\(\Delta ABC \sim \Delta LMN\). In \(\Delta ABC\), AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct \(\Delta ABC\) and \(\Delta LMN\) such that \(\dfrac{BC}{MN} = \dfrac{5}{4}\).[Figure: none; construct both triangles.]

Answer: MN = 4.8 cm, LM = 4.4 cm, LN = 3.6 cm.

Since Δ ABC ~ Δ LMN, \(\dfrac{AB}{LM} = \dfrac{BC}{MN} = \dfrac{CA}{LN} = \dfrac{5}{4}\).

∴ \(MN = \dfrac{4}{5} \times 6 = 4.8\) cm, \(LM = \dfrac{4}{5} \times 5.5 = 4.4\) cm, \(LN = \dfrac{4}{5} \times 4.5 = 3.6\) cm.

Construction :

1. Draw seg BC = 6 cm. With B as centre and radius 5.5 cm, and with C as centre and radius 4.5 cm, draw arcs intersecting at A. Join AB and AC; Δ ABC is constructed.
2. Draw seg MN = 4.8 cm. With M as centre and radius 4.4 cm, and with N as centre and radius 3.6 cm, draw arcs intersecting at L. Join LM and LN; Δ LMN is the required triangle similar to Δ ABC.

(Alternatively, draw a ray BX making an acute angle with BC, mark 5 equal parts, take the 4th point, and draw a line parallel to the third side through it.)