The quadrilateral formed by the lines $\sqrt{3}x+y=0,\sqrt{3}y+x=0,\sqrt{3}x+y=1,\sqrt{3}y+x=1is$

Given  lines

$\begin{array}{l}\sqrt{3}x+y=0--\left(1\right)\\ x+\sqrt{3}y=0--\left(2\right)\\ \sqrt{3x}+y-1=0--\left(3\right)\\ x+\sqrt{3}y-1=0--\left(4\right)\end{array}$

(1) & (3) are parallel

Distance b/w (1) & (3) $=\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{\left|0+1\right|}{\sqrt{3+1}}=\frac{1}{2}$

(2) & (4) are parallel

Distance b/w (2) & (4) $=\frac{\left|0+1\right|}{\sqrt{3+1}}=\frac{1}{2}$

And the Adjacent sides are not perpendicular

The Quadrilateral is a Rhombus