Sum of Perimeters of Triangles in Square

Area of square $= 4 \implies$ side $= 2$. Perimeter of square $= 4 \times 2 = 8$. Diagonal length $= \sqrt{2^2 + 2^2} = 2\sqrt{2}$. Each triangle contains one outer side of the square and two half-diagonals. Each interior segment (half-diagonal) is shared between two triangles, so each of the $4$ half-diagonals gets counted twice in the total perimeter calculation. Sum of perimeters $= \text{Perimeter of square} + 2 \times (\text{Sum of 2 diagonals}) = 8 + 2 \times (2 \times 2\sqrt{2}) = 8 + 8\sqrt{2} = 8(1 + \sqrt{2})$.