Asked in: March 2024 board paper (same SSS type with ratio 2 : 3 in March 2025)\(\Delta ABC \sim \Delta LMN\). 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}{…

Answer: \(\Delta LMN\) has LM = 4.4 cm, MN = 4.8 cm, LN = 3.6 cm.

Analysis: \(\dfrac{AB}{LM} = \dfrac{BC}{MN} = \dfrac{CA}{NL} = \dfrac{5}{4}\). So \(MN = \dfrac{4 \times 6}{5} = 4.8\) cm, \(LM = \dfrac{4 \times 5.5}{5} = 4.4\) cm, \(LN = \dfrac{4 \times 4.5}{5} = 3.6\) cm.

Steps of construction:

1. Draw seg BC = 6 cm. With B as centre radius 5.5 cm and C as centre radius 4.5 cm draw arcs meeting at A. Join AB and AC to get \(\Delta ABC\).
2. Draw seg MN = 4.8 cm. With M as centre radius 4.4 cm and N as centre radius 3.6 cm draw arcs meeting at L. Join LM and LN to get \(\Delta LMN\).

Alternative (ray method): draw a ray from B making an acute angle with BC on the side opposite A; mark 5 equal parts \(B_1, \ldots, B_5\); join \(B_5C\); through \(B_4\) draw a line parallel to \(B_5C\) meeting BC at N; through N draw a line parallel to CA meeting BA at L. Then \(\Delta LBN\) (with M at B) is the required triangle.

[Figure: \(\Delta ABC\) with sides 5.5, 6, 4.5 cm and the smaller similar \(\Delta LMN\) with sides 4.4, 4.8, 3.6 cm.]