\(\Delta PQR \sim \Delta PMN\). In \(\Delta PQR\), QR = 5.4 cm, \(\angle Q = 60^\circ\), \(\angle R = 45^\circ\), and \(\dfrac{QR}{MN} = \dfrac{3}{4}\). Construct \(\Delta PQR\) and \(\Delta PMN\).

Answer: \(\Delta PMN\) has MN = 7.2 cm, \(\angle M = 60^\circ\), \(\angle N = 45^\circ\).

Analysis: Corresponding angles of similar triangles are equal: \(\angle M = \angle Q = 60^\circ\), \(\angle N = \angle R = 45^\circ\), \(\angle P\) common (75°). \(MN = \dfrac{4 \times 5.4}{3} = 7.2\) cm. Since \(\angle M = \angle Q\), MN is parallel to QR.

Steps of construction:

1. Draw seg QR = 5.4 cm. At Q draw a ray making 60° with QR, and at R draw a ray making 45° with RQ, on the same side. They meet at P: this is \(\Delta PQR\).
2. Draw a ray from P making an acute angle with PQ (away from R) and mark 4 equal parts \(P_1, \ldots, P_4\).
3. Join \(P_3Q\). Through \(P_4\) draw a line parallel to \(P_3Q\), meeting ray PQ (extended) at M.
4. Through M draw a line parallel to QR, meeting ray PR (extended) at N. \(\Delta PMN\) is the required triangle.

Check: MN should measure 7.2 cm; \(PM = \dfrac{4}{3} PQ\).

[Figure: \(\Delta PQR\) with base QR = 5.4 cm and base angles 60° and 45°; sides PQ and PR extended to M and N with MN parallel to QR.]