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

Because the quadrilateral is cyclic, we may draw the diagonal that joins the two given adjacent sides. This diagonal splits the quadrilateral into two triangles that share the same diagonal.

Call the given adjacent sides $$a=2$$ and $$b=5$$, and let the angle between them be $$\theta=60^\circ$$. The diagonal that joins these sides will be denoted by $$d$$.

Area of the first triangle (with sides 2, 5 and included angle 60°):

$$ \text{Area}_1=\dfrac12\,ab\sin\theta =\dfrac12\,(2)(5)\sin60^\circ =5\cdot\dfrac{\sqrt3}{2} =\dfrac{5\sqrt3}{2}. $$

The whole quadrilateral has area $$4\sqrt3$$, so the second triangle must have area

$$ \text{Area}_2=4\sqrt3-\dfrac{5\sqrt3}{2} =\dfrac{8\sqrt3-5\sqrt3}{2} =\dfrac{3\sqrt3}{2}. $$

Let the other two sides of the quadrilateral (adjacent to each other and opposite the first angle) be $$c$$ and $$d_2$$. Their included angle is the angle opposite $$\theta$$. In a cyclic quadrilateral opposite angles sum to $$180^\circ$$, so this angle is

$$ 180^\circ-60^\circ=120^\circ. $$

The second triangle therefore has sides $$c$$ and $$d_2$$ with included angle $$120^\circ$$ and area

$$ \text{Area}_2=\dfrac12\,c\,d_2\sin120^\circ =\dfrac12\,c\,d_2\cdot\dfrac{\sqrt3}{2} =\dfrac{\sqrt3}{4}\,c\,d_2. $$

Equating this to the area already found,

$$ \dfrac{\sqrt3}{4}\,c\,d_2=\dfrac{3\sqrt3}{2}\;\;\;\Longrightarrow\;\;\; c\,d_2=6. \quad -(1) $$

Next, find the length of the common diagonal using the first triangle and the Law of Cosines:

$$ d^2=a^2+b^2-2ab\cos\theta =2^2+5^2-2\cdot2\cdot5\cos60^\circ =4+25-20\cdot\dfrac12 =29-10 =19. $$ Hence $$d=\sqrt{19}.$$

The same diagonal appears in the second triangle, so again by the Law of Cosines, now with sides $$c$$ and $$d_2$$ and included angle $$120^\circ$$,

$$ d^2=c^2+d_2^{\,2}-2c\,d_2\cos120^\circ. $$ Because $$\cos120^\circ=-\dfrac12,$$ this becomes

$$ 19=c^2+d_2^{\,2}-2c\,d_2\left(-\dfrac12\right) =c^2+d_2^{\,2}+c\,d_2. \quad -(2) $$

Substitute the product from (1) into (2):

$$ 19=c^2+d_2^{\,2}+6 \;\;\;\Longrightarrow\;\;\; c^2+d_2^{\,2}=13. \quad -(3) $$

Let the sum $$S=c+d_2.$$ Then

$$ S^2=(c+d_2)^2=c^2+d_2^{\,2}+2c\,d_2 =13+2\cdot6 =13+12 =25, $$ so $$S=5.$$

The perimeter $$P$$ of the quadrilateral is

$$ P=a+b+c+d_2 =2+5+S =2+5+5 =12. $$

Hence, the correct answer is Option D.