Two sides of a triangle are of lengths $5\, cm$ and $1.5\, cm$. The length of the third side of the triangle cannot be (in $cm$):

In triangles $ABC$ and $PQR$,$\angle A = \angle Q$ and $\angle B = \angle R$. Which side of $\triangle PQR$ should be equal to side $BC$ of $\triangle ABC$ so that the two triangles are congruent? Give reason for your answer.

Write the measures of the sides of $\Delta XYZ$ in ascending order,given that $\angle X = 80^{\circ}$ and $\angle Y = 30^{\circ}$.

In triangles $ABC$ and $DEF$,$AB = FD$ and $\angle A = \angle D$. The two triangles will be congruent by the $SAS$ axiom if:

The line segment joining the mid-points $M$ and $N$ of the parallel sides $AB$ and $DC$ respectively of a trapezium $ABCD$ is perpendicular to both the sides $AB$ and $DC$. Prove that $AD = BC$.

In $\Delta PQR$,$X$ is the midpoint of $QR$. $XY$ and $XZ$ are altitudes from $X$ on $PQ$ and $PR$ respectively. If $XY = XZ$,then prove that $\Delta PQR$ is an isosceles triangle.