Construct a triangle similar to a given triangle with sides 5 cm, 12cm and 13 cm and whose sides are of the corresponding sides of the given triangle.

Step 1: Draw a line segment AB = 13 cm.

Step 2: Taking A as center, draw an arc of radius 5 cm. Then, take B as center and draw an arc of radius 12 cm. These arcs will intersect at point C (say). Join AC and BC, such that AC = 5 cm and BC = 12 cm. Thus, ∆ABC is the required triangle.

Step 3: Draw a ray AX making acute angle with straight line AB to the opposite side of vertex C.

Step 4: Locate 5 points on ray AX, namely A ₁ , A ₂ , A ₃ , A ₄ and A ₅ (since 5 is greater among 3 and 5) such that, AA ₁ = A ₁ A ₂ = A ₂ A ₃ = A ₃ A ₄ = A ₄ A ₅ .

Step 5: Join BA ₅ and then draw a line through A ₃ such that it is parallel to BA ₅ and intersects on AB at B’.

Step 6: Now draw a line through B’ such that it is parallel to BC and intersects on AC at C’.

∆AB’C’ is the required triangle with sides 3/5 of the corresponding sides of ∆ABC.