Let ABCD be the square with the side of the length "a" in a coordinate plane,

    A = (0,0),  B = (a,0),  C = (a,a)  and  D = (0,a).


Let (x,y) be the point inside the square ABCD with the distance 4 from A, 5 from D  and  6 from B.


Thus we have these three equations ("distances")


    x^2     + y^2     = 4^2,      (1)

    (a-x)^2 + y^2     = 6^2,      (2)

    x^2     + (y-a)^2 = 5^2.      (3)


Making FOIL in equations (2) and (3), I can re-write them in this form


    x^2             + y^2 = 16,   (4)    (= same as (1) )

    a^2 - 2ax + x^2 + y^2 = 36,   (5)

    x^2 + y^2 - 2ay + a^2 = 25.   (6)


Replacing  x^2 + y^2 by 16  in equations (5) and (6), I obtain new equations instead of them


    a^2 - 2ax = 20                (7)

    a^2 - 2ay =  9                (8)


From equations (7) and (8),  x = ,  y = .

Substituting these expressions for x and y into equation (4), you get


     +  = ,

or, simplifying

     +  = ,

     = 0.


From this bi-quadratic equation, you get for , by applying the quadratic formula

     =  = .


The smaller value does not work for "a" (as it is easy to check), leaving the larger value


     = 

as the only meaningful.


Thus  a =  = 7.534 (approximately).