Trigonometry Bashing...

I'm not sure about if this is the only triangle satisfying the conditions above:a 30-60-90 triangle with side length $4,4\sqrt3,8$ ,so $x=8\sqrt3,(\frac x2)^2=48$

P.S.The other three choices seems impossible.I think choices in the form of $2^m\times3^n$ would be better choices,and it will avoid guesses.Just a suggestion,nevermind.

Let $\overline{AB}=c,\overline{AC}=b,\overline{BC}=a$ . so, it is given that $b-c=4\implies b=c+4$

Let $O$ be the circumcentre of $\triangle ABC$ . So, $\overline{OA}=\overline{OB}=\overline{OC}=4 ~~ (\text{given})$

Observe that $2\angle A=\angle BOC \implies \angle BOC=120^\circ$

Observe that $\triangle BOC$ is isoceles. So, $\angle OBC=\angle OCB=30^\circ$

So, by Sine rule in $\triangle BOC$ we have:

$\dfrac{\sin(30)}{\overline{OB}}=\dfrac{\sin(120)}{a} \implies a=\dfrac{\sin(120)\cdot 4}{\sin30}=\dfrac{\dfrac{\sqrt{3}}{2}\cdot 4}{\dfrac{1}{2}}=4\sqrt{3}$

Now, by Cosine rule in $\triangle ABC$ we have:

$\cos(A)=\cos(60)=\dfrac{1}{2}=\dfrac{b^2+c^2-(4\sqrt{3})^2}{2bc} \implies b^2+c^2-bc=48 - (1)$

by replacing the value of $b$ with $c+4$ in $(1)$ we get:

$\begin{aligned}(c+4)^2+c^2-(c+4)c=48\\ \implies c^2+4c-32=0\\ \implies (c+8)(c-4)=0 \end{aligned}$ .

But since, $c\in\mathbb{N}; c=4$ . $\therefore b=8$

So, now by Heron's formula on $\triangle ABC$ we get the area of $\triangle ABC=8\sqrt{3}=x$

Thus, $\left(\dfrac{x}{2}\right)^2=\left(\dfrac{8\sqrt{3}}{2}\right)^2=\boxed{48}$