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

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Analysis:

As shown in the figure, Let R – P – L and R – Q – T.

∆PQR ~ ∆LTR … [Given]

∴ ∠PRQ ≅ ∠LRT … [Corresponding angles of similar triangles]

PQ/LT = QR/TR = PR/LR ....(i) [Corresponding sides of similar triangles]

But, PQ/LT = 3/4 ...(ii) [Given]

∴ PQ/LT = QR/TR = PR/LR = 3/4 ...[From (i) and (ii)]

∴ sides of LTR are longer than corresponding sides of ∆PQR.

If seg QR is divided into 3 equal parts, then seg TR will be 4 times each part of seg QR.

So, if we construct ∆PQR, point T will be on side RQ, at a distance equal to 4 parts from R. 

Now, point L is the point of intersection of ray RP and a line through T, parallel to PQ.

∆LTR is the required triangle similar to ∆PQR.

Steps of construction:

i. Draw ∆PQR of given measure. Draw ray RS making an acute angle with side RQ.

ii. Taking convenient distance on the compass, mark 4 points R1, R2, R3, and R4, such that RR1 = R1R2 = R2R3 = R3R4.

iii. Join R3Q. Draw line parallel to R3Q through R4 to intersects ray RQ at T.

iv. Draw a line parallel to side PQ through T. Name the point of intersection of this line and ray RP as L. 

∆LTR is the required triangle similar to ∆PQR.

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