If $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. If $\text{Area}(\Delta ABC) = 72$,$BC = 6$,and $YZ = 10$,then $\text{Area}(\Delta XYZ) = \dots$

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. $\overline{AD}$ and $\overline{PM}$ are medians of these triangles. Prove that $AB \times PM = PQ \times AD$.

In the given figure,$\angle BAC = 90^{\circ}$ and $AD \perp BC$. Then,

In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Prove the following:
$(i)$ The correspondence $AMB \leftrightarrow ABC$ is a similarity.
(ii) The correspondence $BMC \leftrightarrow ABC$ is a similarity.
(iii) The correspondence $AMB \leftrightarrow BMC$ is a similarity.

$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow ZXY$. If $AB = 12, BC = 8, CA = 10$ and $ZX = 10$,then $XY + YZ = \ldots$

Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?