In $\Delta PQR$,$m\angle Q = 90^\circ$. If $PQ = 8$ and $QR = 15$,then $PR = \ldots$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AC = 20$,then $BM = \ldots$

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow QPR.$ If $m \angle P : m \angle Q : m \angle R = 2 : 3 : 4,$ then $m \angle A : m \angle B : m \angle C = \ldots$

In $\Delta PQR$ and $\Delta XYZ$,if $\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ} = \frac{2}{5}$,then the ratio of their areas $\text{Area}(\Delta PQR) : \text{Area}(\Delta XYZ) = \ldots$

In the figure,if $\angle ACB = \angle CDA$,$AC = 8 \, cm$ and $AD = 3 \, cm$,find $BD$ (in $cm$).

In $\Delta PQR$,$m \angle Q = 90^{\circ}$. If $PQ = 33$ and $PR = 65$,find $QR$.