The two adjacent sides a cyclic quadrilateral are 2 and 5 and the angle between them is 60°. If the area of the quadrilateral is $4\sqrt{3}$, then the perimeter of the quadrilateral is :

12

$\cos 60°=\frac{4+25-c^2}{2.2.5}⇒\frac{1}{2}=\frac{29-c^2}{20}$

$⇒10=29-c^2⇒c^2=19⇒c=\sqrt{19}$

Now, $\cos 120°=\frac{a^2+b^2-c^2}{2ab}⇒-\frac{1}{2}=\frac{a^2+b^2-19}{2ab}$

$⇒a^2+b^2-19=-ab⇒a^2+b^2+ab=19$

Area of quadrilateral = $\frac{1}{2}×2×5×\sin 60°+\frac{1}{2}ab\sin 120°=4\sqrt{3}$

$⇒\frac{5\sqrt{3}}{2}+\frac{ab\sqrt{3}}{4}=4\sqrt{3}⇒\frac{ab}{4}=4-\frac{5}{2}=\frac{3}{2}$

⇒ ab = 6 $∴ a^2 + b^2 = 13$ ∴ a = 2, b = 3

Perimeter of quadrilateral = 2 + 5 + 2 + 3 = 12