In $\Delta PQR$,$PQ = PR$ and $\angle R = 40^{\circ}$,then $\angle P = \dots$ (in $^{\circ}$)

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC).$

In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM$ is an altitude to the hypotenuse $AC$. If $AB = 12 \, cm$ and $BC = 16 \, cm$,then find the length of $BM$ in $cm$.

In $\Delta XYZ$,$\angle Y = 90^{\circ}$ and $XY = YZ$,then $\angle X = \dots$ (in $^{\circ}$)

$S$ is any point on side $QR$ of a $\triangle PQR$. Show that: $PQ + QR + RP > 2 \, PS$.

In $\Delta PQR$,$\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$,then find $\angle R$. (in $^{\circ}$)