Draw a triangle ABC, right-angled at B, such that AB=3 cm, BC=4 cm. Now construct a triangle PBQ similar to triangle ABC, each of whose sides is 7/5 times the corresponding side of triangle ABC.
Solution:
BQ = 7/5 BC = 7/5 * 4 = 28/5 = 5.6 PB = 7/5 AB = 7/5 * 3 = 21/5 = 4.2 On BA extended mark P at 4.2 cm from B on same side of A On BC extended mark Q at 5.6 cm from B on same side of C Join P, Q ∴ triangle PBQ is formed.