Equivalent Quadrilaterals

$a^2 = xy \\ 2(x+y) = 58\\ x+y=29 \\ xy = a^2 \\ x < 29, y le 29 \\ x = 1, 2, 3, 4, 5, 6, \ldots, 28 ~~ ; ~~ y = 1, 2, 3, 4, 5, 6, \ldots, 29\\ xy = a^2 ~~ \text{ie, a perfect square} \\ \implies \text{The possible value of (x, y) for (x+y)=29 is : } \\ (1, 28), (2, 27), (3, 26), (4, 25), \ldots, (14, 15)\\ \text{The only solution for (x, y), where xy is a perfect square is (4, 25).} \\ \boxed{\therefore \text{Area of square = rectangle = }4 \times 25 = 100 \\ \implies \text{side of square = } \sqrt[]{100} = 10cm}$