Draw a ∆ABC in which BC = 6 cm, CA=5cm and AB=4cm. Construct triangle similar to it and of scale factor 5/3.

Thinking process:

Here scale factor i.e., m>n then the triangle to be construct is larger than the given triangle. Use this concept and then constant the required triangle.

Steps of construction:

1. Draw the line segment BC=6cm.

2. Taking B and C as centres, draw two arcs of radii 4 cm and 5cm respectively intersecting each other at A.

3. Join BA and CA. ∆ABC is the required triangle.

4. From B, draw any ray BD downwards making at acute angle.

5. Mark five points B₁,B_2,B₃,B₄ and B₅ on BD, such that

BB₁=B₁B₂=B₂B₃=B₃B₄=B₄B₅

6. Join B₃C and from B₅ draw B₅M||B₃C intersecting the extended line segment BC at M.

7. From point M draw MN||CA intersecting the extended line segment BA at N.

8. Then, ∆NBM is the required triangle whose sides are equal to 5/3 of the corresponding sides of the ∆ABC.

Hence, ∆NBM is the required triangle.

Justification:

As per the construction, ∆MNB is the required triangle

Let

BB₁=B₁B₂=B₂B₃= B₃B₄₌ B₄B₅₌x

From triangles ∆BCB₃ and ∆BMB₅, we can say that they are similar

[by AA congruency criteria, ∠ B is common in both triangles and ∠BCB₃ =∠BMB₅ and are corresponding angles of the same transverse as B₅D || B₃C]

---- (i)

Similarly,

∆MBN and ∆ABC are similar.

[by AA congruency criteria, ∠ B is common in both triangles and ∠BAC=∠BNM and are corresponding angles of the same transverse as AC || MN]

Hence

---- (ii)

From (i) and (ii) the constructed triangle ∆MNB is of scale times of the ∆ABC.