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.


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