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(D) Let the sides of the triangle be $a$ and $b$. These are the roots of the equation $x^2-2 \sqrt{3} x+2=0$. From the properties of quadratic equatio

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$\sqrt{6}+\sqrt{3}$

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(D) Let the sides of the triangle be $a$ and $b$. These are the roots of the equation $x^2-2 \sqrt{3} x+2=0$. From the properties of quadratic equations,we have $a+b = 2 \sqrt{3}$ and $ab = 2$. The third side $c$ is given by the Law of Cosines: $c^2 = a^2+b^2-2ab \cos(\frac{\pi}{3})$. Since $a^2+b^2 = (a+b)^2-2ab$,we have $a^2+b^2 = (2 \sqrt{3})^2-2(2) = 12-4 = 8$. Thus,$c^2 = 8-2(2)(\frac{1}{2}) = 8-2 = 6$,which implies $c = \sqrt{6}$. The perimeter of the triangle is $a+b+c = 2 \sqrt{3}+\sqrt{6}$.

If the lengths of two sides of a triangle are the roots of the equation $x^2-2 \sqrt{3} x+2=0$ and the angle between these sides is $\frac{\pi}{3}$,then the perimeter of the triangle is

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