Question

The length of the diagonal of a square is $\sqrt{2}$ times its area. Then find its perimeter.

Answer 1

Let the side of square be s

Area$=s\times s=s^2$

$AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{2}s$

$=\sqrt{2}\times s$

$\sqrt{2}s=\sqrt{2}\times s^2$$s=s^2$

$s=1$

$\therefore$Perimeter $=4s$

$=4\times 1=4$