Question 1143644
.


            Take a page of paper and a pen or a pencil and plot it on your own.


            We are not an art studio to make plots for you.


            Instead of plotting the figure,  I will solve the problem, from the beginning to the end.    OK ?



.
<pre>
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 = {{{(a^2-16)/2a}}},  y = {{{(a^2 - 7)/2a}}}.

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


    {{{(a^2-16)^2}}} + {{{(a^2 -7)^2}}} = {{{(4a^2)*9}}},

or, simplifying

    {{{a^4 - 32a^2 + 256}}} + {{{a^4 - 14a^2 + 49}}} = {{36a^2}}},

    {{{2a^4 - 82a^2 + 305}}} = 0.


From this bi-quadratic equation, you get for {{{a^2}}}, by applying the quadratic formula

    {{{a^2}}} = {{{(82 +- sqrt(82^2 - 4*2*305))/(2*2)}}} = {{{(82 +- sqrt(4284))/4}}}.


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


    {{{a^2}}} = {{{(82 + sqrt(4284))/4}}}

as the only meaningful.


Thus  the area of the square is  {{{a^2}}} = {{{(82 + sqrt(4284))/4}}} = 36.863 square units (approximately).
</pre>

Solved.



=============


It is not for the first time such problem comes to the forum.



Some time ago I solved similar problem here under this link


<A HREF=://www.algebra.com/algebra/homework/Parallelograms/Parallelograms.faq.question.1135915.htm>https://www.algebra.com/algebra/homework/Parallelograms/Parallelograms.faq.question.1135915.html</A>


https://www.algebra.com/algebra/homework/Parallelograms/Parallelograms.faq.question.1135915.html



/\/\/\/\/\/\/\/


Thanks for posting this interesting problem.


It was a pleasure for me to solve it again &nbsp;(even for the second time), &nbsp;because the solution is nice.