Square is a quadrilateral.

Let $ABCD$ be a square. An arc of a circle with $A$ as centre and $AB$ as radius is drawn inside the square joining the points $B$ and $D$. Points $P$ on $AB, S$ on $AD, Q$ and $R$ on arc $BD$ are taken such that $PQRS$ is a square. Further suppose that $PQ$ and $RS$ are parallel to $AC$. Then, $\frac{\text{ area }PQRS}{\text{ area }ABCD}$ is

Find the length of the diagonal of a square whose area is $128 {\mathrm{cm}}^2$.  Also, find its perimeter.

In the adjoining figure, $\mathrm{ABCD}$ is a square and $\mathrm{\Delta EDC}$ is an equilateral triangle. Prove that

$\mathrm{AE}=\mathrm{BE}$

Here is a square is drawn for you. Is $\angle ADO=\angle CDO$?

Diagonals of a square $\mathrm{IJKL}$ intersects at point $\mathrm{M}$, Find the measures of $\angle \mathrm{IMJ}$, $\angle \mathrm{JIK}$and $\angle \mathrm{LJK}$.

In the given figure, $\mathrm{ABCD}$ is a square and $\angle \mathrm{PQR}=90°$. If $\mathrm{PB}=\mathrm{QC}=\mathrm{DR}$, prove that

 $\mathrm{QB}=\mathrm{RC}$