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 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 draw B_{5}M||B_{3}C intersecting the extended line segment BC at M.

7. 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.

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