The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares. The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares.
I HAVE TRIED IT AS BELOW BUT ANSWER IS NOT CORRECT.......CHECK - HELP!
Let side of 1st square=x cm.
∴ Area of 1st square=x²cm²
GIVEN,
Sum of areas =400cm² 
∴ Area of 2nd square=(400-x²)cm²
AND side of 2nd square=√[(20-x)²  i.e.20-x                   .......(1)
Difference of perimeters=16cm.
THEN-
4x-4(20-x)=16  (ASSUMING THAT 1ST SQUARE HAS LARGER SIDE)
X=12
HENCE - SIDE OF 1ST SQUARE = 12CM ;
SIDE OF 2ND SQUARE=20-12=8CM.             [FROM (1)]
WHICH IS NOT THE REQUIRED ANSWER AS SUM OF AREAS OF SQUARES OF FOUNDED SIDES IS NOT 400CM²

$x^2+y^2=400$
$4*x-4*y=16$
could you continue?
ok we have  $x^2+y^2=400$ and $x-y=4$
from there  $x=4+y$
$(4+y)^2+y^2=400$
$16+8*y+y^2+y^2=400$
$2*y^2+8*y-384=0$
$y^2+4*y-192=0$
$D=4+192=196$
$y_1=(-2+\sqrt{196})$
$y_2=(-2-\sqrt{196})$
now calculate $x_1$ and $x_2$
$y=12$
$x=16$
now check   $16*16+12*12=256+144=400$
$64-48=16$
