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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

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    $\begingroup$ The error is where you say that $(400-x^2)=(20-x)^2$. $\endgroup$ – D.L. Jan 15 '14 at 14:04
  • $\begingroup$ Use $a^2$ + $b^2=400$ and $4(a-b)=16$. Thus a=4+b and substitute that in the first term $\endgroup$ – Bernd Jan 15 '14 at 14:05
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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?

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  • $\begingroup$ But squareroot and square cancel each other? $\endgroup$ – Vanshaj Jan 15 '14 at 14:08
  • $\begingroup$ Squaring eliminates a square root. Taking the square root of a non-negative number's square eliminates the squaring. However, neither of these things is done in your erroneous step. $400-x^2$ is not a perfect square, but rather a difference of squares. $\endgroup$ – Cameron Buie Jan 15 '14 at 14:10
  • $\begingroup$ I got it. Thanks for help. $\endgroup$ – Vanshaj Jan 15 '14 at 14:28
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$4a-4b=16\rightarrow a-b=4\rightarrow a=4+b$

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

check $12^2+16^2=144+256=400$

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$x^2+y^2=400$

$4*x-4*y=16$

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 $16*16+12*12=256+144=400$

$64-48=16$

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  • $\begingroup$ I am asking for 1 variable.! $\endgroup$ – Vanshaj Jan 15 '14 at 14:07
  • $\begingroup$ there are calculated both variable $\endgroup$ – dato datuashvili Jan 15 '14 at 14:17

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