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

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

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $PR$. If $QR = 9$ and $PR = 13.5$,find $RM$.

Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ is a median in $\Delta ABC$ and $\overline{PM}$ is a median in $\Delta PQR.$ Prove that $\frac{AD}{PM} = \frac{AB}{PQ}.$

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 $^{\circ}$)

Below are given the measures of sides $\overline{PQ}$,$\overline{QR}$ and $\overline{PR}$ of $\Delta PQR$. In each case,determine whether $\Delta PQR$ is a right-angled triangle or not. If it is a right-angled triangle,state which angle is a right angle: $PQ = 8, QR = 6, PR = 12$.