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

$4a-4b=16\rightarrow a-b=4\rightarrow a=4+b$
$a^2+b^2=400\rightarrow a^2+(a+4)^2=2a^2+8a+16=400\rightarrow 2a^2+8a=384\rightarrow a(a+4)=192\rightarrow a=12,b=16$
check $12^2+16^2=144+256=400$
