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 constructed is larger than the given triangle. Use this concept and then constant the required triangle.

Steps of construction

1. Draw the line BC=6cm.

2. B and C as centers, draw two arcs of radius 4 cm and 5cm intersecting each other at A.

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

4. Draw any ray BX downwards making at acute angle.

5. Mark five points B1, B2, B3, B4 and B5 on BX, such that


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

7. Draw MN||CA intersecting the extended line segment BA at N.


∆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.