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

$A$ square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

$ABCD$ is a rhombus such that $\angle ACB = 40^{\circ}$. Then $\angle ADB$ is (in $^{\circ}$)

In a rhombus $ABCD$,$\angle B = 80^{\circ}$,then $\angle ADB = \ldots$ (in $^{\circ}$)

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 $.0$)

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