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, B_{1}, B_{2}, B_{3} and B_{4} such that, BB_{1} = BB_{2} = BB_{3} = BB_{4}.

7. Join CB_{4}.

8. From B_{3}, draw B_{3}D such that, B_{3}D∥CB_{3} where D is a point that lies on the line BC.

9. From D, draw ED such that, ED∥AC 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, DE**∥**CA*

*∴* *∆EBD~∆ABC*

*Hence,*

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