Here; cos$$\theta $$ = $${{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

and  $$\theta $$ = 60^o

$$ \Rightarrow $$   cos 60^o = $${{4 + 25 - {c^2}} \over {2.2.5}}$$

$$ \Rightarrow $$    10 = 29 $$-$$ c²

$$ \Rightarrow $$   c² = 19

$$ \Rightarrow $$   c = $$\sqrt {19} $$

also;  cos$$\theta $$ = $${{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

and $$\theta $$ = 120^o

$$ \Rightarrow $$   $$-$$ $${1 \over 2}$$ = $${{{a^2} + {b^2} - 19} \over {2ab}}$$

$$ \Rightarrow $$   a² + b² $$-$$ 19 = $$-$$ ab

$$ \Rightarrow $$   a² + b² + ab = 19

$$ \therefore $$   Area = $${1 \over 2} \times 2 \times 5$$ sin 60 + $${1 \over 2}$$ ab sin 120^o = 4$$\sqrt 3 $$

$$ \Rightarrow $$   $${{5\sqrt 3 } \over 2} + {{ab\sqrt 3 } \over 4}$$ = $$4\sqrt 3 $$

$$ \Rightarrow $$   $${{ab} \over 4}$$ = 4 $$-$$ $${5 \over 2}$$ = $${3 \over 2}$$

$$ \Rightarrow $$    ab = 6

$$ \therefore $$   a² + b² = 13

$$ \Rightarrow $$   a = 2, b = 3

Perimeter = Sum of all sides

= 2 + 5 + 2 + 3 = 12