$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XYZ$. If $PQ : QR : PR = 3 : 5 : 7$ and the perimeter of $\Delta XYZ$ is $22.5$,then $YZ = \ldots$

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,$2 AB = PQ$,and $BC = 10$,then $QR = \dots$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $CM = 3$,then $BM = \dots$

In the figure,two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6 \, cm$,$PB = 3 \, cm$,$PC = 2.5 \, cm$,$PD = 5 \, cm$,$\angle APB = 50^{\circ}$ and $\angle CDP = 30^{\circ}$. Then,$\angle PBA$ is equal to: (in $^{\circ}$)

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $AB : AC = 24 : 25$. If $BC = 14$,find the perimeter of $\Delta ABC$.

In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$,$AB = 7$,$AC = 5$ and $AD = 5$. Then $BC = \ldots$