In $\Delta ABC$,$m\angle B = 90^\circ$. If $AB = 12$ and $BC = 5$,then $AC = \dots$

In $\Delta ABC$ and $\Delta PQR$,$\angle A \cong \angle P$ and $\angle B \cong \angle Q$. If $\frac{AB}{PQ} = \frac{4}{5}$ and $AC = 20$,find $PR$.

If the measures of the sides of a triangle are $\ldots \ldots \ldots,$ then it is not a right-angled triangle.

In $\Delta ABC$,$\angle B$ is a right angle. Then,according to Pythagoras' theorem,$\ldots \ldots \ldots$ holds good.

In $\Delta ABC$,$m\angle B = 90^{\circ}$. $D$ is the midpoint of $\overline{BC}$ and $F$ is the midpoint of $\overline{AB}$. Prove that $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.

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