Asked in: July 2024 board paper\(\Delta LMN \sim \Delta LQP\). 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: \(\Delta LQP\) with LQ = 6.3 cm, \(\angle L = 50^\circ\), LP = 7.35 cm.

Analysis: \(\dfrac{LM}{LQ} = \dfrac{LN}{LP} = \dfrac{4}{7}\). So \(LQ = \dfrac{7 \times 3.6}{4} = 6.3\) cm and \(LP = \dfrac{7 \times 4.2}{4} = 7.35\) cm. \(\angle L\) is common.

Steps of construction:

1. Draw seg LM = 3.6 cm. At L draw ray LX making an angle of 50° with LM. On ray LX mark N with LN = 4.2 cm. Join MN: this is \(\Delta LMN\) (rough or actual).
2. Draw a ray from L making an acute angle with LM, on the side opposite N, and mark 7 equal parts \(L_1, \ldots, L_7\).
3. Join \(L_4M\). Through \(L_7\) draw a line parallel to \(L_4M\) meeting ray LM at Q (LQ = 6.3 cm).
4. Through Q draw a line parallel to MN meeting ray LN at P. \(\Delta LQP\) is the required triangle.

[Figure: \(\angle L = 50^\circ\); \(\Delta LMN\) inside the larger \(\Delta LQP\), with QP parallel to MN.]

(Since 7.35 cm is awkward to measure, the ray method is better here. LQ can also be drawn directly as 6.3 cm and then QP drawn parallel to MN.)