Asked in: March 2025 board paper\(\Delta ABC \sim \Delta PQR\). In \(\Delta ABC\), AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of \(\Delta ABC\) and \(\Delta PQR\) are in the ratio 2 : 3. Construct \(\Delta ABC\) and \(\Delta PQR\…

Answer: \(\Delta PQR\) has PQ = 5.4 cm, QR = 6 cm, PR = 6.3 cm.

Analysis: \(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{2}{3}\). So \(PQ = \dfrac{3 \times 3.6}{2} = 5.4\) cm, \(QR = \dfrac{3 \times 4}{2} = 6\) cm, \(PR = \dfrac{3 \times 4.2}{2} = 6.3\) cm.

Steps of construction:

1. Draw seg BC = 4 cm. With B as centre radius 3.6 cm and C as centre radius 4.2 cm draw arcs meeting at A. Join AB and AC: this is \(\Delta ABC\).
2. Draw seg QR = 6 cm. With Q as centre radius 5.4 cm and R as centre radius 6.3 cm draw arcs meeting at P. Join PQ and PR: this is \(\Delta PQR\).

Alternative (ray method): on a ray from B (opposite side to A), mark 3 equal parts; join \(B_2C\); through \(B_3\) draw a line parallel to \(B_2C\) meeting ray BC at R; through R draw a line parallel to CA meeting ray BA at P. \(\Delta PBR\) (Q at B) is the required triangle.

[Figure: \(\Delta ABC\) (3.6, 4, 4.2 cm) and the larger similar \(\Delta PQR\) (5.4, 6, 6.3 cm).]