If G is the centroid of $$\triangle^{le}$$ ABC and A = ( 1,3), B = (3, 7) and G = ( 4,5) then the perimeter of the $$\triangle$$ ABC is

A

$$2\sqrt{5} + \sqrt{29} + \sqrt{53}$$

B

$$3\sqrt{5} + \sqrt{13} + \sqrt{53}$$

C

$$\sqrt{5} + 3\sqrt{5} + \sqrt{53}$$

D

$$2\sqrt{5} + 9\sqrt{5} + \sqrt{53}$$