SOLUTION: A point O is inside a square lot. If the distances from point O to the three successive corners of the square lot are 5m, 3m, 4m respectively. Determine the area of the square lot.

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 3 from A, 4 from D  and  5 from B.


Then we have these three equations ("distances")


    x^2     + y^2     = 3^2,      (1)

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

    x^2     + (y-a)^2 = 4^2.      (3)


Making FOIL in equations (2) and (3), I can re-write (1), (2) and (3) in this form


    x^2             + y^2 =  9,   (4)    (= same as (1) )

    a^2 - 2ax + x^2 + y^2 = 25,   (5)

    x^2 + y^2 - 2ay + a^2 = 16.   (6)


Replacing  x^2 + y^2 by 9  in equations (5) and (6), I obtain new equations instead of them


    a^2 - 2ax = 16                (7)

    a^2 - 2ay =  7                (8)


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

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


     +  = ,

or, simplifying

     +  = {{36a^2}}},

     = 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  the area of the square is   =  = 36.863 square units (approximately).