In $\Delta ABC$,$m\angle B = 90^{\circ}$. If $AB : BC = 3 : 4$,find $AB : AC$.

$A$ person begins his journey from point $A$. From $A$,he moves $800 \, m$ towards north to reach point $B$. From $B$,he moves $500 \, m$ towards east to reach point $C$. From $C$,he moves $400 \, m$ towards north to reach point $D$. Find the direct distance in $m$ from $A$ to $D$.

$\Delta ABC$ is an acute-angled triangle and $\overline{AM}$ is an altitude. Prove that $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.

If $\triangle ABC \sim \triangle QRP$,$\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle QRP)} = \frac{9}{4}$,$AB = 18 \, cm$ and $BC = 15 \, cm$,then $PR$ is equal to (in $cm$):

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $PR$. If $PD = 25 DR$,prove that $PQ = 5 QR$.

In $\Delta ABC$ and $\Delta PQR$,$\angle A \cong \angle P$ and $\angle B \cong \angle R$. If $AB = 8$,$PQ = 7.5$ and $AC = 6$,find $PR$.