Question 1148799
.


            The solution by @josgarithmetic is not precisely correct,  producing partly absurdist result.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So I came to bring the &nbsp;<U>CORRECT &nbsp;solution</U>.



<pre>
Let x be the side length of the larger square and y be that of the smaller square.


    4x + 4y = 80    (1)

    x^2 = 3y^2      (2)

===========================>   

    x + y = 20      (3)

    x^2 = 3y^2      (4)



From (1),  y = 20-x.  Substitute it into (4).  You will get


    x^2 = 3*(20-x)^2

    x^2 = 3*(400 - 40x + x^2)

    x^2 = 1200 - 120x + 3x^2

    2x^2 - 120x + 1200 = 0

    x^2  -  60x + 600 = 0

    x = {{{(60 +- sqrt(60^2 - 4*600))/2}}} = {{{(60 +- sqrt(1200))/2}}} = {{{(60 +- 20*sqrt(3))/2}}} = {{{30 +- 10*sqrt(3)}}}.



Although the quadratic equation has 2 roots, only smaller value  x= {{{30 +- 10*sqrt(3)}}}  is the solution to the problem,
since the larger value  {{{30 + 10*sqrt(3)}}}  produces NEGATIVE value of "y", due to (3).



<U>ANSWER</U>.  The larger square has the side length  {{{30 -10*sqrt(3)}}} = 12.68 inches, approximately.

         The smaller square has the side length  {{{20 - (30-10*sqrt(3))}}} = {{{-10 + 10*sqrt(3)}}} = 7.32 in (approximately).
</pre>

Solved.