Question 1202975
<pre>
x = length of each side of first square
y = length of each side of second square

We get:

(1)  4x + 4y = 80            <--- from length information  
(2)  x^2 + y^2 = 300         <--- from area information

Eq (1) can be solved for y;  4y = 80-4x ==>  y=20-x

Substitute "20-x" for "y" in (2), to get:

{{{ x^2 + (20-x)^2 = 300 }}}

{{{ x^2 + (x^2-40x+400) = 300 }}}
{{{  2x^2 - 40x + 400 = 300 }}}
{{{  2x^2 - 40x + 100 = 0 }}}
{{{   x^2 - 20x + 50 = 0 }}}

 From WolframAlpha:
{{{  (x-5sqrt(2)-10)(x+5sqrt(2)-10) = 0 }}}

  x = {{{ 10+5sqrt(2) }}} or  x = {{{ 10-5sqrt(2) }}}

If x = {{{ 10+5sqrt(2) }}}, we get  y = {{{20-10-5sqrt(2) = 10-5sqrt(2)}}}
(which makes sense, it is the other choice for x above, as the assignments of x and y are arbitrary).  

Answer:
  The sides of the two squares measure {{{ 10+5sqrt(2)}}} and {{{10-5sqrt(2)}}}