Let $ABCD$ be a square of a side length $1$. Let $P,Q,R,S$ be points in the interiors of the sides $AD,BC,AB,CD$ respectively, such that $PQ$ and $RS$ intersect at right angles. If $PQ=\cfrac { 3\sqrt { 3 } }{ 4 }$ then $RS$ equals

$PQ\bot RS\Rightarrow$

$c-a=b-d.......(1)$

$PQ=\cfrac { 3\sqrt { 3 } }{ 4 }$

${ PQ }^{ 2 }=\cfrac { 27 }{ 16 }$

$1+{ (a-c) }^{ 2 }=\cfrac { 27 }{ 16 } .....(2)$

$RS=\sqrt { { (b-d) }^{ 2 }+1 } ....(3)$

By equation (1),(2),(3)

$RS=\cfrac { 3\sqrt { 3 } }{ 4 }$