18. A triangle $\triangle A B C$ is inscribed in a circle of radius $4 \mathrm{~cm}$. Suppose that $\angle A=60^{\circ}$, $A C-A B=4 \mathrm{~cm}$, and the area of $\triangle A B C$ is $x \mathrm{~cm}^{2}$. Find the value of $x^{2}$.
18. Ans: 192
Let $A B=c \mathrm{~cm}, B C=a \mathrm{~cm}$ and $C A=b \mathrm{~cm}$. By the cosine rule,
$$
a^{2}=b^{2}+c^{2}-2 b c \cos 60^{\circ}=b^{2}+c^{2}-b c=(b-c)^{2}+b c=4^{2}+b c=16+b c .
$$

Let $O$ be the centre of the circle. Then $\angle B O C=2 \angle A=120^{\circ}$. Moreover, $O B=$ $O C=4 \mathrm{~cm}$. Therefore, by the cosine rule,
$$
a^{2}=4^{2}+4^{2}-2 \times 4 \times 4 \cos 120^{\circ}=48 \Longrightarrow a=4 \sqrt{3} \text {. }
$$

Then $b c=a^{2}-4=48-16=32$. Thus,
$$
\text { area }(\triangle A B C)=\frac{1}{2} b c \sin 60^{\circ}=\frac{1}{2} \times 32 \times \frac{\sqrt{3}}{2}=8 \sqrt{3} \text {. }
$$

Therefore, $x=8 \sqrt{3}$ and $x^{2}=(8 \sqrt{3})^{2}=192$.