Q.4 (2) Δ ABC ~ Δ PQR. In Δ ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of Δ ABC and Δ PQR are in the ratio 2 : 3, construct Δ ABC and Δ PQR.

Answer: PQ = 5.4 cm, QR = 6 cm, PR = 6.3 cm; construct both triangles from their sides.

\(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{2}{3}\).

PQ = \(\tfrac32 \times 3.6\) = 5.4 cm, QR = \(\tfrac32 \times 4\) = 6 cm, PR = \(\tfrac32 \times 4.2\) = 6.3 cm.

Construction:

1. Draw 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.
2. Draw 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.

(Alternatively, construct Δ PQR on the same base by extending BA and BC with the ratio 2 : 3.)