Answer :

Given:

Sides (other than hypotenuse) = 8 cm and 6 cm


Steps of Construction:


1. Draw a line segment BC = 8 cm.


2. Construct B = 90°.


3. From B extend the line till A, to form BA.


4. Join AC. ∆ABC is the required triangle.


5. Now make an acute CBX, below BC.


6. Along BX, mark 4 ( 4 ˃ 3 in 3/4) points, B1, B2, B3 and B4 such that, BB1 = BB2 = BB3 = BB4.


7. Join CB4.


8. From B3, draw B3D such that, B3DCB3 where D is a point that lies on the line BC.


9. From D, draw ED such that, EDAC where E is a point that lies on the line.


∆EBD is the required triangle with the sides that are 3/4 of corresponding sides of ∆ABC.



Justification:


Cleary by construction,









And, DECA


∆EBD~∆ABC


Hence,



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