In trapezium $ABCD$,$AB || CD$. Points $P$ and $Q$ are the midpoints of $AD$ and $BC$ respectively. If $AB = 18 \, cm$ and $PQ = 15 \, cm$,then $CD = \dots \, cm$.

$(1)$ If no three points out of four coplanar points are collinear,then a $\ldots \ldots \ldots$ figure formed by joining these four points in order is called a quadrilateral.
$(2)$ $A$ quadrilateral has $\ldots \ldots \ldots$ pairs of opposite sides.

In quadrilateral $ABCD$,$\angle A = 100^{\circ}$,$\angle B = 80^{\circ}$,and $\angle C = 120^{\circ}$. Find $\angle D$. (in $^{\circ}$)

In the given figure,$AX$ and $CY$ are respectively the bisectors of the opposite angles $A$ and $C$ of a parallelogram $ABCD$. Show that $AX \parallel CY$.

In $\Delta ABC$,$P$,$Q$,and $R$ are the midpoints of $AB$,$BC$,and $CA$ respectively. If $AB = 8 \text{ cm}$,$BC = 6.6 \text{ cm}$,and $CA = 5.4 \text{ cm}$,then find the perimeter of $\Delta PQR$ in $\text{cm}$. (in $.0$)

In rhombus $PQRS$,$PR = 40 \, cm$ and $QS = 42 \, cm$,then $PQ = \dots \dots \dots \, cm$.