*} 3\theta^2-13\theta+12=&\,0\\ (3\theta-4)(\theta-3)=&\,0. \end{align*}

Hence

\begin{align*} \theta=\frac43 \quad\text{or}\quad \theta=3. \end{align*}

Using $r=\frac{10}{2+\theta}$ :

\begin{align*} \theta=\frac43&\Rightarrow r=3,\\ \theta=3&\Rightarrow r=2. \end{align*}

Therefore

\begin{align*} r=3,\ \theta=\frac43 \quad\text{or}\quad r=2,\ \theta=3. \end{align*}