In parallelogram $ABCD$,the diagonals intersect at $P$. If $PA = 3.8 \, cm$ and $PB = 5.2 \, cm$,then $BD = \dots \dots \dots cm$.

Diagonals of a parallelogram $ABCD$ intersect at $O.$ If $\angle BOC = 90^{\circ}$ and $\angle BDC = 50^{\circ},$ then $\angle OAB$ is (in $^{\circ}$)

In parallelogram $ABCD$,$AB = 12.5 \, cm$ and $BC = 7 \, cm$,then find the perimeter of $ABCD$ in $cm$.

Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $O$ such that $OA : OC = 3 : 2$. Is $ABCD$ a parallelogram? Why or why not?

In trapezium $ABCD$,$AB \parallel CD$. If $\angle A = 70^{\circ}$ and $\angle B = 85^{\circ}$,then $\angle D = \ldots$ (in $^{\circ}$)

In $\Delta ABC$,$P$,$Q$,and $R$ are the midpoints of the sides. If the perimeter of $\Delta ABC$ is $23 \, cm$,then find the perimeter of $\Delta PQR$ in $cm$.