Δ ABC ~ Δ PQR. In Δ ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm and \(\dfrac{AB}{PQ} = \dfrac{3}{2}\). Construct Δ ABC and Δ PQR.

Answer: PQ = 3.6 cm, QR = 2.8 cm, PR = 4.0 cm; construct both triangles with these sides.

Δ ABC ~ Δ PQR, so \(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{3}{2}\).

PQ \(= \dfrac{2}{3} \times 5.4 = 3.6\) cm, QR \(= \dfrac{2}{3} \times 4.2 = 2.8\) cm, PR \(= \dfrac{2}{3} \times 6.0 = 4.0\) cm.

Construction of Δ ABC:

1. Draw seg BC = 4.2 cm.
2. With B as centre and radius 5.4 cm, draw an arc above BC.
3. With C as centre and radius 6.0 cm, draw an arc cutting the first arc at A.
4. Join AB and AC.

Construction of Δ PQR:

1. Draw seg QR = 2.8 cm.
2. With Q as centre and radius 3.6 cm, draw an arc.
3. With R as centre and radius 4.0 cm, draw an arc cutting the first arc at P.
4. Join PQ and PR. Δ PQR is the required triangle.