∆ PQR ~ ∆ LTR. In ∆ PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆ PQR and ∆ LTR, such that

_{Steps of construction:}

i. Draw a triangle PQR, with PQ = 4.2 cm, QR = 5.4 cm and PR = 4.8 cm, choosing QR = 5.4 cm as base.

ii. Below QR, draw an acute angle ∠QRX.

iii. Mark four points R₁, R₂, R₃ and R₄ on RX, such that RR₁ = R₁R₂ = R₂R₃ = R₃R₄. [As ratio is 4:3, we choose 4 points]

iv. Join QR₄ and Draw TR₃ || QR₄

v. Draw LT || PQ.