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, \[ \triangle NML \sim \triangle QPR \]
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]