All the three quadrilaterals $\text{ADEC}$, $\text{ABIH}$ and $\text{BCGF}$ are squares and $\angle ABC=\text{9}0°$. If the area of $\text{ADEC}=x^2\text{\hspace{0.17em}units}$ and the area of ${\text{AHIB=y}}^2\text{ units}(x^2>y^2)$. then the area of $\text{BCGF}$ is

Given (1) Area of square $ADEC=x^2$

${\left(AC\right)}^2=x^2$

$AC=x$

(2) area of square $AHIB=y^2$

${\left(AB\right)}^2=y^2$

$AB=y$

In $\Delta ABC,\overline{)ABC}=90°$

$A{C}^2=A{B}^2+B{C}^2$

$x^2=y^2+B{C}^2$

$x^2-y^2=B{C}^2$

$BC=\sqrt{x^2-y^2}$

Area of square $BCGF={\left(\sqrt{x^2-y^2}\right)}^2$

$=x^2-y^2\text{\hspace{0.17em}sq.units}$