Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm. Construct a triangle similar to it and of scale factor `5/3`.
Steps of construction:
1. Draw a line segment BC = 6 cm.
2. Taking B and C as centres, draw two arcs of radii 4 cm and 5 cm intersecting each other at A.
3. Join BA and CA. ∆ABC is the required triangle.
4. From B, draw any ray BX downwards making at acute angle ∠CBX
5. Mark five points B1, B2, B3, B4 and B5 on BX, 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.
Then, ∆NBM is the required triangle
Whose sides is equal to `5/3` of the corresponding sides of the ∆ABC.
Hence, ∆NBM is the required triangle.