The length of a diagonal of a square is $6$. Then, the length of its sides is $\ldots \ldots \ldots$ (in $\sqrt{2}$)

$O$ is the point of intersection of the diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$. Through $O$,a line segment $PQ$ is drawn parallel to $AB$ meeting $AD$ in $P$ and $BC$ in $Q$. Prove that $PO = QO$.

In $\Delta XYZ$,$m \angle Y = 90^{\circ}$ and $XY : XZ = 15 : 17$. If $YZ = 4$,find the perimeter of $\Delta XYZ$.

In the figure,if $LM \parallel CB$ and $LN \parallel CD$,prove that $\frac{AM}{AB} = \frac{AN}{AD}$.

$A$ and $B$ are points on the sides $PQ$ and $PR$ of a triangle $PQR$ respectively,such that $PQ = 12.5 \, cm$,$PA = 5 \, cm$,$BR = 6 \, cm$,and $PB = 4 \, cm$. Is $AB \parallel QR$? Give reasons for your answer.

In $\Delta ABC$,$\overline{AD}$ and $\overline{BE}$ are medians. Line $m$ through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. If $EK = 2.5$,then $AC = \ldots$