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`.

#### Solution

**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 B_{1}, B_{2}, B_{3}, B_{4} and B_{5} on BX, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.**6.** Join B_{3}C and from B_{5} draw B_{5}M ॥ B_{3}C 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.