Question

∆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
 

Answer 1

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 ...[Form (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 form 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 = R₂ R₃ = 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.