The problem is into a textbook of high school of 15 years old.
In a square triangle $ABC$, of hypotenuse $BC$, $A\hat B C=60°$. Knowing that the perimeter is $6(\sqrt3+1)$, find the area of the triangle.
In the text there is a suggestion: Put $\overline{AB}=x$ and solve the equation which formalizes the problem, is found $x=2\sqrt3$.
Solution: I have not taken the suggestion of the textbook and I have used the goniometry.
I have taken, instead $\overline{BC}=x$ and $$\overline{AB}=\overline{BC}\cos 60°=\frac x2, \quad \overline{AC}=\overline{BC}\cos 30°=\frac{\sqrt{3}}{2}x$$ Hence
$$\text{perimeter}=\frac x2+\frac{\sqrt{3}}{2}x+x=6(\sqrt3+1)\iff x=4\sqrt{3}$$ $\overline{AB}=2\sqrt{3}$, $\overline{AC}=\frac{\sqrt{3}}{2}\cdot 4\sqrt{3}=2\sqrt{3}$ and the area $$\mathcal A=\frac{2\sqrt{3}\cdot 2\sqrt{3}}{2}=2\sqrt{3}\ne 6\sqrt{3}$$ that is not the solution of the exercise $6\sqrt3$.
What am I doing wrong?