Official solution
Here; cos\[\theta \] = \[{{{a^2} + {b^2} - {c^2}} \over {2ab}}\]
and \[\theta \] = 60o
\[ \Rightarrow \] cos 60o = \[{{4 + 25 - {c^2}} \over {2.2.5}}\]
\[ \Rightarrow \] 10 = 29 - c2
\[ \Rightarrow \] c2 = 19
\[ \Rightarrow \] c = \[\sqrt {19} \]
also; cos\[\theta \] = \[{{{a^2} + {b^2} - {c^2}} \over {2ab}}\]
and \[\theta \] = 120o
\[ \Rightarrow \] - \[{1 \over 2}\] = \[{{{a^2} + {b^2} - 19} \over {2ab}}\]
\[ \Rightarrow \] 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 120o = 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