\(\Delta LMN \sim \Delta LQP\). In \(\Delta LMN\), LM = 3.6 cm, \(\angle L = 50^\circ\), LN = 4.2 cm and \(\dfrac{LM}{LQ} = \dfrac{4}{7}\). Construct \(\Delta LQP\).

Answer: LQ = 6.3 cm, LP = 7.35 cm and \(\angle QLP = 50^\circ\); construct the triangle from these.

From the similarity L ↔ L, M ↔ Q, N ↔ P, so \(\dfrac{LM}{LQ} = \dfrac{LN}{LP} = \dfrac{4}{7}\).

\(LQ = \dfrac{7}{4}\times 3.6 = 6.3\) cm, \(LP = \dfrac{7}{4}\times 4.2 = 7.35\) cm, and \(\angle QLP = \angle MLN = 50^\circ\).

Construction:

1. Draw seg LQ = 6.3 cm.
2. At L draw a ray making an angle of 50° with ray LQ.
3. On this ray mark point P such that LP = 7.35 cm.
4. Join PQ. \(\Delta LQP\) is the required triangle.