Draw a right triangle in which the sides (other than hypotenuse) are 8 cm and 6 cm. Then construct another triangle whose sides are times the (corresponding) sides of given triangle.3/4

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, B₃D∥CB₃ 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,