In the figure, in $\triangle MNL$ and $\triangle PQR$, $\angle M = \angle Q = 70^\circ$, $MN = 3$ cm, $ML = 4.5$ cm, $PQ = 2$ cm, and $QR = 3$ cm. Then, the following correct relation will be: 
Step 1: Given data.
For $\triangle MNL$: $MN = 3$ cm, $ML = 4.5$ cm, and $\angle M = 70^\circ$.
For $\triangle PQR$: $PQ = 2$ cm, $QR = 3$ cm, and $\angle Q = 70^\circ$.
Step 2: Compare sides including the equal angles.
\[ \dfrac{MN}{QR} = \dfrac{3}{3} = 1, \quad \dfrac{ML}{PQ} = \dfrac{4.5}{2} = 2.25 \] These are not equal, but let’s check other possible corresponding sides. If we consider $\triangle NML$ and $\triangle QPR$: \[ \dfrac{MN}{QP} = \dfrac{3}{2} = 1.5, \quad \dfrac{ML}{QR} = \dfrac{4.5}{3} = 1.5 \] The sides are in the same ratio and included angles are equal (\(70^\circ\)).
Step 3: Conclusion.
Hence, by the SAS similarity criterion, \[ \triangle NML \sim \triangle QPR \]
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: