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

Well, your mistake came about in the step $$\sqrt{400-x^2}=20-x,\tag{$\star$}$$ which isn't true in general. But why can't we draw this conclusion? Observe that if we let $y=-x,$ then $y^2=x^2,$ so $400-y^2=400-x^2.$ But then we can use the same (erroneous) reasoning to conclude that $$20-y \overset{(\star)}{=} \sqrt{400-y^2} = \sqrt{400-x^2} \overset{(\star)}{=} 20-x,$$ from which we can conclude that $y=x.$ But $y=-x,$ so the only way we can have $y=x$ is if $x=y=0.$ Hence, $(\star)$ is true if and only if $x=0,$ and we certainly can't have $x=0$ in this context.

Instead, note that since the difference in perimeters is $16$ cm, then the smaller of the two squares must have sides that are $4$ cm shorter than those of the larger square's sides. That is, if $x$ is the length of the larger squares sides, we need $$x^2+(x-4)^2=400.$$ Can you expand that and take it from there?
