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


BB1=B1B2=B2B3=B3B4=B4B5



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.


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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