Perimeter of rectangle $PQRS$ is $40\, cm$. If $PQ : QR = 3 : 5$,then $RS = \ldots \ldots \ldots \, cm$.

$A$ diagonal of a rectangle is inclined to one side of the rectangle at $25^{\circ}$. The acute angle between the diagonals is (in $^{\circ}$)

$P$ is the mid-point of the side $CD$ of a parallelogram $ABCD$. $A$ line through $C$ parallel to $PA$ intersects $AB$ at $Q$ and $DA$ produced at $R$. Prove that $DA = AR$ and $CQ = QR$.

$ABCD$ is a parallelogram. If its diagonals are equal,then find the value of $\angle ABC$. (in $^{\circ}$)

In quadrilateral $ABCD$,$AC = BD$ and $AC \perp BD$. If $P, Q, R$ and $S$ are the midpoints of the sides $AB, BC, CD$ and $DA$ respectively,prove that $PQRS$ is a square.

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