ΔPQR ~ ΔLTR. In ΔPQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ΔPQR and ΔLTR, such that PQ/LT = 3/4.

Answer: Sides of ΔLTR are LT = 5.6 cm, TR = 7.2 cm, LR = 6.4 cm.

Analysis: \(\dfrac{PQ}{LT}=\dfrac{QR}{TR}=\dfrac{PR}{LR}=\dfrac34\), so ΔLTR is bigger. The vertex R is common, so T lies on ray RQ and L lies on ray RP.

\(LT = 4.2\times\tfrac43 = 5.6\) cm, \(TR = 5.4\times\tfrac43 = 7.2\) cm, \(LR = 4.8\times\tfrac43 = 6.4\) cm.

Steps of construction:

1. Draw seg QR = 5.4 cm. With centre Q radius 4.2 cm and centre R radius 4.8 cm draw arcs meeting at P. Join PQ, PR. This is ΔPQR.
2. Draw ray RX making an acute angle with RQ, on the side opposite to P. Mark 4 equal parts \(R_1, R_2, R_3, R_4\) on it.
3. Join \(R_3Q\). Through \(R_4\) draw a line parallel to \(R_3Q\), meeting ray RQ (extended) at T.
4. Through T draw a line parallel to QP, meeting ray RP (extended) at L.
5. ΔLTR is the required triangle (RT = 7.2 cm, RL = 6.4 cm, LT = 5.6 cm).