Question

Sum of the areas of two squares is 400 $c{m}^2$. If the difference of their perimeters is 16 cm. The absolute difference of the sides of two squares is ----- cm.

Answer 1

$\Rightarrow$ Let the sides of the two squares be $xcm$ and $ycm$, where $x>y$

$\Rightarrow$ Then, their areas are $x^2$ and $y^2$ and their perimeters are $4x$ and $4y.$

By given condition,

$\Rightarrow$ $x^2+y^2=400$ ----- ( 1 )

$\Rightarrow$ And $4x-4y=16$

$\Rightarrow$ $4(x-y)=16$

$\Rightarrow$ $x-y=4$

$\Rightarrow$ $x=y+4$

Substituting the value of $x$ from ( 2 ) in ( 1 ), we get,

$\Rightarrow$ $(y+4{)}^2+y^2=400$

$\Rightarrow$ $y^2+16+8y+y^2=400$

$\Rightarrow$ $2y^2+16+8y=400$

$\Rightarrow$ $y^2+4y-192=0$

$\Rightarrow$ $y^2+16y-12y-192=0$

$\Rightarrow$ $y(y+16)-12(y+16)=0$

$\Rightarrow$ $(y+16)(y-12)=0$

$\therefore$ $y=-16$ or $12$

Since, $y$ cannot be negative, $y=12$

$\Rightarrow$ $x=y+4=12+4=16$

$\therefore$ Sides of the two squares are $16cm$ and $12cm$.

$\Rightarrow$ Required difference $=16-12=4cm$