Draw a ΔABC, right-angled at B such that AB = 3 cm and BC = 4 cm. Now, construct a triangle similar to ΔABC, each of whose sides is 7/5 times the corresponding side of ΔABC.

Steps of Construction:

1. First we have to draw a triangle of given dimensions.

2. Draw a line segment of length BC of length 4cm.

3. Make an angle of 90° at B.

4. cut an arc of radius 3 cm taking B as center on the ray BY. The arc cut point is A.

5. Join AC. ABC is the right angled triangle with the given dimensions.

6. Now we have to make a triangle which 7/5 times of this triangle, that is bigger than this triangle.

7. So we extend the line BC along C to X and BA along A to Y.

8. Draw a ray BZ making an acute angle with BC.

9. Make 7 equal arcs along BZ starting from B then B1 and so on till B7.

10. Join B5C.

11. From B7 draw a ray parallel to B5C cutting the BX at C’.

12. From C’, draw another ray parallel to CA cutting BY at A’.

13. Then A’BC’ is our required triangle.